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Calculations with polarized and un-polarized light will be used to illustrate the difference between pure states and mixtures, and to demonstrate the utility of the density operator.
In Heisenberg’s matrix mechanics pure states are represented by vectors and operators by matrices. For example, light polarized at an angle 2 relative to the vertical is represented by the following Dirac ket vector,
$| \theta \rangle = \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} \nonumber$
The matrix operator for a polarizer oriented at an angle 2 relative to the vertical in Dirac notation is represented by a ket-bra product (outer product).
$\hat{ \theta} = | \theta \rangle \langle \theta | = \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} \cos \theta ~~~ \sin \theta = \begin{pmatrix} \cos ^2 \theta & \cos \theta \sin \theta\ \cos \theta \sin \theta & \sin ^2 \theta \end{pmatrix} \nonumber$
The state vector is normalized because the bra-ket product, or inner product, is unity.
$\langle \theta | \rangle \theta = \cos \theta ~~~ \sin \theta \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} = \cos ^2 \theta + \sin ^2 \theta = 1 \nonumber$
Clearly the state vector is an eigenstate of the operator with unit eigenvalue.
$\widehat{ \theta} | \theta \rangle = | \theta \rangle \langle \theta | \theta \rangle = | \theta \rangle \nonumber$
Un-polarized light from an incandescent bulb is incident upon a polarizing film oriented at an angle $\theta$ relative to the vertical producing 2-polarized light, $| \theta \rangle$. State preparation, such as this, is an essential first step in any quantum mechanical analysis.
As shown below, the expectation value for the passage of this 2-polarized light through a vertical polarizer is $\cos ^2 \theta$.
$\langle V \rangle = \langle \theta | \hat{V} | \theta \rangle = \cos \theta ~~~ \sin \theta \begin{pmatrix} 1 & 0\ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} = \cos ^2 \theta \nonumber$
We can also argue that $\theta$-polarized light is a coherent superposition of vertically and horizontally polarized light.
$| \theta \rangle = | v \rangle \langle v | \theta \rangle + |h \rangle \langle H | \theta \rangle = \cos \theta | v \rangle + \sin \theta | h \rangle = \cos \theta \begin{pmatrix} 1\ 0 \end{pmatrix} + \sin \theta \begin{pmatrix} 0\ 1 \end{pmatrix} \nonumber$
where
$|v \rangle \langle v | + |h \rangle \langle h | = 1 \nonumber$
is the identity operator. In other words, $|v \rangle$ and $|h \rangle$ are an orthonormal basis for linear polarization. Any linear polarization state can be expressed as a superposition of these basis vectors.
The probability that a $\theta$-polarized photon will pass a vertical polarizer in this approach is the probability amplitude for that event squared, $| \langle v | \theta \rangle |^2 = \cos ^2 \theta$.
The calculation for the expectation value $\langle \theta | \hat{V} | \theta \rangle$ can also be re-cast in terms of $\hat{ \theta}$ by exploiting the idempotency of $\hat{V}$.
$\hat{V} \hat{V} = \hat{V} |v \rangle \langle v | = | v \rangle \langle v | v \rangle \langle v | = | v \rangle \langle v | = \hat{V} \nonumber$
Using $\hat{V} = \hat{V} \hat{V}$ we find
$\langle V \rangle = \langle \theta | \hat{V} \hat{V} | \theta \rangle = \langle \theta \hat{V} | v \rangle \langle v | \theta \rangle = \langle v | \theta \rangle \langle \theta | \hat{V} | v \rangle = \langle v | \hat{ \theta } \hat{V} | v \rangle = Trace ( \hat{ \theta} \hat{V} \nonumber$
where Trace signifies the sum of the diagonal elements of the product matrix. By substituting $\hat{ \theta}$ and $\hat{V}$ we find agreement with the previous method of calculating the expectation value.
$\langle V \rangle = Trace \left[ \begin{pmatrix} \cos ^2 \theta & \cos \theta \sin \theta\ \cos \theta \sin \theta & \sin ^2 \theta \end{pmatrix} \begin{pmatrix} 1 & 0\ 0 & 0 \end{pmatrix} \right] = \cos ^2 \theta \nonumber$
This approach, while mathematically interesting, is non-intuitive compared with the previous traditional method. One might question its utility in the face of the more direct method of calculating expectation values. The utility of this method is apparent when one considers the calculation of the expectation values for a mixture – such as un-polarized light. Un-polarized light is an even mixture of all possible polarization states. However, it can be treated as a 50-50 mixture of any two orthogonal polarization states. Such a mixture cannot be represented by a single wavefunction,
$| \psi \rangle = \frac{1}{ \sqrt{2}} \left( | \theta \rangle + \left| \theta \right. + \frac{ \pi}{2} \bigg \rangle \right) \nonumber$
because in quantum mechanical notation this represents a pure state, a coherent superposition of two orthogonal polarization states.
Density Operators
Rather, a mixture must be represented by a density operator (sometimes calld a density matrix), which is a weighted sum of the operators representing the states making up the mixture. The density operator for un-polarized light is,
$\hat{ \rho} = \frac{1}{2} | \theta \rangle \langle \theta | + \frac{1}{2} \bigg| \theta + \frac{ \pi}{2} \bigg \rangle \bigg \langle \theta + \frac{ \pi}{2} \bigg| \nonumber$
The expectation value for un-polarized light to pass a vertical polarizer according to the last method is
$\langle V \rangle = Trace \left[ \frac{1}{2} \left( \begin{pmatrix} \cos ^2 \theta & \cos \theta \sin \theta\ \cos \theta \sin \theta & \sin ^2 \theta \end{pmatrix} + \begin{pmatrix} \cos ^2 ( \theta + \frac{ \pi}{2}) & \cos ( \theta + \frac{ \pi}{2}) \sin ( \theta + \frac{ \pi}{2})\ \cos ( \theta + \frac{ \pi}{2}) \sin ( \theta + \frac{ \pi}{2}) & \sin^2 ( \theta + \frac{ \pi}{2}) \end{pmatrix} \right) \begin{pmatrix} 1 & 0\ 0 & 0 \end{pmatrix} \right] = \frac{1}{2} \nonumber$
This method can be used to derive an alternative method for calculating the expectation value for a mixture.
\begin{align} \langle V \rangle &= Trace \Big[ \hat{ \rho} \hat{V} \Big] = \langle | \hat{ \rho} \hat{V} | v \rangle \[4pt] &= \langle v | \left( \frac{1}{2} | \theta \rangle \langle \theta | + \frac{1}{2} \bigg| \theta + \frac{ \pi}{2} \bigg \rangle \bigg \langle \theta + \frac{ \pi}{2} \bigg| \right) \hat{V} | v \rangle = \frac{1}{2} \langle v | \theta \rangle \langle \theta | v \rangle + \frac{1}{2} \bigg \langle v \bigg| \theta + \frac{ \pi}{2} \bigg \rangle \bigg \langle \theta + \frac{ \pi}{2} \bigg| v \bigg \rangle \[4pt] &= \frac{1}{2} \langle \theta | v \rangle \langle v | \theta \rangle + \frac{1}{2} \bigg \langle \theta + \frac{ \pi}{2} \bigg| v \bigg \rangle \bigg \langle v \bigg| \theta + \frac{ \pi}{2} \bigg \rangle = \frac{1}{2} \langle \theta | \hat{V} | \theta \rangle + \frac{1}{2} \bigg \langle \theta + \frac{ \pi}{2} \bigg| \hat{V} \bigg| \theta + \frac{ \pi}{2} \bigg \rangle \[4pt] &= \frac{1}{2} \cos ^2 \theta + \frac{1}{2} \cos ^2 ( \theta + \frac{ \pi}{2} ) = \frac{1}{2} \end{align} \nonumber
In general, for a mixture the expectation value is calculated as
$\langle V \rangle = \sum_i p_i \langle \psi _i | \hat{V} | \psi_i \rangle \nonumber$
where $p_i$ is the probability associated with the state $| \psi _i \rangle$.
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Starting with the traditional expression for the calculation of the expectation value, the identity operator is inserted between the measurement operator and the ket containing the wave function. Rearranging terms gives the trace function operating on the product of the state's density operator and the measurement operator.
\begin{align*} \langle \psi | \hat{O} | \psi \rangle &= \sum_i \langle \psi | \hat{O} | i \rangle \langle i | \psi \rangle \[4pt] &= \sum_i \langle i | \psi \rangle \langle \psi | \hat{O} | i \rangle \[4pt] &= Trace \left( | \psi \rangle \langle \psi | \hat{O} \right) \end{align*}
where
$\sum_i | i \rangle \langle i | = identity \nonumber$
Next this transition is carried out in detail using matrix algebra.
$\begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix}\begin{pmatrix} a\ b \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0\ 0 & 1 \end{pmatrix}\begin{pmatrix} a\ b \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \left[ \begin{pmatrix} 1\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} + \begin{pmatrix} 0\ 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \end{pmatrix} \right] \begin{pmatrix} a\ b \end{pmatrix} = 2ab \nonumber$
$\begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\ 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a\ b \end{pmatrix} = \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0\ 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \end{pmatrix}\begin{pmatrix} a\ b \end{pmatrix} = 2ab \nonumber$
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a \ b \end{pmatrix} \begin{pmatrix} a & b \end{pmatrix}\begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\ 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} a\ b \end{pmatrix} \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0\ 1 \end{pmatrix} = 2ab \nonumber$
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a^2 & ab \ ab & b^2 \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\ 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} a^2 & ab\ ab & b^2 \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0\ 1 \end{pmatrix} = 2ab \nonumber$
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} ab & a^2 \ b^2 & ab \end{pmatrix} \begin{pmatrix} 1\ 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} ab & a^2\ b^2 & ab \end{pmatrix} \begin{pmatrix} 0\ 1 \end{pmatrix} = ab + ab = Trace \begin{pmatrix} ab & a^2 \ b^2 & ab\ \end{pmatrix} \nonumber$
$\begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} a\ b \end{pmatrix} \rightarrow 2ab ~~~ \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0\ 0 & 1 \end{pmatrix} \begin{pmatrix} a\ b \end{pmatrix} \rightarrow 2ab \nonumber$
$\begin{pmatrix} 1\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} + \begin{pmatrix} 0\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0\ 0 & 1 \end{pmatrix} \nonumber$
$\begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a\ b \end{pmatrix} + \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} a \ b\end{pmatrix} \rightarrow 2ab \nonumber$
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} a \ b \end{pmatrix} \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\ 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} a\ b \end{pmatrix} \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow 2ab \nonumber$
$tr \left[ \begin{pmatrix} a \ b \end {pmatrix} \begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} 0 & 1 \ 1 & 0 \ \end{pmatrix} \right] \rightarrow 2ab ~~~ tr \left[ \begin{pmatrix} 0 & 1 \ 1 & 0 \ \end{pmatrix} \begin{pmatrix} a \ b \end {pmatrix} \begin{pmatrix} a & b \end{pmatrix} \right] \rightarrow 2ab \nonumber$
The last calculation on the right is justified by the following:
$\langle \psi | \hat{O} | \psi \rangle = \sum_i \langle \psi | i \rangle \langle i | \hat{O} | \psi \rangle = \sum_i \langle i | \hat{O} | \psi \rangle \langle \psi | i \rangle = Trace \left( \hat{O} | \psi \rangle \langle \psi | \right) \nonumber$
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The superposition principle is a deep and difficult quantum mechanical concept. There is no classical analog to lean on in probing its meaning, because it is impossible to simulate it with classical models. And yet because it pervades quantum mechanics we must find ways to present it to students that have some probability of success in revealing its significance.
The title of the first chapter in Dirac’s famous treatise The Principles of Quantum Mechanics (Clarendon Press, Oxford) is The Principle of Superposition. On page 12 of the 4th edition he wrote,
The nature of the relationships which the superposition principle requires to exist between the states of any system is of a kind that cannot be explained in terms of familiar physical concepts. One cannot in the classical sense picture a system being partly in each of two states and see the equivalence of this to the system being completely in some other state. There is an entirely new idea involved, to which one must get accustomed and in terms of which one must proceed to build up an exact mathematical theory, without having any detailed classical picture.
Perhaps the easiest way to illustrate the superposition principle mathematically is with demonstrations using polarizing films. Suppose unpolarized light illuminates a polarizer oriented at 45o to the vertical creating a beam of diagonally polarized photons, $|D \rangle$. If the diagonally polarized beam now illuminates a second polarizer oriented in the vertical or horizontal direction, 50% of the photons are transmitted.
The quantum mechanical explanation for these results is to write the wave function of a diagonally polarized photon as a superposition in the V‐H basis, $\langle v | v \rangle = \langle H | H \rangle = 1; \langle v | H \rangle \langle H |V \rangle = 0$.
$|D \rangle = \frac{1}{ \sqrt{2}} \left[ |V \rangle + |H \rangle \right] \nonumber$
This superposition is illustrated graphically as follows.
Using either equation or figure we have, $| \langle V |D \rangle |^2 = | \langle H | D \rangle |^2 = \frac{1}{2}$, in agreement with the demonstration.
The same result would occur if the diagonally polarized beam was assumed to be a 50‐ 50 mixture of vertical and horizontal photons rather than a superposition. However, according to quantum mechanical principles, a mixture is represented by a density operator rather than by a wave function. In this case the appropriate density operator is
$\hat{ \rho}_{mix} = \frac{1}{2} |V \rangle \langle V | + \frac{1}{2} |H \rangle \langle H| \nonumber$
The probability (expectation value) that a 50‐50 mixture of V‐H photons will pass a vertical or horizontal polarizer is easily calculated.
$\langle V| \hat{ \rho}_{mix} |V \rangle = \langle H | \hat{ \rho}_{mix} |H \rangle = \frac{1}{2} \nonumber$
At this point, the superposition and mixture models give the same result for what happens in the presence of the second (H or V) polarizer. However, it is not difficult to demonstrate that diagonally polarized light is not a mixture of vertical and horizontally polarized photons. If the second polarizer is replaced with another diagonally oriented polarizer, 100% of the photons are transmitted. This is because a diagonally polarized photon is an eigenstate of the diagonal polarization measurement operator.
The expectation value for a 50‐50 mixture is not in agreement with this outcome. It predicts that only 50% of the photons will pass the diagonal oriented polarizing film.
$\langle D | \hat{ \rho}_{mix} | D \rangle = \frac{1}{2} \langle D | V \rangle \langle V | D \rangle + \frac{1}{2} \langle D | H \rangle \langle H | D \rangle = \frac{1}{2} \left( \frac{1}{ \sqrt{2}} \right)^2 + \frac{1}{2} \left( \frac{1}{ \sqrt{2}} \right) ^2 = \frac{1}{2} \nonumber$
This result will seem more plausible when written in the following equivalent form.
$\langle D | \hat{ \rho}_{mix} | D \rangle = \frac{1}{2} \langle D | V \rangle \langle V | D \rangle + \frac{1}{2} \langle D | H \rangle \langle H | D \rangle = \frac{1}{2} \langle V | D \rangle \langle D | V \rangle + \frac{1}{2} \langle H | D \rangle \langle D | H \rangle = \frac{1}{2} \langle V | \hat{D} | H \rangle = \frac{1}{2} \nonumber$
It is instructive to repeat these calculations using matrix mechanics. In this approach to quantum mechanics states are represented by vectors and operators by matrices. For example, the vector representations for the vertical, horizontal and diagonal polarization states are given below.
$|V \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~~ |H \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} ~~~ |D \rangle = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \nonumber$
With vector algebra it is easy to show that a diagonally polarized photon is a weighted superposition of the vertical and horizontal polarization states.
$|D \rangle = \frac{1}{ \sqrt{2}}\begin{pmatrix} 1 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \nonumber$
The operators representing the various orientations of the polarizing films are as follows.
$\hat{V} = |V \rangle \langle V | = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
$\hat{H} = |H \rangle \langle H | = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\hat{D} = |D \rangle \langle D | = \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \nonumber$
With these vectors (polarization states) and matrices (measurement operators) it is straight forward to reproduce the previous calculations. For example, the probability that a diagonally polarized photon will pass a vertical polarizer is calculated as follows.
$\langle D | \hat{V} | D \rangle = \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} \end{pmatrix} = \frac{1}{2} = | \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} \end{pmatrix} |^2 = | \langle V|D \rangle | ^2 = \langle D|V \rangle \langle V|D \rangle = \langle D | \hat{V} |D \rangle \nonumber$
The density operator representing the mixture is constructed as follows,
$\hat{\rho}_{mix} = \frac{1}{2}|V \rangle \langle V | + \frac{1}{2} | H \rangle \langle H | = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{pmatrix} \nonumber$
Therefore the expectation value for a 50‐50 V‐H mixture to pass a diagonal polarizer is as determined previously, ½.
$\langle D | \hat{ \rho}_{mix} | D \rangle = \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} \end{pmatrix} = \frac{1}{2} \nonumber$
With regard to our desire for classical pictures,
Dirac said, ...the main object of physical science is not the provision of pictures, but is the formulation of laws governing phenomena and application of these laws to the discovery of new phenomena. If a picture exists, so much the better; but whether a picture exists or not is a matter of only secondary importance. In the case of atomic phenomena no picture can be expected to exist in the usual sense of the word ‘picture’ by which is meant a model functioning essentially on classical lines.
In the August 2008 (page 66) issue of Scientific American, Monroe and Wineland attempted to provide a “picture” of the superposition with a figure of the “ambiguous” cube.
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Readily available and inexpensive polarizing films can be used to illustrate many fundamental quantum mechanical concepts. The purpose of this tutorial is to use polarized light to illustrate one of quantum theory’s deepest and most challenging concepts - the linear superposition. According to Richard Feynman the superposition principle “has in it the heart of quantum mechanics” and is its “only mystery.” (1)
Three Polarizer “Paradox”
If two polarizing films are aligned in the same direction light from the first polarizer passes through the second. If the polarizers are opposed at a 90° angle, the polarized light from the first polarizer is stopped by the second. If a third polarizer is sandwiched between the two opposed polarizers at a 45° angle some light gets through the last polarizer.(2, 3) The analysis that follows is based on the quantum mechanical superposition principle. A graphical representation is shown below.
Quantum Mechanical Analysis
A photon polarized at an angle 2 with respect to the vertical can be written as a linear combination (superposition) of a vertically polarized photon, $|v \rangle$, and a horizontally polarized photon, $|h \rangle$. We say $|v \rangle$ and $|h \rangle$ are the polarization basis states, which means $\langle v | v \rangle = \langle h|h \rangle = 1$ and $\langle v | h \rangle = \langle h|v \rangle = 0$.
$| \Theta \rangle = | v \rangle \langle v | \Theta \rangle + | h \rangle \langle h | \Theta \rangle \nonumber$
From this figure we see that the projections of $| \Theta \rangle$, onto $|v \rangle \langle \langle v | \Theta \rangle \rangle$ and $|h \rangle \langle \langle h | \Theta \rangle \rangle$ are $\cos \Theta$ and $\sin \Theta$, respectively.
$|\Theta \rangle = | v \rangle \cos \Theta + |h \rangle \sin \Theta \nonumber$
The probability that a photon polarized at an angle $\Theta$ will pass a vertical polarizer is
$| \langle v | \Theta \rangle |^2 = \cos^2 \Theta \nonumber$
The light incident on the first polarizer is unpolarized, but the photons that pass the vertical polarizer are vertically polarized. In other words the photons are eigenfunctions of the measurement operator, which in this case is a vertically oriented linear polarizer. At this point only two experiments have definite outcomes.
1. The probability that vertically polarized photons will pass a second filter that is also vertically polarized is one. It is certain that a vertically polarized photon will pass a vertically polarized filter. $| \langle v | v \rangle | ^2 = \cos^2 0 = 1 \nonumber$
2. The probability that vertically polarized photons will pass a second filter that is horizontally polarized is zero. It is certain that a vertically polarized photon will not pass a horizontally polarized filter.
$| \langle h | v \rangle |^2 = \cos^2 90 = 0 \nonumber$
For any other orientation of the second filter, quantum mechanics can only predict the probability that a vertically polarized photon will pass, and that probability is, of course,
$| \langle \Theta | v \rangle |^2 = \cos^2 \Theta \nonumber$
Now a vertically polarized photon may be written as a linear superposition of any other orthogonal basis states, for example $|45^o \rangle$, and $|-45^o \rangle$.
\begin{align} |v \rangle &= | 45^o \rangle \langle 45^o | v \rangle +|- 45^o \rangle \langle -45^o | v \rangle \[4pt] &= | 45^o \rangle \cos 45^o +|- 45^o \rangle \cos 45^o \[4pt] &= | 45^o \rangle 0.707 +|- 45^o \rangle 0.707 \end{align} \nonumber
If a 45° polarizer is inserted between the vertical and horizontal polarizers photons get through the horizontal polarizer that stopped them previously. Here is the quantum mechanical explanation. The probability that a vertically polarized photon will get through a polarizer oriented at an angle of 45° is, by eqns (7), 1/2.
$| \langle 45^o | v \rangle |^2 = \cos^2 45^o = \frac{1}{2} \nonumber$
At this point the photon is in the state $|45^o \rangle$, and according to the superposition principle a photon in this state can be written as a linear combination of $|v \rangle$ and $|h \rangle$.
$|45^o \rangle = | v \rangle \langle v|45^o \rangle + |h \rangle \langle h + 45^o \rangle \nonumber$
$|45^o \rangle = | v \rangle \cos45^o + |h \rangle \sin 45^o \nonumber$
Therefore, the probability that this photon will pass the final horizontally oriented polarizer is
$| \langle h | 45^o \rangle |^2 = \sin^2 45^o = \frac{1}{2} \nonumber$
Alternatively, the probability that a photon emerging from the vertical polarizer will pass through the final horizontal polarizer in the presence of an intermediate 45° polarizer can be calculated as follows:
$| \langle h | 45^o \rangle \langle 45^o | v \rangle |^2 = | \sin^2 45^o \cos^2 45^o |^2 = \frac{1}{4} \nonumber$
In other words, half of the photons that emerge from the vertical polarizer pass the 45° polarizer, and half of those pass the final horizontal polarizer. So 25% of the photons that pass the initial vertical polarizer also pass the final horizontal polarizer.
Unpolarized Light
The unpolarized light illuminating the first polarizer is a 50-50 mixture of any orthogonal set of polarizer angles – (0°, 90°), (-45°, 45°), or in the general case ($\left(\Theta, \Theta + \frac{ \pi}{2} \right)$.(4) A mixture cannot be represented by a single wave function, so the expectation value for a mixture of unpolarized photons to pass a vertical polarizer is calculated as follows:
$\langle V \rangle = \sum_i p_i \langle \Psi_i | \hat{V} | \Psi_i \rangle \nonumber$
where V represents the operator associated with the vertical polarizer, and pi is the fraction of photons in the state $| \Psi_i \rangle$.
For the general case of a beam of unpolarized light incident on a vertical polarizer the expectation value for transmission is,
$\langle V \rangle = \frac{1}{2} \left\langle \Theta \right| \hat{V} \left| \Theta \right\rangle + \frac{1}{2} \left\langle \Theta + \frac{ \pi}{2} \right| \hat{V} \left| \Theta + \frac{ \pi}{2} \right\rangle \label{267.16}$
Since $\hat{V} = | v \rangle \langle v |$, Equation \ref{267.16} can be written as,
$\langle V \rangle = \frac{1}{2} \left\langle \Theta | v \right\rangle \left\langle v | \Theta \right\rangle + \frac{1}{2} \left\langle \Theta + \frac{ \pi}{2} | v \right\rangle \left\langle v | \Theta + \frac{ \pi}{2} \right\rangle \nonumber$
This expression simplifies to,
$\langle V \rangle = \frac{1}{2} \left[ \cos^2 \Theta + \cos^2 \left( \Theta + \frac{ \pi}{2} \right) \right] = \frac{1}{2} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.12%3A_Polarized_Light_and_Quantum_Mechanics.txt
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A beam of unpolarized light illuminates a vertical polarizer and 50% of the light emerges vertically polarized. This light beam encounters a diagonal polarizer oriented at a 45 degree angle to the original vertical polarizer and 50% of it emerges as diagonally polarized light. Finally 50% of the diagonally polarized light passes a horizontally oriented polarizer. In other words 12.5% of the light illuminating the first vertical polarizer passes the final horizontal polarizer. However, if the diagonal polarizer sandwiched between the vertical and horizontal polarizers is removed, no light emerges form the final horizontal polarizer.
Using the figure below vector algebra will be used to analyze this so-called "three-polarizer paradox." The paradox being that it is surprising that the insertion of the diagonal polarizer between crossed polarizers allows photons to pass the final horizontal polarizer.
Eigenstate for a $\Theta$-polarized photon:
$\Theta ( \theta) = \begin{pmatrix} \cos \theta \sin \theta \end{pmatrix} ~~~ \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} simplify \rightarrow 1 \nonumber$
Eigenstates for vertically, horizontally and diagonally polarized light (S (slant) represents D's orthogonal partner, just as V and H are orthogonal partners, or basis states):
$V = \Theta (0) \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$H = \Theta (0) \left( \frac{ \pi}{2}\right) \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$D = \Theta (0) \left( \frac{ \pi}{4}\right) float,~4~ \rightarrow \begin{pmatrix} 0.7071 \ 0.7071 \end{pmatrix} \nonumber$
$S = \Theta (0) \left( - \frac{ \pi}{4}\right) float,~4~ \rightarrow \begin{pmatrix} 0.7071 \ -0.7071 \end{pmatrix} \nonumber$
Confirm with vector addition the superpositions shown in the figure:
$\frac{D + S}{ \sqrt{2}} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\frac{D - S}{ \sqrt{2}} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\frac{V + H}{ \sqrt{2}} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} \nonumber$
$\frac{V - H}{ \sqrt{2}} = \begin{pmatrix} 0.707 \ -0.707 \end{pmatrix} \nonumber$
The polarizers are operators, and their matrices are the outer products of their polarization vectors:
$| \rightarrow \rangle \langle \rightarrow | ~~~ H H^T = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$| \nearrow \rangle \langle \nearrow | ~~~ D D^T = \begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 \end{pmatrix} \nonumber$
$| \uparrow \rangle \langle \uparrow | ~~~ V V^T = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Unpolarized light is an even mixture of all polarization angles between 0 and $\frac{ \pi}{2}$ radians. Probability is the absolute magnitude squared of the probability amplitude. The fraction of a beam of unpolarized light that will pass a vertically oriented polarizer is 0.5. To confirm this result we must integrate over all possible polarization angles, summing the squares of the probability amplitudes for each angle . The factor $\frac{2}{ \pi}$ [( $\pi$/2)-1] normalizes the calculation.
Probability Amplitude:
$\langle V | \Theta \rangle ~~~ V^T \Theta ( \theta) \rightarrow \cos \theta \nonumber$
Probability:
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} | \langle V | \Theta \rangle |^2 d \Theta \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( V^T \Theta ( \theta) \right)^2 d \theta = 0.5 \nonumber$
As is well known, and easy to demonstrate, the probability that unpolarized light (or light of any polarization) will pass two crossed polarizing films (vertical followed by horizontal for example) is 0.
Probability Amplitude:
$\langle H | V \rangle \langle V | \Theta \rangle ~~~ H^T V V^T \Theta ( \theta) \rightarrow 0 \nonumber$
Probability:
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} | \langle H | V \rangle \langle V | \Theta \rangle |^2 d \Theta \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( H^T V V^T \Theta ( \theta) \right)^2 d \theta = 0 \nonumber$
However, if a polarizing film oriented diagonally at a 45 degree angle is inserted between the crossed polarizers light gets through the final horizontal filter. The following calculation shows that 12.5% of the unpolarized light illuminating the initial vertical filter gets through this arrangement of polarizing films in agreement with the figure above and experience.
Probability Amplitude:
$\langle H | V \rangle \langle D | V \rangle \langle V | \Theta \rangle ~~~ H^T D D^T V V^T \Theta ( \theta) ~ float,~1 \rightarrow 0.5 \cos \theta \nonumber$
Probability:
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} | \langle H | V \rangle \langle D | V \rangle \langle V | \Theta \rangle |^2 d \Theta \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \right| H^T D D^T V V^T \Theta ( \theta) \left| \right)^2 d \theta = 0.125 \nonumber$
The three probabilities can be also calculated by an equivalent method that explicitly treats the polarizers as measurement operators. From above we see that, $V_{op} = VV^T$, $H_{op} = HH^T$, $D_{op} = D D^T$, and write the probability calculations as follows.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( V^T V_{op} \Theta ( \theta) \right)^2 d \theta = 0.5 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( H^T V_{op} \Theta ( \theta) \right)^2 d \theta = 0 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( H^T D_{op} V_{op} \Theta ( \theta) \right)^2 d \theta = 0.125 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.13%3A_The_Three-Polarizer_Paradox.txt
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It is convenient to use matrix mechanics to describe experiments with polarized light. In this tutorial we will restrict our attention to plane polarized light. However, it would be just as easy to use matrix mechanics to describe the behavior of circularly polarized light.
The Three Polarizer Paradox
If two polarizing films are aligned in the same direction light from the first polarizer passes through the second. If the polarizers are opposed at a 90o angle, the polarized light transmitted from the first polarizer is absorbed by the second. If a third polarizer is sandwiched between the two opposed polarizers at a 45o angle some light gets through the last polarizer. This is illustrated below using Dirac notation for both the photons and the polarizers.
In what follows this demonstration is analyzed using the principles of matrix mechanics. Light polarized in the vertical and horizontal directions, and at an angle θ relative to the vertical can be represented by the following vectors,
$| v \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~~ |h \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} | \Theta \rangle = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
The graphical representation between $|v \rangle$, $|h \rangle$ and $| \theta \rangle$ is shown below.
As can be seen in this figure, | v ⟩ and | h ⟩ represent an orthonormal basis set, and | θ ⟩ represents a general vector in the space defined by | v ⟩ and | h ⟩. As is shown below it is a normalized state vector.
$\begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} = \cos^2 \theta + \sin^2 \theta = 1 \nonumber$
Naturally | θ ⟩ can be written as a linear superposition of the polarization base states:
$| \Theta \rangle = | v \rangle \langle v | \Theta \rangle + |h \rangle \langle h | \Theta \rangle = |v \rangle \cos \theta + |h \rangle \sin \theta \nonumber$
A polarizing filter (film) oriented in the veritcal direction in this basis can be represented by the following matrix operator.
$\hat{V} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Note the results of the following measurements on the three previously defined polarization states using this measurement operator.
$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \nonumber$
$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \theta \ 0 \end{pmatrix} \nonumber$
Vertically polarized light passes the vertical polarizer, horizontally polarized light is absorbed (annihilated), and θ-polarized light, if it passes, becomes vertically polarized with reduced intensity. The probability that it will pass the vertical polarizer is cos2θ.
The operators associated with a horizontal filter and a filter oriented at an angle θ relative to the vertical are given below.
$\hat{H} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\hat{ \Theta} = \begin{pmatrix} \cos ^2 \theta & \sin \theta \cos \theta \ \sin \theta \cos \theta & \sin^2 \theta \end{pmatrix} \nonumber$
The operator associated with a filter oriented at an angle θ relative to the vertical is generated as shown below.
$\hat{ \theta} = | \theta \rangle \langle \theta | = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} = \begin{pmatrix} \cos ^2 \theta & \cos \theta \sin \theta \ \sin \theta \cos \theta & \sin^2 \theta \end{pmatrix} \nonumber$
We now have the mathematical apparatus in place to examine the three-polarizer paradox using the basic principles of matrix mechanics. Unpolarized light, such as that coming from an incandecent light bulb, might be considered to be a mixture of all polarization angles between 0 and π radians. Assuming this, the probability amplitude that a θ-polarized photon will pass a vertically oriented filter is,
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
The probability is the square of the absolute magnitude of the probability amplitude. Thus the fraction of a beam of unpolarized photons that will pass a vertically oriented polarizer is 0.5. To achieve this result we must integrate over all possible polarization angles. The normalization constant for this calculation is 2/π.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left| \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix}\right| ^2 d \theta = 0.5 \nonumber$
As is well known, and easy to demonstrate, the probability that unpolarized light (or light of any polarization) will pass two crossed polarizing films (vertical followed by horizontal, for example) is zero.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left| \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix}\right| ^2 d \theta = 0 \nonumber$
However, if a polarizing film oriented at a 45o angle is inserted between the crossed polarizers light gets through the final horizontal filter. The operator for a 45o polarizer is obtained as follows:
$\theta = \frac{ \pi}{4} \nonumber$
$\hat{ \Theta} = \begin{pmatrix} \cos^2 \theta & \sin \theta \cos \theta \ \sin \theta \cos \theta & \sin^2 \theta \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \nonumber$
The following calculation shows that 12.5% of the unpolarized light illuminating the initial vertical polarizer is transmitted through this arrangement of polarizing films.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left| \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix}\right| ^2 d \theta = 0.125 \nonumber$
It is easy to see that if the positions of the diagonal and horizontal polarizers are interchanged no light would be transmitted because the first and second polarizers would be crossed. This illustrates that the horizontal and diagonal polarization measurements do not commute; the outcome is dependent on the order of the measurements.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left| \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix}\right| ^2 d \theta = 0 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.14%3A_Matrix_Mechanics_Approach_to_Polarized_Light.txt
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It is convenient and illustrative of quantum mechanical principles to use matrix mechanics to describe experiments with polarized light. In this tutorial we will restrict our attention to plane polarized light. However, it would be just as easy to use matrix mechanics to describe the behavior of circularly polarized light (see appendix).
In matrix mechanics we use vectors to represent states and matrices to represent measurement operator Light polarized in the vertical and horizontal directions will serve as base states and will be represented the following vectors. As will be shown shortly any other polarization state can be written as a linear superposition of these basis vectors.
$v = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~~ h = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow 1 ~~~ \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow 1 ~~~ \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow 0 \nonumber$
$v^T v \rightarrow 1 ~~~ h^T h \rightarrow 1 ~~~ v^T h \rightarrow 0 \nonumber$
Note from above that |v> and |h> form an orthonormal basis set for this two-dimensional vector space. Light polarized at an angle $\theta$ relative to the verticle can be written as
$\Theta ( \theta) = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
It is easy to show that this state is normalized
$\begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \rightarrow \cos \theta^2 + \sin \theta^2 simplify \rightarrow 1 \nonumber$
In addition, $| \theta \rangle$ can be written as a linear superposition of the polarization base states:
$| \theta \rangle = | v \rangle \langle v | \theta \rangle + | h \rangle \langle h | \theta \rangle = | v \rangle \cos \theta + | h \rangle \sin \theta \nonumber$
$v \cos \theta + h \sin \theta \rightarrow \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
A polarizing filter oriented in the vertical position in this basis can be represented by the following mat operator - |v><v|.
$V_{op} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}$ which can also be accomplished this way $v v^T \rightarrow \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}$
Note the results of the following measurements on the three defined polarization states using this operator
$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ 0 \end{pmatrix} \nonumber$
$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \rightarrow \begin{pmatrix} \cos \theta \ 0 \end{pmatrix} \nonumber$
$V_{op} v \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$V_{op} h \rightarrow \begin{pmatrix} 0 \ 0 \end{pmatrix} \nonumber$
$V_{op} \Theta ( \theta) \rightarrow \begin{pmatrix} \cos \theta \ 0 \end{pmatrix} \nonumber$
Vertically polarized light passes the vertical polarizer, horizontally polarized light is absorbed (anihilated and $\theta$-polarized light, if it passes, becomes vertically polarized with reduced intensity. The probability will pass the vertical polarizer is $\cos^2 \theta$. The operators associated with a horizontal filter and a filter oriented at an angle $\theta$ relative to the vertical shown below. The latter is, of course, general and can be used to represent any operator by putting in the appropriate value for $\theta$ as is shown below.
$H_{op} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\Theta_{op} ( \theta) = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \rightarrow \begin{pmatrix} \cos \theta^2 & \cos \theta \sin \theta \ \cos \theta \sin \theta & \sin \theta^2 \end{pmatrix} \nonumber$
The operators for vertical, diagonal and horizontal polarizers are shown below using the general operator
$\Theta_{op} (0) \rightarrow \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
$\Theta_{op} \left( \frac{ \pi}{4} \right) ~float,~1~ \rightarrow \begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 \end{pmatrix} \nonumber$
$\Theta_{op} \left( \frac{ \pi}{2} \right) \rightarrow \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
We now examine the three polarizer paradox illustrated in the figure below. Unpolarized light illuminates vertical polarizer and 50% of the light emerges as vertically polarized light. This light encounters a diagonal polarizer oriented at a 45o angle to the original vertical polarizer and 50% of it emerges as diagonally polarized light. Finally 50% of the diagonally polarized light passes a horizontally oriented polarizer. In other words 12.5% of the light illuminating the first vertical polarizer passes the final horizontal polarizer. However, if the diagonal polarizer sandwiched between the vertical and horizontal polarizers is remove no light emerges form the final horizontal polarizer.
Matrix mechanics will now be used to examine the so-called "three polarizer paradox." Please note that figure offers an alternative analysis based on Dirac bracket notation and the superposition principle.
Unpolarized light, such as that coming from an incandescent light bulb, might be considered to be an be mixture of all polarization angles between 0 and $\frac{ \pi}{2}$ radians. The probability amplitude that a $\theta$-polarized photon will pass a vertical filter is
$\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
The probability is the square of the absolute magnitude of the probability amplitude. Thus the fraction of a beam of unpolarized light that will pass a vertically oriented polarizer is 0.5. To achieve this result we must integrate over all possible polarization angles. The factor $2/ \pi$ [($\pi$/2)-1] normalizes the calculation.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left[ \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \right] ^2 d \theta = 0.5 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( v^T V_{op} \Theta ( \theta) \right) ^2 d \theta = 0.5 \nonumber$
As is well known, and easy to demonstrate, the probability that unpolarized light (or light of any polarization) will pass two crossed polarizing films (vertical followed by horizontal for example) is 0.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left[ \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \right] ^2 d \theta = 0 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( v^T H_{op} V_{op} \Theta ( \theta) \right) ^2 d \theta = 0.5 \nonumber$
However, if a polarizing film oriented diagonally at a 45o angle is inserted between the crossed polarizers light gets through the final horizontal filter. The operator for a 45o polarizer is obtained from $\Theta_{op}$.
$\Theta_{op} \frac{ \pi}{4} = \begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 \end{pmatrix} \nonumber$
The following calculation shows that 12.5% of the unpolarized light illuminating the initial vertical filter gets through this arrangement of polarizing films in agreement with the figure above and experience.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left[ \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \right] ^2 d \theta = 0.125 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( h^T \Theta_{op} \frac{ \pi}{2} \Theta_{op} \frac{ \pi}{4} \Theta_{op} (0) \Theta ( \theta) \right) ^2 d \theta \rightarrow \frac{1}{8} \nonumber$
In addition, the non-commutivity rule can be demonstrated by switching the second and third filters. When this is done no photons emerge from the apparatus for obvious reasons; the first two filters are crossed.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left[ \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 \end{pmatrix} \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \right] ^2 d \theta = 0 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( h^T \Theta_{op} \frac{ \pi}{4} \Theta_{op} \frac{ \pi}{2} \Theta_{op} (0) \Theta ( \theta) \right) ^2 d \theta \rightarrow 0 \nonumber$
Appendix:
The base states for circularly polarized light are given below, along with their superpositions in the h-v representation.
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} ~~~ \frac{1}{ \sqrt{2}} (v + ih) = \begin{pmatrix} 0.707 \ 0.707i \end{pmatrix} \nonumber$
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} ~~~ \frac{1}{ \sqrt{2}} (v - ih) = \begin{pmatrix} 0.707 \ -0.707i \end{pmatrix} \nonumber$
Left and right circularly polarized light also form an orthonormal basis set.
$\overline{L^T} L \rightarrow 1 \nonumber$
$\overline{R^T} R \rightarrow 1 \nonumber$
$\overline{L^T} R \rightarrow 0 \nonumber$
$\overline{R^T} L \rightarrow 0 \nonumber$
The appropriate operators are:
$L_{op} = L \overline{L^T} ~ float,~1 \rightarrow \begin{pmatrix} 0.5 & -0.5i \ 0.5i & 0.5 \end{pmatrix} \nonumber$
$R_{op} = R \overline{R^T} ~ float,~1 \rightarrow \begin{pmatrix} 0.5 & 0.5i \ -0.5i & 0.5 \end{pmatrix} \nonumber$
Vertically and horizontally polarized light can be written as superpositions of circularly polarized light
$v = \frac{1}{ \sqrt{2}} (L+R) \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$h = \frac{i}{ \sqrt{2}} (R-L) \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Earlier unpolarized light was considered to be an even mixture of all polarization angles between 0o and Another approach is to consider it to be a 50-50 mixture of any two orthogonal polarization angles. The expectation value for unpolarized light passing a vertical polarizer under this model is outlined below.
$\langle V \rangle = \sum_i p_i \left\langle \psi_i \right| \hat{V} \left| \psi_i \right\rangle = \frac{1}{2} \langle \Theta | \hat{V} | \Theta \rangle + \frac{1}{2} \left\langle \Theta + \frac{ \pi}{2} \right| \hat{V} \left| \Theta + \frac{ \pi}{2} \right\rangle = \frac{1}{2} \langle \Theta | v \rangle \langle v | \Theta \rangle + \frac{1}{2} \left\langle \Theta + \frac{ \pi}{2} | v \right\rangle \left\langle v | \Theta + \frac{ \pi}{2} \right\rangle \nonumber$
Just as calculated earlier 50% of the unpolarized light passes the vertical polarizer.
$\frac{1}{2} \Theta ( \theta) ^T V_{op} \Theta ( \theta) + \frac{1}{2} \Theta \left( \theta + \frac{ \pi}{2} \right) V_{op} \Theta \left( \theta + \frac{ \pi}{2} \right) ~ simplify \rightarrow \frac{1}{2} \nonumber$
Unpolarized light can also be represented by the density operator shown below.
$\hat{ \rho} = \frac{1}{2} \left| \Theta \right\rangle \left\langle \Theta \right| + \frac{1}{2} \left| \Theta + \frac{ \pi}{2} \right\rangle \left\langle \Theta + \frac{ \pi}{2} \right| \nonumber$
$\rho = \frac{1}{2} \Theta_{op} ( \theta) + \frac{1}{2} \Theta_{op} \left( \theta + \frac{ \pi}{2} \right)~simplify~ \rightarrow \begin{pmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{pmatrix} \nonumber$
It is not difficult to show that the expectation value for unpolarized light passing a verticle polarizer can be calculated as the trace of the product of this operator with the vertical polarization operator.
$Tr \left[ \left( \frac{1}{2} \left| \theta \right\rangle \langle \Theta \left| \Theta + \frac{ \pi}{2} \right\rangle \left\langle \Theta + \frac{ \pi}{2} \right| \right) \hat{V} \right] = \frac{1}{2} \nonumber$
$tr( \rho V_{op}) ~simplify~ \rightarrow \frac{1}{2} \nonumber$
Of course there is nothing special about the vertical direction. Unpolarized light has a 50% probability passing a polarizer of any other orientation.
$tr( \rho H_{op}) = 0.5 \nonumber$
$tr( \rho L_{op}) = 0.5 \nonumber$
$tr( \rho R_{op}) = 0.5 \nonumber$
$tr \left( \rho \Theta _{op} \left( \frac{ \pi}{3} \right) \right) = 0.5 \nonumber$
$tr \left( \rho \Theta _{op} \left( \frac{ \pi}{4} \right) \right) = 0.5 \nonumber$
$tr \left( \rho \Theta _{op} \left( \frac{ \pi}{8} \right) \right) = 0.5 \nonumber$
In other words, the intensity of light passing a polarizer from an unpolarized source is independent of t orientation of the polarizer. Finally, the result of the three-polarizer demonstration is recalculated using the density matrix approach.
$D_{op} = \Theta \left( \frac{ \pi}{4} \right) \nonumber$
$tr(H_{op} D_{op}) tr(D_{op} V_{op}) tr( V_{op} \rho) \rightarrow \frac{1}{8} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.15%3A_Matrix_Mechanics_Approach_to_Polarized_Light_-_Version_2.txt
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Eigenstates and operators are provided for a series of matrix mechanics exercises involving polarized light.
Eigenstate for a $\theta$-polarized light:
$\Theta ( \theta) = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} ~ simplify~ \rightarrow 1 \nonumber$
$| \theta \rangle = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\langle \theta | \theta \rangle = \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} = 1 \nonumber$
Operator for a $\theta$-oriented polarizer:
$\Theta_{op} ( \theta) = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix}\begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \rightarrow \begin{pmatrix} \cos \theta^2 & \cos \theta \sin \theta \ \cos \theta & \sin \theta^2 \end{pmatrix} \nonumber$
$\hat{ \Theta} | \theta \rangle \langle \theta | = \begin{pmatrix} \cos \theta \ \hline \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \nonumber$
Eigenstates for vertically, horizontally, and diagonally polarized light:
Vertically:
$V = \Theta (0) \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$V^T \rightarrow \begin{pmatrix} 1 & 0 \end{pmatrix} \nonumber$
$|V \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~ \langle V | = \begin{pmatrix} 1 & 0 \end{pmatrix} \nonumber$
Horizontally:
$H = \Theta \left( \frac{ \pi}{2} \right) \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$H^T \rightarrow \begin{pmatrix} 0 & 1 \end{pmatrix} \nonumber$
$|H \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} ~~ \langle H | = \begin{pmatrix} 0 & 1 \end{pmatrix} \nonumber$
Diagonally:
$D = \Theta \left( \frac{ \pi}{4} \right) ~float,~4~ \rightarrow \begin{pmatrix} 0.7071 \ 0.7071 \end{pmatrix} \nonumber$
$D^T \rightarrow \begin{pmatrix} 0.7071 & 0.7071 \end{pmatrix} \nonumber$
$|D \rangle = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} ~~ \langle D | = \begin{pmatrix} 0.707 & 0.707 \end{pmatrix} \nonumber$
Operators for vertically, horizontally, and diagonally oriented polarizers:
Vertically:
$V_{op} = \Theta _{op} (0) \rightarrow \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
$\hat{V} = |V \rangle \langle V| \nonumber$
Horizontally:
$H_{op} = \Theta _{op} \left( \frac{ \pi}{2} \right) \rightarrow \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\hat{H} = |H \rangle \langle H| \nonumber$
Diagonally:
$D_{op} = \Theta _{op} \left( \frac{ \pi}{4} \right) ~ float,~1~ \rightarrow \begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 \end{pmatrix} \nonumber$
$\hat{D} = |D \rangle \langle D| \nonumber$
Demonstrate that a $\theta$‐polarized photon is an eigenfunction of a $\theta$‐oriented polarizer, with eigenvalue 1.
$\begin{pmatrix} \cos \theta^2 & \cos \theta \sin \theta \ \cos \theta \sin \theta & \sin \theta^2 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} ~simplify \rightarrow \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\hat{ \Theta} | \theta \rangle = | \theta \rangle \nonumber$
or
$\Theta_{op} ( \theta) \Theta ( \theta) simplify~ \rightarrow \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
Demonstrate that a $\theta$‐polarized photon is a linear superposition of the vertical and horizontal polarization states.
$\cos \theta \begin{pmatrix} 1 \ 0 \end{pmatrix} + \sin \theta \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
Demonstrate that a vertically polarized photon is a linear superposition of the $\pi$/4 and -$\pi$/4 polarization states.
$\frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} \cos \frac{ \pi}{4} \ \sin \frac{ \pi}{4} \end{pmatrix} + \begin{pmatrix} \cos \frac{ \pi}{4} \ \sin \frac{- \pi}{4} \end{pmatrix} \right] \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Calculate the probability amplitude and probability that a $\pi$/3 (60 degree) polarized photon will pass a vertical polarizer.
$V^T \Theta \frac{ \pi}{3} = 0.5 \nonumber$
$\left( \left| V^T \Theta \left( \frac{ \pi}{3} \right) \right| \right) = 0.25 \nonumber$
or
$\Theta \left( \frac{ \pi}{3} \right)^T V_{op} \Theta \left( \frac{ \pi}{3} \right) = 0.25 \nonumber$
$\left\langle V| \frac{ \pi}{3} \right\rangle = 0.5 \nonumber$
$\left| \left\langle V| \frac{ \pi}{3} \right\rangle \right|^2 = 0.5 \nonumber$
$\left\langle \frac{ \pi}{3} \left| \hat{V} \right| \frac{ \pi}{3} \right\rangle = \left\langle \frac{ \pi}{3} | V \right\rangle \left\langle V | \frac{ \pi}{3} \right\rangle = \left| \left\langle V | \frac{ \pi}{3} \right\rangle \right|^2 = 0.25 \nonumber$
Calculate the probability amplitude and probability that vertically polarized photon will pass a $\pi$/3 (60 degree) polarizer.
$\Theta \left( \frac{ \pi}{3} \right) V = 0.5 \nonumber$
$\left( \left| \Theta \left( \frac{ \pi}{3} \right)^T V \right| \right) ^2 = 0.25 \nonumber$
or
$V^T \Theta_{op} \left( \frac{ \pi}{3} \right) V = 0.25 \nonumber$
Calculate the probability amplitude and probability that a $\pi$/3 (60 degree) polarized photon will pass a diagonal polarizer.
$D^T \Theta \left( \frac{ \pi}{3} \right) = 0.966 \nonumber$
$\left( \left| D^T \Theta \left( \frac{ \pi}{3} \right) \right| \right)^2 = 0.933 \nonumber$
or
$\Theta \left( \frac{ \pi}{3} \right) D_{op} \Theta \left( \frac{ \pi}{3} \right) = 0.933 \nonumber$
Calculate the probability that a $\pi$/3 (60 degree) polarized photon will pass the following sequence of polarizers: vertical, diagonal, horizontal.
$\left( \left| H^T H_{op} D_{op} V_{op} \Theta \left( \frac{ \pi}{3} \right) \right| \right)^2 = 0.063 \nonumber$
$\left| \langle H | \hat{H} \hat{D} \hat{V} \left| \frac{ \pi}{3} \right\rangle \right|^2 = 0.063 \nonumber$
Calculate the probability that a $\pi$/3 (60 degree) polarized photon will pass the following sequence of polarizers: vertical, horizontal, diagonal. Explain the result.
$\left( \left| D^T D_{op} H_{op} V_{op} \Theta \left( \frac{ \pi}{3} \right) \right| \right)^2 = 0 \nonumber$
With this sequence the first two polarizers are crossed (have a 90 degree relative angle). Thus the vertically polarized photon emerging from the first polarizer is stopped by the second polarizer.
Confirm the results shown in the diagram shown below. In other words, show that 12.5% of the incident unpolarized light will pass an arrangement of vertical, diagonal and horizontal polarizers.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \left| H^T H_{op} D_{op} V_{op} \Theta ( \theta0 \right| \right)^2 d \theta = 0.125 \nonumber$
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} | \langle H | \hat{H} \hat{D} \hat{V} | \theta \rangle |^2 d \theta \nonumber$
This calculation can also be performed assuming that unpolarized light is a 50‐50 mixture of vertically and horizontally polarized light.
$\frac{1}{2} ( | H^T H_{op} D_{op} V_{op} V |)^2 + \frac{1}{2} (| H^T H_{op} D_{op} V_{op} H |)^2 = 0.125 \nonumber$
Now a $\pi$/6 polarizer is placed between the vertical and diagonal polarizers, and a $\pi$/3 polarizer is placed between the diagonal and horizontal polarizer. Calculate the fraction of light that emerges from the final horizontal polarizer and explain the result.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \left| H^T H_{op} \Theta_{op} \left( \frac{ \pi}{3} \right) D_{op} \Theta_{op} \left( \frac{ \pi}{6} \right) V_{op} \Theta ( \theta) \right| \right) ^2 d \theta = 0.245 \nonumber$
This calculation can also be performed assuming that unpolarized light is a 50‐50 mixture of vertically and horizontally polarized light.
$\frac{1}{2} \left( \left| H^T H_{op} \Theta_{op} \left( \frac{ \pi}{3} \right) D_{op} \Theta_{op} \left( \frac{ \pi}{6} \right) V_{op} V \right| \right)^2 + \frac{1}{2} \left( \left| H^T H_{op} \Theta_{op} \left( \frac{ \pi}{3} \right) D_{op} \Theta_{op} \left( \frac{ \pi}{6} \right) V_{op} H \right| \right)^2 = 0.245 \nonumber$
The initial and final polarizers are crossed and will not transmit light. A single p/4 polarizer sandwiched in between allows light through for the reasons presented earlier. The addition of two more polarizers increases the fraction of transmitted light because the relative angles between successive polarizers has been reduced. Significantly more light gets through each set of polarizers because the angle between them is smaller.
Calculate the probability that unpolarized light will pass the following sequence of polarizers: vertical, $\pi$/3 (60 degree), diagonal.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \left| D^T D_{op} \Theta_{op} \left( \frac{ \pi}{3} \right) V_{op} \Theta ( \theta) \right| \right) ^2 d \theta = 0.117 \nonumber$
or
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \left| D^T \Theta_{op} \left( \frac{ \pi}{3} \right) V_{op} \Theta ( \theta) \right| \right) ^2 d \theta = 0.117 \nonumber$
Calculate the probability that unpolarized light will pass the following sequence of polarizers: vertical, diagonal, $\pi$/3 (60 degree). Explain the difference in the results.
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \left| \Theta \left( \frac{ \pi}{3} \right)^T \Theta_{op} \left( \frac{ \pi}{3} \right) D_{op} V_{op} \Theta ( \theta) \right| \right) ^2 d \theta = 0.233 \nonumber$
or
$\frac{2}{ \pi} \int_{0}^{ \frac{ \pi}{2}} \left( \left| \Theta \left( \frac{ \pi}{3} \right)^T D_{op} V_{op} \Theta ( \theta) \right| \right) ^2 d \theta = 0.233 \nonumber$
The operators representing the measurement of diagonal and 60 degree polarization do not commute.
Calculate the polarization of the incident photon such that the probability it will pass three polarizers (vertical, horizontal, diagonal) is 0.10
$\theta = 75 ~deg \nonumber$
Given
$\left( \left| H^T H_{op} D_{op} V_{op} \Theta ( \theta) \right| \right)^2 = .10 \nonumber$
Find ($\theta$) = 50.768 deg
The next few exercises involve circularly polarized light.
The base states for circularly polarized light are:
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
Show that they form an orthonormal basis set:
$\overline{L^T} L \rightarrow 1 \nonumber$
$\overline{R^T} R \rightarrow 1 \nonumber$
$\overline{L^T} R \rightarrow 0 \nonumber$
$\overline{R^T} L \rightarrow 0 \nonumber$
Show that they are linear superpositions of the vertical and horizontal polarization states:
$\frac{1}{ \sqrt{2}}\ (V + iH) = \begin{pmatrix} 0.707 \ 0.707i \end{pmatrix}] \[ \frac{1}{ \sqrt{2}}\ (V - iH) = \begin{pmatrix} 0.707 \ -0.707i \end{pmatrix}] Show that vertically and horizontally polarized light can be written as superpositions of circularly polarized light: \[ \frac{1}{ \sqrt{2}}\ (L + R) = \begin{pmatrix} 1 \ 0 \end{pmatrix}] \[ \frac{i}{ \sqrt{2}}\ (R - L) = \begin{pmatrix} 0 \ 1 \end{pmatrix}] The angular momentum operator in atomic units is: \[ Pang = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \nonumber$
Calculate the expectation value for angular momentum for a vertical, horizontal and diagonal polarized photon.
$V^T Pang (V) = 0 \nonumber$
$H^T Pang (H) = 0 \nonumber$
$D^T Pang (D) = 0 \nonumber$
Calculate the expectation value for angular momentum for a $\theta$ polarized photon.
$\Theta ( \theta)^T Pang \Theta ( \theta) = 0 \nonumber$
Calculate the expectation value for angular momentum for left and right circularly polarized photons.
$\overline{L^T} Pang (L) = 1 \nonumber$
$\overline{R^T} Pang (R) = -1 \nonumber$
The remaining exercises deal with the effects of half and quarter wave plates.
The remaining exercises deal with the effects of half and quarter wave plates.
$W_2 = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \nonumber$
$W_2 ( \theta) = \begin{pmatrix} \cos (2 \theta) & \sin (2 \theta) \ \sin (2 \theta) & -\cos (2 \theta) \end{pmatrix} \nonumber$
Quarter wave plate and $\pi$/4 rotated quarter wave plate (which has the same effect as a 50‐50 beam splitter):
$W_4 = \begin{pmatrix} 1 & 0 \ 0 & -i \end{pmatrix} \nonumber$
$BS = \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} \nonumber$
Show that,
When a half wave plate is placed between aligned polarizers all the light gets through.
$\left( \left| V^T W_2 (0) H \right| \right)^2 = 0 \nonumber$
$\left( \left| V^T W_2 (0) V \right| \right)^2 = 1 \nonumber$
When a half wave plate is placed between crossed polarizers no light gets through.
$\left( \left| V^T W_2 (0) H \right| \right)^2 = 0 \nonumber$
When the half wave plate is rotated by an angle of $\pi$/4 all the light gets through.
$\left( \left| V^T W_2 \left( \frac{ \pi}{4} \right) H \right| \right)^2 = 1 \nonumber$
When the half wave plate is rotated by an additional angle of $\pi$/4 no light gets through.
$\left( \left| V^T W_2 \left( \frac{ \pi}{2} \right) H \right| \right)^2 = 0 \nonumber$
When a half wave plate rotated by $\pi$/4 is placed between two vertical or horizontal polarizers no light gets through.
$\left( \left| V^T W_2 \left( \frac{ \pi}{4} \right) V \right| \right)^2 = 0 \nonumber$
$\left( \left| H^T W_2 \left( \frac{ \pi}{4} \right) H \right| \right)^2 = 0 \nonumber$
There is no effect when a quarter wave plate is inserted between either aligned or crossed polarizers.
$\left( \left| V^T W_4 V \right| \right)^2 = 1 \nonumber$
$\left( \left| H^T W_4 V \right| \right)^2 = 1 \nonumber$
If the quarter wave plate is rotated by $\pi$/4 50% of the light gets through.
$\left( \left| V^T BS (V) \right| \right)^2 = 0.5 \nonumber$
$\left( \left| H^T BS (V) \right| \right)^2 = 0.5 \nonumber$
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The Linear Superposition
Unpolarized light consists of photons of all possible polarization angles. A photon polarized at an angle $\theta$ with respect to the vertical can be written as a linear superposition of a vertically polarized photon, $| v \rangle$, and a horizontally polarized photon, $| h \rangle$. $| v \rangle$ and $| h \rangle$ are the polarization basis states.
$| \theta \rangle = | v \rangle \langle v | \theta \rangle + | h \rangle \langle h | \theta \rangle \nonumber$
From the figure above it can be seen that the projection of $| \theta \rangle$ onto $| v \rangle$ and $| h \rangle$ are $\cos \theta$ and $\sin \theta$ respectively.
$| \theta \rangle = | v \rangle \cos \theta + | h \rangle \sin \theta \nonumber$
The probability that a photon polarized at an angle $\theta$ will pass a vertical polarizer is
$| \langle | \theta \rangle | ^2 = \cos^2 \theta \nonumber$
By integrating this function over all possible angles we find that half of the incident light passes through a vertical polarizer. See the figure below.
$\frac{ \int_{0}^{2 \pi} \cos^2 \theta d \theta}{2 \pi} = 0.5 \nonumber$
The photons that pass the vertical polarizer are now vertically polarized. That is they are eigenfunctions of that measurement operator.
The probability that a vertically polarized photon will pass a second filter that is vertically polarized is one.
$| \langle v | v \rangle |^2 = \cos^2 0^o = 1 \nonumber$
The probability that a vertically polarized photon will pass a second filter that is horizontally polarized is zero.
$| \langle h | v \rangle |^2 = \cos^2 90^o = 0 \nonumber$
The vertically polarized photon can be written as linear superposition of any other set of orthogonal basis states, for example $|45^o \rangle$ and $| -45^o \rangle$.
$|v \rangle = |45^o \rangle \langle 45^o | v \rangle + | -45^o \rangle \langle -45^o | v \rangle \nonumber$
$|v \rangle = |45^o \rangle \cos 45^o + | -45^o \rangle \cos -45^o \nonumber$
$|v \rangle = |45^o \rangle 0.707 + | -45^o \rangle 0.707 \nonumber$
Now if a 45o polarizer is inserted in between the vertical and horizontal polarizers photons get through the horizontal polarizer that stopped them previously (see the last figure).
Here's the explanation. The probability that a vertically polarized photon will get through a polarizer oriented at an angle of 45o is 0.5. See the figure below.
$| \langle 45^o | v \rangle |^2 = \cos^2 45^o = 0.5 \nonumber$
Now the photon is in the state of $|45^o \rangle$ which can be written as a linear superposition of $|v \rangle$ and $\h \rangle$.
$|45^o \rangle = |v \rangle \langle v | 45^o \rangle + | h \rangle \langle h | 45^o \rangle \nonumber$
$|45^o \rangle = |v \rangle \cos 45^o + | h \rangle \sin 45^o \nonumber$
$|45^o \rangle = |v \rangle 0.707 + | h \rangle 0.707 \nonumber$
Thus, the probability that this photon will pass the final horizontally oriented polarizer is
$| \langle H | 45^o \rangle |^2 = \sin^2 45^o = 0.5 \nonumber$
See the figure below.
In this last figure the intensity of the light emerging from the final horizontal polarizing filter can be calculated compactly as
$\frac{ \int_{0}^{2 \pi} | \langle h | 45^o \rangle \langle 45^o | v \rangle \langle v | \theta \rangle|^2 d \theta}{2 \pi} = \frac{1}{8} \nonumber$
The term inside the integral is the probability that a photon with polarization θ will pass through the three filters. This expression is integrated over all values of θ.
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The following paragraph appears in an encyclopedia entry on neutron optics.(1) A description of the original experiment is available in the physics research literature. (2) It is the purpose of this tutorial to work through the quantum math of the neutron interferometry experiment described here.
One of the most important properties of quantum theory is the superposition principle, according to which the wave functions of coherent states combine additively. One of the most simple, yet most profound examples of this involves the spin for spin-½ particles. Such particles have two spin states, spin “up” (spin is directed upward along the z axis, sz = + ½), and spin “down” (sz = −½). If the z component of spin is measured, it is found that the particle is either spin up or spin down. Yet a spin along the x axis can be described as a linear combination of these two states. By a judicious use of magnetic fields, the neutron was polarized so that it was spin up in one leg of the interferometer and spin down in the other. Then when these two beams of equal amplitude, each polarized along the z axis, were recombined after passing through the interferometer, it was seen that they were polarized in the x-y plane, exactly according to the laws of quantum theory. This polarization was perpendicular to that of either of the beams separately, which proved both the coherence of the beam and the superposition principle for spin. This is the first time a beam of spin-½ particles has ever been spatially separated and then coherently recombined in this fashion.
The figure shows a simplified schematic of the neutron interferometer and the spin states of the neutron in each leg of the interferometer. The labels U and D refer to the upper and lower arms of the interferometer. The interferometer is fabricated out of a perfect single crystal and the Bragg planes of the crystal function as beam splitters, permitting both transmission and reflection. The middle element, which is labeled as a mirror, is actually another beam splitter; the neutrons transmitted in the upper and lower arms are discarded. The following conventions and simplifications are adopted:
1. It is assumed that the diagram describes the history of an individual neutron.
2. The paths to the detectors are of equal distance, meaning that the neutron beam is recombined at the final beam splitter in such a way that the labels U and D become meaningless at the detectors. They are retained in the diagram so the reader can more readily follow the math.
3. At the beam splitters reflection is assigned a 90° (= π/2 = i) phase shift relative to transmission. (Reminder; the mirror is actually another beam splitter.)
4. In the interest of mathematical economy normalization constants are omitted.
In this experiment neutrons polarized spin-up in the z-direction illuminate a beam splitter creating a superposition of being transmitted (D) and reflected (U). A second beam splitter (mirror) redirects the beam to a third beam splitter which recombines the D and U branches before detection at A and B. But prior to recombination, the lower branch spin state is flipped to spin-down in the z-direction. Recombination at the final beam splitter leads to the superposition states at A and B shown in the figure below.
It is easy to show that these superposition states at A and B are actually spin-down and spin-up in the x-direction. This is because x-direction spin states can be written as linear superpositions of the z-direction spin states. As previously, normalization constants are omitted.
$| \uparrow_x \rangle = | \uparrow_z \rangle + | \downarrow_z rangle \nonumber$
and
$| \downarrow_x \rangle = - | \uparrow_z \rangle + | \downarrow_z rangle \nonumber$
Using these superpositions it is easy to show that the detected spin polarizations are $| \downarrow_x \rangle$ at A and $| \uparrow_x \rangle$ at B.
$|A \rangle = i | \downarrow_x \rangle \nonumber$
and
$|B \rangle = - | \uparrow_x \rangle \nonumber$
In summary, this experiment is a “textbook” illustration of the quantum mechanical superposition principle.
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The illustration of the concept of interaction-free measurement requires the use of an interferometer. A simple illustration employs a Mach-Zehnder interferometer (MZI) like the one shown here.
This equal-arm MZI consists of two 50-50 beam splitters (BS1, BS2), two mirrors (M) and two detectors (D1, D2). A source emits a photon which interacts with BS1 producing the following superposition. (By convention a 90 degree (i) phase shift is assigned to reflection.
$S = \frac{1}{ \sqrt{2}} (T + iR) \label{1}$
The transmitted and reflected branches are united at BS2 by the mirrors, where they evolve into the following superpositions in the basis of the detectors.
$T = \frac{1}{ \sqrt{2}} (i D_1 + D_2) \label{2}$
$R = \frac{1}{ \sqrt{2}} (D_1 + i D_2) \label{3}$
Substitution of Equations \ref{2} and \ref{3} into Equation \ref{1} reveals that the output photon is always registered at $D_1$. There are two paths to each detector and constructive interference occurs at D1 and destructive interference at $D_2$.
\begin{align} S &= \frac{1}{ \sqrt{2}} (T + iR) \bigg|_{substitute,~ R = \frac{1}{ \sqrt{2}} (D_1 + i D_2)}^{substitute~ T = \frac{1}{ \sqrt{2}} (i D_1 + D_2)} \rightarrow \[4pt] & = D_{1} i \end{align} \nonumber
Probability at D1: $(|i|)^2 = 1$
The MZI provides a rudimentary method of determining whether an obstruction is present in its upper arm without actually interacting with it. As we shall see, it is not an efficient method, but it does clearly illustrate the principle involved which then can be used in a more elaborate and sophisticated interferometer to yield better results.
In the presence of the obstruction, Equation \ref{2} becomes $T = \gamma_{Absorbed}$. This leads to the following result at the detectors.
\begin{align} S &= \frac{1}{ \sqrt{2}} (T + iR) \bigg|_{substitute,~ R = \frac{1}{ \sqrt{2}} (D_1 + i D_2)}^{substitute~ T = \gamma_{Absorbed}} \rightarrow \[4pt] & = \frac{ \sqrt{2} \gamma_{Absorbed}}{2} - \frac{D_2}{2} + \frac{D_1 i}{2} \end{align} \nonumber
Quantum mechanics predicts that for a large number of experiments 50% of the photons will be absorbed by the obstruction, 25% will be detected at D1 and 25% will be detect at D2. This later result is the signature of interaction-free measurement. Even if the photon is not absorbed, the mere presence of the obstruction causes the probability of detection at D2 to go from zero to 25%. The photon's arrival at D2 signals the presence of an obstruction in the upper arm of the MZI, and the obstruction is detected without an interaction.
Of course, 25% is not great efficiency, so this is "a proof of principle" example. However, with a little ingenuity the probability of interaction-free detection can be increased dramatically. To see how this can be accomplished read "Quantum Seeing in the Dark" by Kwiat, Weinfurter and Zeilinger in the November 1996 issue of Scientific American.
However, it is possible to improve performance significantly by using a system of nested MZIs which is only slightly more complicated than the simple MZI used earlier. To simplify analysis Feynman's "sum over histories" approach to quantum mechanics will be used. The probability amplitudes for transmission and reflection at the beam splitters are required.
$T = \frac{1}{ \sqrt{2}} \nonumber$
$R = \frac{i}{ \sqrt{2}} \nonumber$
Placing an additional BS before the original MZI and another BS before D2 and renaming it D3, plus an additional mirror and new detector D2, yields a nested interferometer configuration that significantly increases the probability of interaction-free detection of the obstruction.
To interact with the obstruction a photon must be transmitted at the first and second beam splitters. In this case there is only one history and the probability of the interaction occurring is the square of its absolute magnitude.
$(| T T |)^2 \rightarrow \frac{1}{4} = 25% \nonumber$
The probabilities of detectors 1, 2 and 3 firing are calculated using the same methodology.
The probability D1 registers a photon:
$(| TRT|)^2 \rightarrow \frac{1}{8} = 12.5% \nonumber$
The probability D2 registers a photon:
$(|TRRR + RT|)^2 \rightarrow \frac{1}{16} = 6.25% \nonumber$
The probability D3 registers a photon:
$(| TRRT + RR|)^2 = \frac{9}{16} = 56.25% \nonumber$
With the modified interferometer detecting the presence of the obstruction without interacting with it increases from 25% to 56.25%.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.19%3A_Interaction_Free_Measurement_-_Seeing_in_the_Dark.txt
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The following device consisting of an equal-arm polarization interferometer, a switchable mirror and a polarization rotator appears in "Quantum Seeing in the Dark" by Kwiat, Weinfurter and Zeilinger in the November 1996 issue of Scientific American. It is used to illustrate the concept of interaction-free measurement. In what follows I analyze its operation using matrix and tensor algebra.
The polarizing beam splitter transmits horizontally polarized photons and reflects those that are vertically polarized. Therefore, vector states which contain direction of motion and polarization information of the photons are required. This is accomplished using vector tensor multiplication.
Vertical motion:
$x = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Horizontal motion:
$x = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Horizontal polarization:
$x = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Vertical polarization:
$x = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Photon direction of propagation and polarization states:
$xh = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$xh = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$xv = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$xv = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$yh = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$yh = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$yv = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$yv = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
Now the matrix which represents a polarizing beam splitter which transmits horizontally polarized photons and reflects vertically polarized photons is constructed.
$PBS = (xh) xh^T + (yv)xv^T + (yh)yh^T + (xv)yv^T \nonumber$
$PBS = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} \nonumber$
The matrix representing the polarization rotator is:
$Rot( \theta) = \begin{pmatrix} \cos \theta & - \sin \theta & 0 & 0 \ \sin \theta & \cos \theta & 0 & 0 \ 0 & 0 & \cos \theta & -\sin \theta \ 0 & 0 & \sin \theta & \cos \theta \end{pmatrix} \nonumber$
It is unnecessary to include matrices representing the mirrors because they simply reflect the photon back to the beam splitter for a second encounter. First consider the case in which there is no obstructing pebble in the y-branch of the interferometer. A horizontally polarized photon enters the system and the polarization rotator rotates the polarization by 15 degrees (π/12) toward the vertical. This results in a superposition of |xh> and |xv>. This photon state enters the interferometer, |xh> and |xv> are temporarily split, and then recombined to reform the entering superposition. After one cycle the photon's polarization state is as follows.
$\left( PBS (PBS) Rot \left( \frac{ \pi}{12} \right) \right)^1 xh = \begin{pmatrix} 0.966 \ 0.259 \ 0 \ 0 \end{pmatrix} \nonumber$
$0.966xh + 0.259xv = \begin{pmatrix} 0.966 \ 0.259 \ 0 \ 0 \end{pmatrix} \nonumber$
The bottom mirror reflects the photon back to the polarization rotator for another pass through the interferometer. After six cycles the switchable mirror at the bottom of the device releases a vertically polarized photon. In other words, if the equal arm interferometer does not have an obstruction in the y-branch, the initial horizontally polarized photon is rotated stepwise to the vertical polarization state.
$\left( PBS (PBS) Rot \left( \frac{ \pi}{12} \right) \right)^6 xh = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$\left[ \left| xv^T \left( PBS (PBS) Rot \left( \frac{ \pi}{12} \right) \right)^6 xh \right| \right]^2 = 1 \nonumber$
Practically this would be confirmed by having a polarizing beam splitter (PBS) at the output channel of the device. As shown below the photon will be detected at the vertical (y-direction) port and not the horizontal (x-direction) port.
$\left[ \left| yv^T PBS \left( PBS (PBS) Rot \left( \frac{ \pi}{12} \right) \right)^6 xh \right| \right]^2 = 1 \nonumber$
$\left[ \left| xh^T PBS \left( PBS (PBS) Rot \left( \frac{ \pi}{12} \right) \right)^6 xh \right| \right]^2 = 0 \nonumber$
The presence of the pebble takes the y-branch of the interferometer "out of play." Under these circumstances the polarizing beam splitter is represented by the following matrix.
$PBS' = (xh) xh^T \nonumber$
$PBS' = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ \end{pmatrix} \nonumber$
Now after six cycles there is a 66% probability that the mirror will release a horizontally polarized photon. The other 34% of the time the photon will be absorbed by the pebble.
$\left( PBS' (PBS') Rot \left( \frac{ \pi}{12} \right) \right)^6 xh = \begin{pmatrix} 0.812 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$\left[ \left| xh^T \left( PBS' (PBS') Rot \left( \frac{ \pi}{12} \right) \right)^6 xh \right| \right]^2 = 0.66 \nonumber$
As noted earlier, the actual device would have a final PBS after the switchable mirror, showing that a vertical photon is never detected (it has been absorbed) and a horizontal photon is recorded with a 0.66 probability.
$\left[ \left| yv^T PBS \left( PBS' (PBS') Rot \left( \frac{ \pi}{12} \right) \right)^6 xh \right| \right]^2 = 0 \nonumber$
$\left[ \left| xh^T PBS \left( PBS' (PBS') Rot \left( \frac{ \pi}{12} \right) \right)^6 xh \right| \right]^2 = 0.66 \nonumber$
Recall that in the absence of the obstruction the mirror releases a vertically polarized photon 100% of the time. The detection of a horizontally polarized photon indicates the presence of the pebble without an interaction with the pebble. With six cycles interaction-free measurement occurs about two thirds of the time. Increasing the number of cycles increases the utility of the device. For thirty cycles (a π/60 polarization rotation per cycle) a horizontally polarized photon is detected 92% of the time: interaction-free measurement occurs 92% of the time.
$\left[ \left| xh^T \left( PBS' (PBS') Rot \left( \frac{ \pi}{60} \right) \right)^{30} xh \right| \right]^2 = 0.92 \nonumber$
$\left[ \left| xh^T PBS \left( PBS' (PBS') Rot \left( \frac{ \pi}{60} \right) \right)^{30} xh \right| \right]^2 = 0.92 \nonumber$
Increasing the number of cycles increases the reliability of the method, but is also experimentally very challenging.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.20%3A_Quantum_Seeing_in_the_Dark_-_A_Matrix-Tensor_Analysis.txt
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Path difference between interferometer arms is δ. Phase accumulated due to path difference: $exp \left( i \frac{2 \pi \delta}{ \lambda} \right)$
S stands for source, D for detector, T for transmitted and R for reflected. The evolution of the photon wave function at various stages is given below.
$S = \frac{1}{ \sqrt{2}} (T + iR) \nonumber$
$T = \frac{ exp \left( i \frac{2 \pi \delta}{ \lambda} \right)}{ \sqrt{2}} (iD + S) \nonumber$
$R = \frac{1}{ \sqrt{2}} (D + iS) \nonumber$
$S = \frac{1}{ \sqrt{2}} (T +iR) |^{substitute,~T = \frac{exp \left( i \frac{2 \pi \delta}{ \lambda}\right)}{ \sqrt{2}} (iD+S)}_{substitute,~R = \frac{1}{ \sqrt{2}} (D + iS)} \rightarrow S = - \frac{S}{2} + \frac{S e^{ \frac{2i \pi \delta}{ \lambda}}}{2} + \frac{D e^{ \frac{2i \pi \delta}{ \lambda}}}{2} + \frac{Di}{2} \nonumber$
The probability the photon will arrive at the detector is the square of the absolute magnitude of the coefficient of D.
$\frac{-1}{2} \left( e^{ \frac{-2 i \pi \delta}{ \lambda}} \right) \frac{1}{2} \left( e^{ \frac{-2 i \pi \delta}{ \lambda}} \right) simplify~ \rightarrow \sin \left( \frac{ \pi \delta}{ \lambda} \right)^2 \nonumber$
The same results are now illustrated using a matrix mechanics approach. Horizontal and vertical motion of the photon are represented by vectors. The source emits a horizontal photon, the detector receives a vertical photon. The beam splitter and the phase shift due to path length difference are represented by matrices. A matrix representation for the mirrors is unnecessary because the simply return the photon to the beam splitter.
Horizontal motion: $\begin{pmatrix} 1 \ 0 \end{pmatrix}$
Vertical motion: $\begin{pmatrix} 0 \ 1 \end{pmatrix}$
Beam splitter:
$BS = \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} \nonumber$
Phase shift:
$A ( \delta) = \begin{pmatrix} e^{2 i \pi \frac{ \delta}{ \lambda}} & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Calculate the probability amplitude and probability that the photon will arrive at the detector
$\begin{pmatrix} 0 & 1 \end{pmatrix} BS \begin{pmatrix} e^{2 i \pi \frac{ \delta}{ \lambda}} & 0 \ 0 & 1 \end{pmatrix} BS \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \frac{e^{ \frac{2i \pi \delta}{ \lambda}}}{2} + \frac{1}{2} i \nonumber$
$\left( \frac{e^{ \frac{-2i \pi \delta}{ \lambda} (-i)}}{2} + \frac{1}{2} (-i) \right) \left( \frac{e^{ \frac{2i \pi \delta}{ \lambda} (i)}}{2} + \frac{1}{2} i \right) simplify \rightarrow \cos \left( \frac{\pi \delta}{ \lambda}\right)^2 \nonumber$
Calculate the probability amplitude and probability that the photon will be returned to the source.
$\begin{pmatrix} 1 & 0 \end{pmatrix} BS \begin{pmatrix} e^{2i \pi \frac{ \delta}{ \lambda}} & 0 \ 0 & 1 \end{pmatrix} BS \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \frac{e ^{ \frac{2 i \pi \delta}{ \lambda}}}{2} - \frac{1}{2} \nonumber$
$\left( \frac{e^{ \frac{-2i \pi \delta}{ \lambda}}}{2} - \frac{1}{2} \right) \left( \frac{e^{ \frac{2i \pi \delta}{ \lambda}}}{2} - \frac{1}{2} \right) simplify \rightarrow \sin \left( \frac{\pi \delta}{ \lambda}\right)^2 \nonumber$
Plotting these results in units of λ yields:
$\delta = 0, .01 .. 1 \nonumber$
7.22: A Quantum Circuit for a Michelson Interferometer
Schematic diagram of a Mach‐Zehnder interferometer (MZI).
The following quantum circuit simulates the MZI.
The arms of the MZI are represented by the following orthonormal basis.
$| 0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$|1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
The matrices representing the Hadamard and phase shift gates are:
$H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
$A ( \theta) = \begin{pmatrix} 1 & 0 \ 0 & e^{i \phi} \end{pmatrix} \nonumber$
Step‐by‐step through the circuit. The first Hadamard gate creates a superposition of the |0 > and |1 > states. The phase shifter operates on the lower arm of the MZI. The final Hadamard gate allows interference between the two arms of the MZI.
$H \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2}}{2} \end{pmatrix} \nonumber$
$A ( \phi) H \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} e^{ \phi i}}{2} \end{pmatrix} \nonumber$
$H A ( \phi) H \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ e^{ \phi i}}{2} + \frac{1}{2} \ \frac{1}{2} - \frac{ e^{ \phi i}}{2} \end{pmatrix} \nonumber$
Probability of detection at the |0 > port:
$P_0 ( \phi) = \left[ \left| \begin{pmatrix} 1 & 0 \end{pmatrix} H A( \phi) H \begin{pmatrix} 1 \ 0 \end{pmatrix} \right| \right]^2 |_{simplify}^{assume,~ \phi = real} \rightarrow \frac{ \cos \phi}{2} + \frac{1}{2} \nonumber$
Probability of detection at the |1 > port:
$P_1 ( \phi) = \left[ \left| \begin{pmatrix} 0 & 1 \end{pmatrix} H A( \phi) H \begin{pmatrix} 1 \ 0 \end{pmatrix} \right| \right]^2 |_{simplify}^{assume,~ \phi = real} \rightarrow \frac{1}{2} - \frac{ \cos \phi}{2} \nonumber$
A graphical representation of the above calculations shows the interference effects as a function of ϕ.
7.23: The Ramsey Atomic Interferometer
The Ramsey interferometer, which closely resembles the Mach-Zehnder interferometer, is constructed using two π/2 Rabi pulses (R1 and R2) separated by a phase shifter in the lower arm, as shown below.
where
$|e \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$|g \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
The matrix representations of the phase shifter and the Rabi elements are as follows:
$PhaseShift ( \phi) = \begin{pmatrix} 1 & 0 \ 0 & e^{i \phi} \end{pmatrix} \nonumber$
$Rabi( \phi) = \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) & - \sin \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) & \cos \left( \frac{ \theta}{2} \right) \end{pmatrix} \nonumber$
$Rabi \left( \frac{ \pi}{2} \right) = \begin{pmatrix} 0.707 & -0.707 \ 0.707 & 0.707 \end{pmatrix} \nonumber$
The input to the interferometer is the upper state |e> of a two-state atom. The first pulse behaves like a Hadamard gate creating a coherent superposition of |e> and the lower state of the atom, |g>.
$Rabi \left( \frac{ \pi}{2} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} 2^{ \frac{1}{2}} \ \frac{1}{2} 2^{ \frac{1}{2}} \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \nonumber$
The phase shifter alters the superposition by adding a phase to |g>.
$PhaseShift( \phi ) Rabi \left( \frac{ \pi}{2} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} 2^{ \frac{1}{2}} \ \frac{1}{2} e^{i \phi} 2^{ \frac{1}{2}} \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ e^{i \phi} \end{pmatrix} \nonumber$
In the absence of a phase shift (ϕ = 0) the two π/2 pulses behave like a not gate yielding |g> at the output channel.
$Rabi \left( \frac{ \pi}{2} \right) PhaseShift(0) Rabi \left( \frac{ \pi}{2} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
However if ϕ = π, the interferometer is equivalent to the identity operator and the output is |e>.
$Rabi \left( \frac{ \pi}{2} \right) PhaseShift( \pi) Rabi \left( \frac{ \pi}{2} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
In general, the result is a superposition of |e> and |g>.
$Rabi \left( \frac{ \pi}{2} \right) PhaseShift ( \phi) Rabi \frac{ \phi}{2} \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} - \frac{1}{2} e^{i \phi} \ \frac{1}{2} +\frac{1}{2} e^{i \phi} \end{pmatrix} \nonumber$
$\begin{pmatrix} e \ g \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 - e^{i \phi} \ 1 + e^{i \phi} \end{pmatrix} \nonumber$
The probabilities of detecting |e> and |g> at the output channel depend on the phase ϕ, exhibiting interference effects as in the Mach-Zehnder interferometer with a phase shifter in the lower arm. This is shown below both algebraically and graphically.
$P_e ( \phi) = \left[ \left| \begin{pmatrix} 1 \ 0 \end{pmatrix}^T Rabi \left( \frac{ \pi}{2} \right) PhaseShift ( \phi) Rabi \left( \frac{ \pi}{2} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} \right| \right]^2 ~ simplify \rightarrow \frac{1}{2} - \frac{1}{2} \cos \phi \nonumber$
$P_g ( \phi) = \left[ \left| \begin{pmatrix} 0 \ 1 \end{pmatrix}^T Rabi \left( \frac{ \pi}{2} \right) PhaseShift ( \phi) Rabi \left( \frac{ \pi}{2} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} \right| \right]^2 ~ simplify \rightarrow \frac{1}{2} + \frac{1}{2} \cos \phi \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.21%3A_Two_Analyses_of_the_Michelson_Interferometer.txt
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Optical activity is the rotation of linearly polarized light as it advances through a chiral medium. The quantum explanation for optical rotation is based on the fact that linearly polarized light can be written as a superposition of left and right circularly polarized light, which possess angular momentum (see Appendix).
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
This is demonstrated for vertical, horizontal and Θ polarized light.
$V = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} (R + L) \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$H = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\frac{i}{ \sqrt{2}} (L-R) \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\Theta = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} (e^{-i \theta} R + e^{i \theta} L) ~simplify \rightarrow \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
To proceed to a quantum interpretation for optical activity, we assume that $| \theta \rangle$ polarized light entering a chiral medium will be in the state $| \theta + \alpha x \rangle$ after traveling a distance x, where α is the optical activity. Our immediate goal is to find the quantum operator for such a process. By analogy with the time‐dependent Schrödinger equation, $i \frac{h}{2 \pi} \frac{d}{dt} \Psi = H \Psi$, R. L. Martin (Basic Quantum Mechanics, page 30) suggests the following differential equation to describe the spatial evolution of polarization in a chiral medium.
$i \frac{d}{dx} \Psi ( \theta , \alpha , x) = K \Psi ( \theta , \alpha , x) \nonumber$
where
$\Psi ( \theta , \alpha , x) = \begin{pmatrix} \cos ( \theta + \alpha x) \ \sin ( \theta + \alpha x) \end{pmatrix} \nonumber$
Since αx is an angle, id/dx is an angular momentum operator and what follows determines its matrix representation. Consulting the Appendix we see that it is α times the photon angular momentum operator.
Substitution of ψ yields,
$i \alpha \begin{pmatrix} - \sin ( \theta + \alpha x) \ \cos ( \theta + \alpha x) \end{pmatrix} = K \begin{pmatrix} \cos ( \theta + \alpha x) \ \sin ( \theta + \alpha x) \end{pmatrix} \nonumber$
which requires
$K = \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} \nonumber$
$i \frac{d}{dx} \begin{pmatrix} \cos ( \theta + \alpha x) \ \sin ( \theta + \alpha x) \end{pmatrix} \rightarrow \begin{pmatrix} - \alpha \sin ( \theta + \alpha x) \ \alpha \cos ( \theta + \alpha x) \end{pmatrix} \nonumber$
$\begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} \begin{pmatrix} \cos ( \theta + \alpha x) \ \sin ( \theta + \alpha x) \end{pmatrix} \rightarrow \begin{pmatrix} - \alpha \sin ( \theta + \alpha x) \ \alpha \cos ( \theta + \alpha x) \end{pmatrix} \nonumber$
Because they play an essential role in understanding optical rotation, we pause briefly to show that |R > and |L > are eigenvectors of K with eigenvalues α and ‐α respectively.
$\begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} R \rightarrow \begin{pmatrix} \frac{ \sqrt{2} \alpha}{2} \ \frac{ \sqrt{2} \alpha i}{2} \end{pmatrix} \nonumber$
$\alpha \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
$\begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} L \rightarrow \begin{pmatrix} - \frac{ \sqrt{2} \alpha}{2} \ \frac{ \sqrt{2} \alpha i}{2} \end{pmatrix} \nonumber$
$- \alpha \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
Integration of the differential equation describing the change in polarization during propagation through an optically active medium yields the sought after polarization spatial evolution operator.
$\Psi_{final} = e^{-iKx} \Psi_{initial} = e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \Psi_{initial} \nonumber$
where
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} ~simplify~ \rightarrow \begin{pmatrix} \cos ( \alpha x) & - \sin ( \alpha x) \ \sin ( \alpha x) & \cos ( \alpha x) \end{pmatrix} \nonumber$
This demonstrates that the spatial evolution operator is a matrix which rotates the plane of polarization by an angle αx. Now the effect of the operator on vertical, horizontal and Θ polarized light is demonstrated.
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} 1 \ 0 \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} \cos ( \alpha x) \ \sin ( \alpha x) \end{pmatrix} \nonumber$
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} 0 \ 1 \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} - \sin ( \alpha x) \ \cos ( \alpha x) \end{pmatrix} \nonumber$
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} \cos ( \theta + \alpha x) \ \sin ( \theta + \alpha x) \end{pmatrix} \nonumber$
Next the circular polarization states are operated on.
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} \frac{ \sqrt{2} e^{ - \alpha x i}}{2} \ \frac{ \sqrt{2} e^{ - \alpha x i} i}{2} \end{pmatrix} \nonumber$
or
$e^{-i \alpha x} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} \frac{ \sqrt{2} e^{ \alpha x i}}{2} \ - \frac{ \sqrt{2} e^{ - \alpha x i} i}{2} \end{pmatrix} \nonumber$
or
$e^{i \alpha x} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
We see that propagation through the chiral medium causes a phase shift in the circular polarization states. This turns out to be of essential importance in understanding the rotation of plane polarized light which occurs during propagation. It was shown previously, that plane polarized light can be expressed as a superposition of left and right circularly polarized light. Letʹs rewrite that superposition using the phase relation just demonstrated. For the sake of computational simplicity we will use α = 1, and because the phase factor is a trigonometric function we will express distance in units of π.
Our example involves a vertically polarized light source illuminating a chiral medium. At the point of entry, x = 0, we show that the light is indeed vertically polarized.
$x = 0$:
$\frac{1}{2} \left[ exp (-i \alpha x) \begin{pmatrix} 1 \ i \end{pmatrix} + exp(i \alpha x) \begin{pmatrix} 1 \ -i \end{pmatrix} \right] = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
This formulation shows that as light penetrates the chiral medium, the vertically polarized photon is rotated.
$x = \frac{ \pi}{4}$:
$\frac{1}{2} \left[ exp (-i \alpha x) \begin{pmatrix} 1 \ i \end{pmatrix} + exp(i \alpha x) \begin{pmatrix} 1 \ -i \end{pmatrix} \right] = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} \nonumber$
$x = \frac{ \pi}{2}$:
$\frac{1}{2} \left[ exp (-i \alpha x) \begin{pmatrix} 1 \ i \end{pmatrix} + exp(i \alpha x) \begin{pmatrix} 1 \ -i \end{pmatrix} \right] = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Of course, the same results are obtained by operating directly on |v >.
$x = 0$:
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} 1 \ 0 \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$x = \frac{ \pi}{4}$:
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} 1 \ 0 \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2}}{2} \end{pmatrix} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} \nonumber$
$x = \frac{ \pi}{2}$:
$e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} 1 \ 0 \end{pmatrix} ~simplify~ \rightarrow = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
The juxtaposition of the two sets of calculations suggests that the rotation of plane polarized light in a chiral medium is due to phase changes in its left, |L >, and right, |R >, circularly polarized components during propagation. Couching the explanation in terms of |L > and |R > is further justified by the fact that they are the eigenstates of the Hermitian K matrix which produces the rotation. A picture of the progression of the angle of rotation as light advances through a chiral medium can be created by plotting the vertical vs horizontal components of the polarization vector as a function of x.
$x = 0, .1 \pi .. 2 \pi$:
$V(x) = e^{-i \begin{pmatrix} 0 & -i \alpha \ i \alpha & 0 \end{pmatrix} x} \begin{pmatrix} 1 \ 0 \end{pmatrix} ~simplify~ \rightarrow \begin{pmatrix} \cos x \ \sin x \end{pmatrix} \nonumber$
for α = 1.
Appendix
The angular momentum operator for photons in atomic units is $M = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix}$, which means, of course, that K = αM. |R > and |L > are eigenvectors of M with eigenvalues +1 and ‐1, respectively. As shown earlier they are eigenvectors of K with eigenvalues α and ‐α, respectively.
$M R = \begin{pmatrix} 0.707 \ 0.707i \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
$M L = \begin{pmatrix} -0.707 \ 0.707i \end{pmatrix} \nonumber$
$-\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
The linearly polarized states are not eigenvectors of M and have angular momentum expectation values of zero.
$\begin{pmatrix} 1 & 0 \end{pmatrix} M \begin{pmatrix} 1 \ 0 \end{pmatrix} = 0 \nonumber$
$\begin{pmatrix} 0 & 1 \end{pmatrix} M \begin{pmatrix} 0 \ 1 \end{pmatrix} = 0 \nonumber$
$\begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} M \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \rightarrow 0 \nonumber$
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The following is a summary of "Quantum Cheshire Cats" by Aharonov, Popescu, Rohrlich and Skrzypczyk which was published in the New Journal of Physics 15, 113015 (2013) and can also be accessed at: arXiv:1202.0631v2.
In the absence of the half-wave plate (HWP) and the phase shifter (PS) a horizontally polarized photon entering the interferometer from the lower left (propagating to the upper right) arrives at D2 with a 90 degree (π/2, i) phase shift. (By convention reflection at a beam splitter introduces a 90 degree phase shift.)
$|R \rangle |H \rangle \xrightarrow{BS_1} \frac{1}{ \sqrt{2}} [ i | L \rangle + |R \rangle ] |H \rangle \xrightarrow{BS_2} i | D_2 \rangle | H \rangle \nonumber$
The state immediately after the first beam splitter is the pre-selected state.
$| Psi \rangle = \frac{1}{ \sqrt{2}} \left[ i | L \rangle + | R \rangle \right] |H \rangle \nonumber$
The post-selected state is,
$| \Phi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle |H \rangle + | R \rangle |V \rangle \right] \nonumber$
The HWP (converts |V> to |H> in the R-branch) and PS transform this state to,
$| \Phi \rangle \xrightarrow[PS]{HWP} \frac{1}{ \sqrt{2}} \left[ | L \rangle + i | R \rangle \right] |H \rangle \nonumber$
which exits the second beam splitter through the left port to encounter a polarizing beam splitter which transmits horizontal polarization and reflects vertical polarization. Thus, the post-selected state is detected at D1. The evolution of the post-selected state is summarized as follows:
$| \Phi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle |H \rangle + |R \rangle |V \rangle \right] \xrightarrow[PS]{HWP} \frac{1}{ \sqrt{2}} \left[ | L \rangle + i | R \rangle \right] |H \rangle \xrightarrow{BS_2} i |L \rangle |H \rangle \xrightarrow{PBS} i | D_1 \rangle |H \rangle \nonumber$
The last term on the right side below is the weak value of A multiplied by the probability of its occurrence for the preselected state Ψ and the post-selected state Φ.
$\langle \Psi | \hat{A} | \Psi \rangle = \sum_j \langle \Psi | \Phi_j \rangle \langle \Phi_j | \hat{A} | \Psi \rangle = \sum_j \langle \Psi | \Phi_j \rangle \langle \Phi_j | \Psi \rangle \frac{ \langle \Phi_j | \hat{A} | \Psi \rangle}{ \langle \Phi_j | \Psi \rangle} = \sum_j p_j \frac{ \langle \Phi_j | \hat{A} | \Psi \rangle}{ \langle \Phi_j | \Psi \rangle} \nonumber$
The weak value calculations are carried out in a 4-dimensional Hilbert space created by the tensor product of the photon's direction of propagation and polarization vectors.
Direction of propagation vectors:
$L = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$R = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Polarization state vectors:
$H = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$V = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Pre-selected state:
$\Psi = \frac{1}{ \sqrt{2}} (iL + R)H = \frac{1}{ \sqrt{2}} \left[ i \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] + \begin{pmatrix} 1 \ 0 \end{pmatrix} = frac{1}{ \sqrt{2}} \begin{pmatrix} i \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} i \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
Post-selected state:
$\Phi = \frac{1}{ \sqrt{2}} (LH + RV) = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \nonumber$
$\Phi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
Direction of propagation operators:
$Left = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
$Right = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Photon angular momentum operator:
$Pang = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \nonumber$
Identity operator:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
The following weak value calculations show that for the pre- and post-selection ensemble of observations the photon is in the left arm of the interferometer while its angular momentum is in the right arm. Like the case of the Cheshire cat, a photon property has been separated from the photon.
$\begin{pmatrix} "" & "Left ~Arm" & "Right ~Arm" \ "Arm" & \frac{ \Phi^T kronecker(Left, ~I) \Psi}{ \Phi^T \Psi} & \frac{ \Phi^T kronecker(Right, ~I) \Psi}{ \Phi^T \Psi} \ "Pang" & \frac{ \Phi^T kronecker(Left, ~Pang) \Psi}{ \Phi^T \Psi} & \frac{ \Phi^T kronecker(Right, ~Pang) \Psi}{ \Phi^T \Psi} \end{pmatrix} = \begin{pmatrix} "" & "Left~Arm" & "Right~Arm \ "Arm" & 1 & 0 \ "Pang" & 0 & 1 \end{pmatrix} \nonumber$
The following shows the evolution of the pre-selected state to the final state at the detectors. The intermediate is the state illuminating BS2. The polarization state at the detectors is ignored.
$| \Psi \rangle \rightarrow \frac{i}{ \sqrt{2}} \left[ |L \rangle |H \rangle + |R \rangle |V \rangle \right] \rightarrow - \frac{1}{2} |D_1 \rangle + \frac{i}{2} | D_3 \rangle + \frac{(i-1)}{2} |D_2 \rangle \nonumber$
Squaring the magnitude of the probability amplitudes shows that the probabilities that D1, D3 and D2 will fire are 1/4, 1/4 and 1/2, respectively. The probability at D1 is consistent with the probability that the post-selected state is contained in the pre-selected state. A photon in the post-selected state has a probability of 1 of reaching D1 and it represents a 25% contribution to the pre-selected state.
$(| \Phi^T \Psi |)^2 \rightarrow \frac{1}{4} \nonumber$
Note that the expectation values for the pre-selected state show no path-polarization separation.
$\begin{pmatrix} "" & "Left~Arm" & "Right~Arm" \ "Arm" & ( \overline{ \Psi})^T kronecker(Left,~I) \Psi & ( \overline{ \Psi})^T kronecker(Right,~I) \Psi \ "Pang" & ( \overline{ \Psi})^T kronecker(Left,~Pang) \Psi & ( \overline{ \Psi})^T kronecker(Right,~Pang) \Psi \ "Hop" & ( \overline{ \Psi})^T kronecker(Left,~HH^T) \Psi & ( \overline{ \Psi})^T kronecker(Right,~HH^T) \Psi \ "Vop" & ( \overline{ \Psi})^T kronecker(Left,~VV^T) \Psi & ( \overline{ \Psi})^T kronecker(Right,~VV^T) \Psi \end{pmatrix} = \begin{pmatrix} "" & "Left~Arm" & "Right~Arm" \ "Arm" & 0.5 & 0.5 \ "Pang" & 0 & 0 \ "Hop" & 0.5 & 0.5 \ "Vop" & 0 & 0 \end{pmatrix} \nonumber$
In addition the following table shows that linear polarization (HV) is not separated from the photon's path.
$HV = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \nonumber$
$\begin{pmatrix} "" & "Left ~Arm" & "Right ~Arm" \ "Arm" & \frac{ \Phi^T kronecker(Left, ~I) \Psi}{ \Phi^T \Psi} & \frac{ \Phi^T kronecker(Right, ~I) \Psi}{ \Phi^T \Psi} \ "HV" & \frac{ \Phi^T kronecker(Left, ~HV) \Psi}{ \Phi^T \Psi} & \frac{ \Phi^T kronecker(Right, ~HV) \Psi}{ \Phi^T \Psi} \end{pmatrix} = \begin{pmatrix} "" & "Left~Arm" & "Right~Arm \ "Arm" & 1 & 0 \ "HV" & 1 & 0 \end{pmatrix} \nonumber$
The "Complete Quantum Cheshire Cat" by Guryanova, Brunner and Popescu (arXiv 1203.4215) provides an optical set-up which achieves complete path-polarization separation for the photon.
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In this experiment a down converter, DC, converts an incident photon into two lower energy photons. One photon takes the upper path and the other the lower path. The results of this experiment are that both photons are detected at either A or B. One photon is never detected at A while the other is detected at B. A quantum mechanical analysis of this phenomena is provided below.
After the down converter the initial photon leaving the source has evolved into a state which is an entangled linear superposition.
$| Source \rangle \frac{[ |up \rangle_1 |down \rangle_2 + |down \rangle_1 |up \rangle_2 ]}{ 2^{ \frac{1}{2}}} \label{eq1}$
This is an entangled state (non-factorable) because it acknowledges that it is unknown which photon takes which path. It also incorporates the fact that photons are bosons and consequently the state function must be symmetric with respect to interchange of the coordinates (paths) of the photons. If photons were fermions, |Source> would have to be antisymmetric with respect interchange of coordinates and the positive sign in Equation \ref{eq.1} would be replaced with a negative sign.
A photon that takes the upper path has a 50% chance of being detected at A or B. To reach A it must be reflected at the beam splitter and to reach B is must be transmitted. Conservation of energy requires a 90 degree phase difference between transmission and reflection, and by convention this phase difference is assigned to reflection. To reach detector A the upper photon must undergo a reflection at the beam splitter and its phase shift is recorded by multiplying |A> by i [(-1)1/2]. Thus in terms of the detector states |A> and |B> the photon taking the upper path evolves into the state shown in Equation \ref{eq2}.
$| up \rangle \frac{[ i |A \rangle + | B \rangle ]}{ 2^{ \frac{1}{2}}} \label{eq2}$
Similar arguments show that the photon taking the lower path will evolve to the state given by Equation \ref{eq3}.
$| down \rangle \frac{[ i |A \rangle + | B \rangle ]}{ 2^{ \frac{1}{2}}} \label{eq3}$
When Equations \ref{eq2} and \ref{eq3} are substituted into Equation \ref{eq1}, the following final state results (A and B represent the detectors, while 1 and 2 designate the photons):
$|Source \rangle \frac{[i | A \rangle_1 |A \rangle_2 + i^2 | A \rangle_1 B \rangle_2 + |B \rangle_1 |A \rangle_2 + i|B \rangle_1 |B \rangle_2 + i | A \rangle_1 |A \rangle_2 + | A \rangle_1 |B \rangle_2 + i^2 |B \rangle_1 |A \rangle_2 + i|B \rangle_1 |B \rangle_2]}{ 2^{ \frac{3}{2}}} \nonumber$
Thus there eight final probability amplitudes, and they come in four pairs as can be seen above. However, two of the pairs destructively interfer (see note below) with each other and the final state is,
$| Source \rangle \frac{[ i|A \rangle_1 |A \rangle_2 + i|B \rangle_1 |B \rangle_2]}{ 2^{ \frac{1}{2}}} \nonumber$
The probability of an outcome is found by taking the square of the absolute magnitude of its probability amplitude. Thus the probability that both photons will be recorded at A or both photons will be recorded at B is calculated as follows:
$P(AA) = \left| \frac{i}{2} ^{ \frac{1}{2}}\right|^2 = \frac{1}{2} \nonumber$
$P(BB) = \left| \frac{i}{2} ^{ \frac{1}{2}}\right|^2 = \frac{1}{2} \nonumber$
It was noted earlier that fermions require anti-symmetric state functions. So if this experiment could be performed with fermions Equation \ref{eq1} would become,
$| Source \rangle \frac{[|up \rangle_1 |down \rangle_2 - |down \rangle_1 |up \rangle_2 ]}{ 2^{ \frac{1}{2}}} \label{eq8}$
After substitution of Equations \ref{eq2} and \ref{eq3} into Equation \ref{eq8}, we find
$| Source \rangle \frac{[ B \rangle_1 |A \rangle_2 - |A \rangle_1 |B \rangle_2]}{ 2^{ \frac{1}{2}}} \nonumber$
In other words the fermions are always detected at different detectors and are never found at the same detector at the same time. In summary, the sociology of bosons and fermions can be briefly stated: bosons are gregarious and enjoy company; fermions are antisocial and prefer solitude.
Note: "The things that interfere in quantum mechanics are not particles. They are probability amplitudes for certain events. It is the fact that probability amplitudes add up like complex numbers that is responsible for all quantum mechanical interferences." Roy J. Glauber, American Journal of Physics, 63(1), 12 (1995).
Reference: Greenberger, D. M.; Horne, M. A.; Zeilinger, A. Physics Today, 1993, 44(8), 22.
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A parametric down converter, PDC, transforms an incident photon into two lower energy photons. One photon takes the upper path and the other the lower path or vice versa. The principles of quantum mechanics require that the wave function for this event be written as the following entangled superposition.
Entangled superposition for bosons:
$\frac{1}{ \sqrt{2}}(U_1 D_2 + D_1 U_2) \nonumber$
At the beam splitter, BS, the probability amplitude for transmission is $\frac{1}{ \sqrt{2}}$ and the probability amplitude for reflection is $\frac{i}{ \sqrt{2}}$. Therefore, for the four possible arrivals at the detectors we have,
$U_1 = \frac{1}{ \sqrt{2}} (i A_1 + B_1) \nonumber$
$D_1 = \frac{1}{ \sqrt{2}} (A_1 + i B_1) \nonumber$
$U_2 = \frac{1}{ \sqrt{2}} (i A_2 + B_2) \nonumber$
$D_2 = \frac{1}{ \sqrt{2}} (A_2 + i B_2) \nonumber$
Bosons are always observed at the same detector.
$\frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2} (i A_1 + B_1 \sqrt{2} (A_2 + iB_2 + \sqrt{2} (A_1 + i B_1 \sqrt{2} (i A_2 + B_2}\right] ~simplify~ \rightarrow \sqrt{2} \left( \frac{A_1 A_2}{2} + \frac{B_1 B_2}{2} \right) i \nonumber$
Entangled superposition for fermions:
$\frac{1}{ \sqrt{2}} U_1 D_2 - D_1 U_2 \nonumber$
Fermions are never observed at the same detector.
$\frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2} (i A_1 + B_1 \sqrt{2} (A_2 + iB_2 - \sqrt{2} (A_1 + i B_1 \sqrt{2} (i A_2 + B_2}\right] ~simplify~ \rightarrow \frac{ \sqrt{2} A_2 B_1}{2} + \frac{ \sqrt{2} A_1 B_2}{2} \nonumber$
In summary, the sociology of bosons and fermions can be briefly stated: bosons are gregarious and enjoy company; fermions are antisocial and prefer solitude.
Another way to do the calculation using Mathcad.
Bosons:
$\frac{1}{ \sqrt{2}} (U_1 D_2 + D_1 U_2)~ \begin{array}{|l} \text{substitute},~ U_1 = \frac{1}{ \sqrt{2}} (iA_1 + B_1) \ \text{substitute},~ D_2 = \frac{1}{ \sqrt{2}} (A_2 + iB_2) \ \text{substitute},~ D_1 = \frac{1}{ \sqrt{2}} (A_1 + iB_1) \ \text{substitute},~ U_2 = \frac{1}{ \sqrt{2}} (iA_2 + B_2) \ \text{simplify} \ \end{array} \rightarrow \sqrt{2} \left( \frac{A_1 A_2}{2} + \frac{B_1 B_2}{2} \right) i \nonumber$
Fermions:
$\frac{1}{ \sqrt{2}} (U_1 D_2 - D_1 U_2)~ \begin{array}{|l} \text{substitute},~ U_1 = \frac{1}{ \sqrt{2}} (iA_1 + B_1) \ \text{substitute},~ D_2 = \frac{1}{ \sqrt{2}} (A_2 + iB_2) \ \text{substitute},~ D_1 = \frac{1}{ \sqrt{2}} (A_1 + iB_1) \ \text{substitute},~ U_2 = \frac{1}{ \sqrt{2}} (iA_2 + B_2) \ \text{simplify} \ \end{array} \rightarrow \frac{ \sqrt{2} A_2 B_1}{2} - \frac{ \sqrt{2} A_1 B_2}{2} \nonumber$
7.28: Analysis of a Two-photon Interferometer
In the two‐photon interferometer shown below, an atom simultaneously emits identical photons to the east and west. Two paths (long and short) are available to detectors A and B in the east and west arms of the interferometer. The short paths are the same on both sides, but the long paths can have different lengths causing phase differences for the east and west directions.
Labeling the photons 1 and 2, and the directions W and E, the entangled photon wave function is:
$\frac{1}{ \sqrt{2}} (W_1 E_2 + E_1 W_2) \nonumber$
At the beam splitters (dashed diagonal lines), the probability amplitude for transmission is $\frac{1}{ \sqrt{2}}$ and the probability amplitude for reflection is $\frac{1}{ \sqrt{2}}$. By convention a π/2 phase shift is assigned to reflection. Armed with this information it in possible to trace the evolution of the initial entangled state. After the first beam splitter in each arm we have:
$W_1 = \frac{1}{ \sqrt{2}} (S_{w1} + i exp(i \phi_w) L_{w1}) \nonumber$
$E_1 = \frac{1}{ \sqrt{2}} (S_{e1} + i exp(i \phi_e) L_{e1}) \nonumber$
$W_2 = \frac{1}{ \sqrt{2}} (S_{w2} + i exp(i \phi_w) L_{w2}) \nonumber$
$E_2 = \frac{1}{ \sqrt{2}} (S_{e2} + i exp(i \phi_e) L_{e2}) \nonumber$
The second beam splitter has the following effect on the eight terms above:
$S_{w1} = \frac{1}{ \sqrt{2}} (A_{w1} + i B_{w1}) \nonumber$
$L_{w1} = \frac{1}{ \sqrt{2}} (i A_{w1} + B_{w1}) \nonumber$
$S_{w2} = \frac{1}{ \sqrt{2}} (A_{w2} + i B_{w2}) \nonumber$
$L_{w2} = \frac{1}{ \sqrt{2}} (i A_{w2} + B_{w2}) \nonumber$
$S_{e1} = \frac{1}{ \sqrt{2}} (A_{e1} + i B_{e1}) \nonumber$
$L_{e1} = \frac{1}{ \sqrt{2}} (i A_{e1} + B_{e1}) \nonumber$
$S_{e2} = \frac{1}{ \sqrt{2}} (A_{e2} + i B_{e2}) \nonumber$
$L_{e2} = \frac{1}{ \sqrt{2}} (i A_{e2} + B_{e2}) \nonumber$
Mathcad facilitates the substitution of these 12 expressions into the original entangled two‐photon wave function as is shown below. To expedite interpretation we choose specific values for the relative phases of the photons in the branches of the interferometer. If the L paths both have ϕ = 0, they have the same length and the same length as the S paths. The photons always arrive at the B detectors. Other simple cases are summarized in the following table. However, most L‐path phase relationships lead to complicated output wave functions.
$\begin{bmatrix} \phi_w & \phi_e & WF \ 0 & 0 & \frac{-1}{2} 2^{ \frac{1}{2}} (B_{w1} B_{e2} + B_{e1} B_{w2}) \ 0 & \pi & \frac{1}{2} i 2^{ \frac{1}{2}} (B_{w1} A_{e2} + A_{e1} B_{w2}) \ \pi & \pi & \frac{1}{2} 2^{ \frac{1}{2}} (A_{w1} A_{e2} + A_{e1} A_{w2}) \end{bmatrix} \nonumber$
$\phi_w = 0 ~~~ \phi_e = 0 \nonumber$
$\frac{1}{ \sqrt{2}} (W_1 E_2 + E_1 W_2)~ \begin{array}{|l} \text{substitute},~ W_1 = \frac{1}{ \sqrt{2}} (S_{w1} + i exp(i \phi_w) L_{w1}) \ \text{substitute},~ W_2 = \frac{1}{ \sqrt{2}} (S_{w2} + i exp(i \phi_w) L_{w2}) \ \text{substitute},~ E_1 = \frac{1}{ \sqrt{2}} (S_{e1} + i exp(i \phi_w) L_{e1}) \ \text{substitute},~ E_2 = \frac{1}{ \sqrt{2}} (S_{e2} + i exp(i \phi_w) L_{e2}) \ \text{substitute},~ S_{w1} = \frac{1}{ \sqrt{2}} (A_{w1} + iB_{w1}) \ \text{substitute},~ L_{w1} = \frac{1}{ \sqrt{2}} (iA_{w1} + B_{w1}) \ \text{substitute},~ S_{w2} = \frac{1}{ \sqrt{2}} (A_{w2} + iB_{w2}) \ \text{substitute},~ L_{w2} = \frac{1}{ \sqrt{2}} (iA_{w2} + B_{w2}) \ \text{substitute},~ S_{e1} = \frac{1}{ \sqrt{2}} (A_{e1} + iB_{e1}) \ \text{substitute},~ L_{e1} = \frac{1}{ \sqrt{2}} (iA_{e1} + B_{e1}) \ \text{substitute},~ S_{e2} = \frac{1}{ \sqrt{2}} (A_{e2} + iB_{e2}) \ \text{substitute},~ L_{e2} = \frac{1}{ \sqrt{2}} (i A_{e2} + B_{e2}) \ \text{simplify} \ \end{array} \rightarrow \frac{-1}{2} 2^{ \frac{1}{2}} (B_{w1} B_{e2} + B_{e1} B_{w2}) \nonumber$
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In this experiment a parametric down converter, PDC, transforms an incident photon into two lower energy photons, one horizontally polarized and the other vertically polarized. One photon takes the upper path and the other the lower path or vice versa. The upper branch of the apparatus contains a polarization rotator, θ. The photon paths are combined at a 50-50 beam splitter, after which they interact with a polarizing beam splitter which separates horizontal (transmitted) and vertical (reflected) polarization. The results of this experiment are that no matter the angle θ both photons are detected at either the upper set of detectors (uh,dv) or the lower set of detectors (uv,dh). One photon is never detected at the upper set of detectors while the other is detected at the lower set of detectors. A quantum mechanical analysis of this phenomena is provided below.
PDC = Parametric Down Converter
BS = Beam Splitter
PBS = Polarizing Beam Splitter
uh = moving up, horizontal polarization
dv = moving down, vertical polarization
uv = moving up, vertical polarization
dh = moving down, horizontal polarization
θ = polarization rotator
The vectors representing the motional and polarization states of the photons are:
Motional states:
$u = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$d = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Polarization states:
$h = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$v = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
There are four motional-polarization photon states after the PDC. These are formed by the tensor product of the appropriate motional and polarization vectors.
Combined motional and polarization states:
$uh = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$uv = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$dh = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$dv = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The optical devices are represented by matrices and operate on the vector states of the photons.
Rotation:
$R( \theta ) \begin{pmatrix} \cos \theta & - \sin \theta \ \sin \theta & \cos \theta \end{pmatrix} \nonumber$
Beam splitter:
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
Polarizing beam splitter:
$PBS = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} \nonumber$
An operator for the mirrors is not necessary, because if it is included it has no effect on the measurement results. The identity and a null matrix are also required as will be seen below.
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Null matrix:
$N = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} \nonumber$
The photon state after the PDC is a even superposition of all possible motional-polarization states.
$| \Psi \rangle _i = \frac{1}{2} [ |uh \rangle |dv \rangle + |uv \rangle |dh \rangle + |dh \rangle |uv \rangle + |dv \rangle |uh \rangle ] = \frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \right] \nonumber$
This state is formed as follows using Mathcad's kronecker command.
$\Psi (a, b) = \text{submatrix} [ \text{kronecker} [ \text{augment} (a, N),~ ( \text{augment}(b, N)), 1,~16,~1,~1] \nonumber$
$\Psi_i = \frac{1}{2} \left( \Psi_{(uh,~dv)} + \Psi_{(uv,~dvh)} + \Psi_{(dh,~uv)} + \Psi_{(dv,~uh)} \right) \nonumber$
$\Psi_i^T = \begin{pmatrix} 0 & 0 & 0 & 0.5 & 0 & 0 & 0.5 & 0 & 0 & 0.5 & 0 & 0 & 0.5 & 0 & 0 & 0 \end{pmatrix} \nonumber$
The final state is calculated as a function of the orientation of the polarization rotator.
$\Psi_f ( \theta) = \text{kronecker} (PBS,PBS) \text{kronecker} (BS, \text{kronecker} (I, \text{kronecker} (BS,~ I))) \text{kronecker} (I,~ \text{kronecker} (R, ( \theta), \text{kronecker} (I,~ I))) \Psi_i \nonumber$
For θ = 0 the detected states are |uh>|dv>, |uv>|dh>, |dh>|uv> and |dv>|uh> each with a 25% probability of occurrence.
$\begin{bmatrix} \left( \left| \Psi (uh, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (uv, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (dh, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (dv, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, dv) \Psi_f ( \theta) \right| \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & 0 & 0 & \frac{1}{4} \ 0 & 0 & \frac{1}{4} & 0 \ 0 & \frac{1}{4} & 0 & 0 \ \frac{1}{4} & 0 & 0 & 0 \end{pmatrix} \nonumber$
For $\theta = \frac{ \pi}{2}$ the detected states are |uh>|uh>, |uv>|uv>, |dh>|dh> and |dv>|dv> each with a 25% probability of occurrence.
$\begin{bmatrix} \left( \left| \Psi (uh, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (uv, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (dh, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (dv, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, dv) \Psi_f ( \theta) \right| \right)^2 \end{bmatrix} = \begin{pmatrix} \frac{1}{4} & 0 & 0 & 0 \ 0 & \frac{1}{4} & 0 & 0 \ 0 & 0 & \frac{1}{4} & 0 \ 0 & 0 & 0 & \frac{1}{4} \end{pmatrix} \nonumber$
For $\theta = \frac{ \pi}{4}$ the results of the previous examples occur with each state being observed with a probability of 12.5%.
$\begin{bmatrix} \left( \left| \Psi (uh, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uh, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (uv, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (uv, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (dh, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dh, dv) \Psi_f ( \theta) \right| \right)^2 \ \left( \left| \Psi (dv, uh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, uv) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, dh) \Psi_f ( \theta) \right| \right)^2 & \left( \left| \Psi (dv, dv) \Psi_f ( \theta) \right| \right)^2 \end{bmatrix} = \begin{pmatrix} \frac{1}{8} & 0 & 0 & \frac{1}{0} \ 0 & \frac{1}{8} & \frac{1}{8} & 0 \ 0 & \frac{1}{8} & \frac{1}{8} & 0 \ \frac{1}{8} & 0 & 0 & \frac{1}{8} \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.29%3A_Two-photon_Interferometry.txt
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In this experiment a down converter, DC, converts an incident photon into two lower energy photons. One photon takes an upper path traveling left or right, and the other photon takes one of the lower paths traveling in the opposite direction. The results of this experiment are that both photons are recorded at either the A detectors or B detectors. One photon is never observed at an A detector while the other is observed at a B detector. A quantum mechanical analysis of this phenomena is provided below.
After the down converter the initial photon has evolved into a state which is an entangled linear superposition.
$|DC \rangle \frac{[ |up \rangle_1 |down \rangle_2 + |down \rangle_1 |up \rangle_2]}{2^{ \frac{1}{2}}} \nonumber$
This is an entangled state (non-factorable) because it acknowledges that it is unknown which photon takes which path. It also incorporates the fact that photons are bosons and consequently the state function must be symmetric with respect to interchange of the coordinates (paths) of the photons. If photons were fermions, |DC> would have to be antisymmetric with respect interchange of coordinates and the positive sign in equation (1) would be replaced with a negative sign.
A photon that takes one of the upper paths has a 50% chance of being recorded at an A or a B detector. To reach A it must be reflected at the beam splitter and to reach B is must be transmitted. Conservation of energy requires a 90 degree phase difference between transmission and reflection, and by convention this phase difference is assigned to reflection. To reach detector A the upper photon must undergo a reflection at the beam splitter and its phase shift is recorded by multiplying |A> by i [(-1)1/2]. Thus in terms of the detector states |A> and |B> a photon taking an upper path evolves into the state shown in equation (2).
$|up \rangle \frac{[ i|A \rangle + |B \rangle ]}{2^{ \frac{1}{2}}} \nonumber$
Similar arguments show that a photon taking one of the lower paths will evolve into the state given by equation (3).
$|down \rangle \frac{[ |A \rangle + i|B \rangle ]}{2^{ \frac{1}{2}}} \nonumber$
When equations (2) and (3) are substituted into equation (1) the following final state results:
$|Source \rangle \frac{[ i|A \rangle_1 |A \rangle_2 + i^2 |A \rangle_1 |B \rangle_2 + |B \rangle_1 |A \rangle_2 + i |B \rangle_1 |B \rangle_2 + i|A \rangle_1 |A \rangle_2 + |A \rangle_1 |B \rangle_2 + i^2 |B \rangle_1 |A \rangle_2 + i |B \rangle_1 |B \rangle_2]}{2^{ \frac{3}{2}}} \nonumber$
Thus there are eight final probability amplitudes, and they come in four pairs as can be seen above. However, two of the pairs destructively interfer (see note below) with each other and the final state is,
$|Source \rangle \frac{[ i|A \rangle_1 |A \rangle_2 + i |B \rangle_1 |B \rangle_2 ]}{2^{ \frac{1}{2}}} \nonumber$
The probability of an outcome is found by taking the square of the absolute magnitude of its probability amplitude. Thus the probability that both photons will be observed at A detectors, or both photons will be observed at B detectors is calculated as follows:
$P(AA) = \left| \frac{1}{2} ^{ \frac{1}{2}} \right| ^2 = \frac{1}{2} \nonumber$
$P(BB) = \left| \frac{i}{2} ^{ \frac{1}{2}} \right| ^2 = \frac{1}{2} \nonumber$
It was noted earlier that fermions require anti-symmetric state functions. So if this experiment could be performed with fermions the results would be P(AA) = P(BB) = 0, and P(AB) = P(BA) = 1/2.
Note: "The things that interfere in quantum mechanics are not particles. They are probability amplitudes for certain events. It is the fact that probability amplitudes add up like complex numbers that is responsible for all quantum mechanical interferences." Roy J. Glauber, American Journal of Physics, 63(1), 12 (1995).
Reference: Greenberger, D. M.; Horne, M. A.; Zeilinger, A. Physics Today, 1993, 44(8), 22.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.30%3A_Another_Two_Photon_Interference_Experiment.txt
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Jim Baggott’s excellent text, The Meaning of Quantum Theory, deals with the fundamentals of quantum mechanics and its various philosophical interpretations. His demonstrations of computational methods (generally using Dirac algebra) are easy to follow and his summaries of the interpretive positions of Bohr, Einstein, Bohm, Everett, etc. are lucid. Of special note is the clarity with which he deals with the importance of the work of John S. Bell and the eponymous Bell’s theorem.
In order to appreciate the significance of Bell’s theorem you need to be willing to do some math. This tutorial deals with recasting the mathematics of Baggott’s treatment of quantum correlations (pages 119 to 127) using tensor algebra. Not because there is any deficiency in his math, but simply to present another, related method.
The experiment analyzed involves the simultaneous release of two entangled photons in opposite directions as shown below, followed by measurement of their polarization state in the vertical/horizontal measurement basis. The experimental set‐up allows for the rotation of the right‐hand detector through an angle of Θ, in order to more thoroughly explore the consequences of photon entanglement.
The experiment involves measurement of the polarization states of the photon pairs (A,B) as diagramed above. The various polarization states involved are presented below in vector notation: v = vertical; h = horizontal; L = left circular; R = right circular.
$|v \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$|v' \rangle = \begin{pmatrix} \cos \theta \ - \sin \theta \end{pmatrix} \nonumber$
$|v \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$|h' \rangle = \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} \nonumber$
$| L \rangle = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
$| R \rangle = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
The initial state (see Baggott for details) is the following entangled superposition. In other words, both photons, the one moving to the right and the one moving to the left are in the same circular polarization state.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] \nonumber$
This initial photon state is recast using the vector tensor product, which is formed as follows.
$\begin{pmatrix} a \ b \end{pmatrix} \otimes \begin{pmatrix} c \ d \end{pmatrix} = \begin{pmatrix} ac \ ad \ bc \ bd \end{pmatrix} \nonumber$
Therefore in the tensor format the initial state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
There are four measurement outcomes for the photon pairs. Both photons can be detected in the vertical state, both in the horizontal state, A in the vertical state and B in the horizontal state, and vice versa. The measurement eigenstates are calculated below in tensor format.
$\psi '_{vv} \rangle = | v \rangle_A |v' \rangle_B = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta \ - \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \theta \ - \sin \theta \ 0 \ 0 \end{pmatrix} \nonumber$
$\psi '_{vh} \rangle = | v \rangle_A |h' \rangle_B = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} = \begin{pmatrix} \sin \theta \ \cos \theta \ 0 \ 0 \end{pmatrix} \nonumber$
$\psi '_{hv} \rangle = | h \rangle_A |v' \rangle_B = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta \ - \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \cos \theta \ - \sin \theta \end{pmatrix} \nonumber$
$\psi '_{hh} \rangle = | h \rangle_A |h' \rangle_B = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \sin \theta \ \cos \theta \end{pmatrix} \nonumber$
The probability amplitudes for these experimental possibilities are found by calculating the projection of their eigenstates onto the initial state vector.
$\langle \psi '_{vv} | \Psi \rangle \begin{pmatrix} \cos \theta & - \sin \theta & 0 & 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \cos \theta \nonumber$
$\langle \psi '_{vh} | \Psi \rangle \begin{pmatrix} \sin \theta & \cos \theta & 0 & 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \sin \theta \nonumber$
$\langle \psi '_{hv} | \Psi \rangle \begin{pmatrix} 0 & 0 & \cos \theta & - \sin \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \sin \theta \nonumber$
$\langle \psi '_{hh} | \Psi \rangle \begin{pmatrix} 0 & 0 & \sin \theta & \cos \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} = - \frac{1}{ \sqrt{2}} \cos \theta \nonumber$
This yields the following probabilities for the four experimental outcomes. (See Baggott’s equation 4.16 on page 127 of his text.)
$P_{vv} ( \theta ) = \left| \langle \psi '_{vv} | \Psi \rangle \right|^2 = \frac{1}{2} \cos^2 \theta \nonumber$
$P_{vh} ( \theta ) = \left| \langle \psi '_{vh} | \Psi \rangle \right|^2 = \frac{1}{2} \sin^2 \theta \nonumber$
$P_{hv} ( \theta ) = \left| \langle \psi '_{hv} | \Psi \rangle \right|^2 = \frac{1}{2} \sin^2 \theta \nonumber$
$P_{hh} ( \theta ) = \left| \langle \psi '_{hh} | \Psi \rangle \right|^2 = \frac{1}{2} \cos^2 \theta \nonumber$
(Note that if Θ = 0, we get the results Baggott reports on page 122.)
Baggott goes on to show how this experiment and its more sophisticated successors, and their quantum mechanical analyses, illuminate the contradiction between the quantum and classical visions of reality. The purpose of this tutorial has been merely to present a somewhat different format for carrying out the quantum mechanical calculations.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.31%3A_Quantum_Correlations_Illuminated_with_Tensor_Algebra.txt
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Photons from separate sources, u and d, arrive simultaneously at the beam splitter shown in the figure below.
Because the photons arrive simultaneously at the beam splitter they are indistinguishable and donʹt possess separate identities. We are forced by quantum mechanical principles to represent their collective state at the beam splitter (BS) by the following entangled wave function. The plus sign in this superposition indicates that photons are bosons; their wave functions are symmetric with respect to the interchange of the photon labels.
$| \Psi_B \rangle = \frac{1}{ \sqrt{2}} \left[ | u \rangle_1 |d \rangle_2 + |d \rangle_1 |u \rangle_2 \right] \nonumber$
The following vector representations are used for the photon motional states.
$|u \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$|d \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Writing $\Psi_B$ in tensor format yields,
$| \Psi _B\rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$\Psi _B = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
Transmission and reflection occur at 50-50 beam splitters. By convention, reflection is assigned a π/2 (exp(iπ/2) = i) phase shift relative to transmission. The matrix representing a 50-50 beam splitter operating on an individual photon is given below, along with its effect on the u and d photon states.
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
$BS \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} i}{2} \end{pmatrix} \nonumber$
$BS \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}i }{2} \ \frac{ \sqrt{2}}{2} \end{pmatrix} \nonumber$
This shows that the beam splitter converts pure states, u and d, into superpositions of u and d.
In this experiment the beam splitter operates on both photons, so in tensor format the operator representing the beam splitter becomes the following 4x4 matrix.
$\hat{BS} = \hat{BS}_2 \otimes \hat{BS}_1 = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} & i \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \ i \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} & 1 \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \nonumber$
There are four possible photon output states after the beam splitter (|uu >, |ud >, |du >, |dd >), which are shown below in tensor format.
$|uu \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$uu = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$|ud \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$ud = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$|du \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$du = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$|dd \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
$dd = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The output state created by the interaction of the initial entangled state with the beam splitter is,
$\Psi_{Bout} = \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2} i}{2} \ 0 \ 0 \ \frac{ \sqrt{2} i}{2} \end{pmatrix} \nonumber$
By inspection we can see that the photons are always observed in the same output channel, either |uu > or |dd >, with equal probabilities of 0.5 (|0.707i|2). Not surprisingly we see bosonic behavior by the photons. After the beam splitter they are in an even superposition of both being at u and both being at d. Unlike fermions bosons can be in the same state at the same time.
$\left( \left| uu^T \Psi _{Bout} \right| \right)^2 \rightarrow \frac{1}{2} \nonumber$
$\left( \left| dd^T \Psi _{Bout} \right| \right)^2 \rightarrow \frac{1}{2} \nonumber$
Naturally this means they are never observed in different output channels, |ud> or |du>.
$\left( \left| ud^T \Psi _{Bout} \right| \right)^2 \rightarrow 0 \nonumber$
$\left( \left| du^T \Psi _{Bout} \right| \right)^2 \rightarrow 0 \nonumber$
Now letʹs pretend that photons are fermions and examine the consequences for this experiment. Fermions have antisymmetric wave functions; the wave function changes sign on interchange of the constituents labels.
$| \Psi _F \rangle = \frac{1}{ \sqrt{2}} \left[ |u \rangle_1 |d \rangle_2 - |d \rangle_1 |u \rangle_2 \right] \nonumber$
Written in tensor format the antisymmetric wave function is
$| \Psi_F \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
$\Psi _F = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
Now the interaction with the beam splitter yields the following result.
$|Psi _{Fout} = \frac{1}{2} \begin{pmatrix} 1 & i & i & 1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ \frac{ \sqrt{2}}{2} \ - \frac{ \sqrt{2}}{2} \ 0 \end{pmatrix} \nonumber$
We see by inspection or calculation that fermionic photons are never observed in the same output channel after interaction with the beam splitter.
$\left( \left| uu^T \Psi_{Fout} \right| \right)^2 \rightarrow 0 \nonumber$
$\left( \left| dd^T \Psi_{Fout} \right| \right)^2 \rightarrow 0 \nonumber$
Naturally this means they are always observed in different output channels, |ud > or |du>.
$\left( \left| ud^T \Psi_{Fout} \right| \right)^2 \rightarrow \frac{1}{2} \nonumber$
$\left( \left| du^T \Psi_{Fout} \right| \right)^2 \rightarrow \frac{1}{2} \nonumber$
In summary, the sociology of bosons and fermions can be briefly stated: bosons are gregarious and enjoy company; fermions are antisocial and prefer solitude.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.32%3A_Two_Photon_Entanglement_-_A_Tensor_Algebra_Analysis.txt
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In this experiment a down converter, DC, transforms an incident photon into two lower energy photons. One photon takes the upper path and the other the lower path or vice versa. The results of this experiment are that both photons are detected at either U or D. One photon is never detected at U while the other is detected at D. A quantum mechanical analysis of this phenomena is provided below.
Orthonormal basis states:
Photon moving in up-direction:
$u = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$u^T u = 1 \nonumber$
Photon moving in down-direction:
$d = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$d^T d = 1 \nonumber$
$u^T d = 0 \nonumber$
Operators:
Operator for interaction with the mirror:
$M = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
Operator for interaction with a 50/50 beam splitter:
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
A 90o phase shift between transmission and reflection at the beam splitter is required to satisfy energy conservation. By convention the phase shift is assigned to reflection.
The down-converter creates the following entangled state:
$|Psi_b \rangle = \frac{ \left[|u \rangle_1 |d \rangle_2 + d \rangle_1 u \rangle_2 \right]}{2 ^ \frac{1}{2}} \nonumber$
This is a symmetric state because photons are bosons.
After creation in the down-converter, both photons interact with a mirror and a beam splitter before reaching a detector, either U or D. To be detected at the U-detector the photon must be moving in the up-direction (photon state = |u>). To be detected at the D-detector the photon must be moving in the down-direction (photon state = |d>). The probabilities for the four possible experimental outcomes are calculated below.
Both photons arrive at the U-detector: $|_1 \langle u |_2 \langle u | \textbf{BS ~ M} | \Psi_b \rangle |^2$.
$\left[ \left| \frac{(u^T BS M u) (u^T BS M d) + (u^T BS M d) (u^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0.5 \nonumber$
Both photons arrive at the D-detector: $|_1 \langle d |_2 \langle d | \textbf{BS ~ M} | \Psi_b \rangle |^2$.
$\left[ \left| \frac{(d^T BS M u) (d^T BS M d) + (d^T BS M d) (d^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0.5 \nonumber$
Photon 1 arrives at the U-detector and photon 2 arrives at the D-detector: $|_1 \langle u |_2 \langle d | \textbf{BS ~ M} | \Psi_b \rangle |^2$.
$\left[ \left| \frac{(u^T BS M u) (d^T BS M d) + (u^T BS M d) (d^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0 \nonumber$
Photon 1 arrives at the D-detector and photon 2 arrives at the U-detector: $|_1 \langle d |_2 \langle u | \textbf{BS ~ M} | \Psi_b \rangle |^2$.
$\left[ \left| \frac{(d^T BS M u) (u^T BS M d) + (d^T BS M d) (u^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0 \nonumber$
If the experiment could be performed with fermions, they would be created in the following anti-symmetric entangled state:
$|Psi_f \rangle = \frac{ \left[|u \rangle_1 |d \rangle_2 - d \rangle_1 u \rangle_2 \right]}{2 ^ \frac{1}{2}} \nonumber$
As the analysis below shows, the results for fermions would be exactly opposite to those for bosons. Two fermions would never arrive at the same detector.
Both fermions arrive at the U-detector: $|_1 \langle u |_2 \langle u | \textbf{BS ~ M} | \Psi_f \rangle |^2$
$\left[ \left| \frac{(u^T BS M u) (u^T BS M d) - (u^T BS M d) (u^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0 \nonumber$
Both fermions arrive at the D-detector: $|_1 \langle d |_2 \langle d | \textbf{BS ~ M} | \Psi_f \rangle |^2$
$\left[ \left| \frac{(d^T BS M u) (d^T BS M d) - (d^T BS M d) (d^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0 \nonumber$
Fermion 1 arrives at the U-detector and fermion 2 arrives at the D-detector: $|_1 \langle u |_2 \langle d | \textbf{BS ~ M} | \Psi_f \rangle |^2$
$\left[ \left| \frac{(u^T BS M u) (d^T BS M d) - (u^T BS M d) (d^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0.5 \nonumber$
Fermion 1 arrives at the D-detector and fermion 2 arrives at the U-detector: $|_1 \langle d |_2 \langle u | \textbf{BS ~ M} | \Psi_f \rangle |^2$
$\left[ \left| \frac{(d^T BS M u) (u^T BS M d) - (d^T BS M d) (u^T BS M u)}{ \sqrt{2}} \right| \right]^2 = 0.5 \nonumber$
The results of this tutorial enable us to formulate a sociology for bosons and fermions: bosons are gregarious and enjoy companionship; fermions are anti-social and prefer solitude.
But why do bosons always end up at the same detector and fermions (hypothetically) always end up at different detectors? Why in both cases do half of the possible outcomes not occur? Are the bosons and fermions interfering with each other directly? Is there a subtle attractive interaction between bosons and an equally subtle, non-electrostatic, repulsive interaction between fermions?
Not according to Roy Glauber who said, "The things that interfere in quantum mechanics are not particles. They are probability amplitudes for certain events. It is the fact that probability amplitudes add up like complex numbers that accounts for all quantum mechanical interferences." [American Journal of Physics 63, 12 (1995)]
The analysis used in this tutorial clearly illustrates Glauber's assertion.
Reference: Greenberger, D.M.; Horne, M.A.; Zeilinger, A. Physics Today, 1993, 44(8), 22.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.33%3A_Two_Photon_Interference_-_Matrix_Mechanics_Approach.txt
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This tutorial provides a simplified analysis of a recent experiment describing the “interference between two indistinguishable electrons from independent sources.”(1) A schematic of the interferometer used in this experiment is shown below.
S1 and S2 are the electron sources, BS labels the four 50‐50 beam splitters and D1, D2, etc. are the electron detectors.
To simplify the analysis it will be assumed that the arms of the interferometer are of equal length. This will make it unnecessary to consider the accumulation of phase differences due to unequal path distances to the detectors.
The sources simultaneously inject electrons into the interferometer. The presence of the beam splitters provides each source beam access to all four detectors. For example, the source wave functions written in the basis of the detectors are,
$|S1 \rangle \rightarrow \frac{1}{2} [ i| D1 \rangle - |D2 \rangle + i|D3 \rangle + |D4 \rangle] \nonumber$
$|S2 \rangle \rightarrow \frac{1}{2} [ i| D1 \rangle + |D2 \rangle + i|D3 \rangle - |D4 \rangle] \nonumber$
Following the usual convention a π/2 phase shift (represented by i) is assigned to reflection at the beam splitter. It should be noted that there is no opportunity for electron self‐interference because the individual electron beams are not recombined. This means the source of interference in this experiment is due to the indistinguishability of the electrons and the fermionic character of the total wave function which is an entangled two‐electron state.
$| \Psi \rangle_{tot} = \frac{1}{ \sqrt{2}} [ |S1 \rangle_a |S2 \rangle_b - |S2 \rangle_a |S1 \rangle_b] \nonumber$
Substitution of the first two equations into the third yields,
$| \Psi \rangle_{tot} = \frac{i \sqrt{2}}{4} [ |D1 \rangle_a |D2 \rangle_b - |D2 \rangle_a |D1 \rangle_b - |D2 \rangle_a |D3 \rangle_b -D1 \rangle_a |D4 \rangle_b - D3 \rangle_a |D4 \rangle_b + |D4 \rangle_a |D1 \rangle_b + |D3 \rangle_a |D2 \rangle_b + |D4 \rangle_a |D3 \rangle_b] \nonumber$
where the subscripts $a$and $b$ label the electrons.
This result predicts that the electrons never arrive at the same detector, and that the D1‐D3 and D2‐D4 coincidences are not observed. This is in agreement with the more general results reported by Neder, et al. (1) See, for example, their Equation 3 for ϕ1 = ϕ2 = ϕ3 = ϕ4 = 0.
Neder, et al. comment as follows on the distinction between single‐ and two‐particle interference.
Very much like the ubiquitous quantum interference of a single particle with itself, quantum interference of two independent, but indistinquishable, particles is also possible. For a single particle, the interference is between the amplitudes of the particle’s wave function, whereas the interference between two particles is a direct result of quantum exchange statistics. Such interference is observed only in the joint probability of finding the particles in two separated detectors, after they were injected from spatially separated and independent sources.
While it is customary to talk about particle interference, as these authors do, Nobel Laureate Roy Glauber recommended more careful language. (2)
The things that interfere in quantum mechanics are not particles. They are probability amplitudes for certain events. It is the fact that probability amplitudes add up like complex numbers that is responsible for all quantum mechanical interferences.
The insightfulness of this observation is clearly revealed in the mathematical underpinnings of this experiment.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.34%3A_Two-electron_Interference.txt
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Photons from separate sources, u and d, arrive simultaneously at the beam splitter shown in the figure below.
Because these photons are indistinguishable they do not possess separate identities, and we are forced by quantum mechanical principles to represent their collective state at the beam splitter (BS) by the following entangled wave function. The plus sign in this superposition indicates that photons are bosons; their wave functions are symmetric with respect to the interchange of the photon labels.
$\Psi = \frac{1}{ \sqrt{2}} (u_1 d_2 + d_1 u_2) \nonumber$
Transmission and reflection occur at the 50‐50 beam splitters. By convention, reflection is assigned a π/2 phase shift relative to transmission. Thus, in the basis of the detectors, a and b, the photons originating in ports u and d evolve as follows.
$u = \frac{1}{ \sqrt{2}} (ia + b) \nonumber$
$d = \frac{1}{ \sqrt{2}} (a + ib) \nonumber$
Adding photon labels we have:
$u_1 = \frac{1}{ \sqrt{2}} (ia_1 + b_1) \nonumber$
$u_2 = \frac{1}{ \sqrt{2}} (ia_2 + b_2) \nonumber$
$d_1 = \frac{1}{ \sqrt{2}} (ia_1 + b_1) \nonumber$
$d_2 = \frac{1}{ \sqrt{2}} (ia_2 + b_2) \nonumber$
Substitution of these states into Ψ yields,
$\Psi = \frac{1}{ \sqrt{2}} (u_1 d_2 + d_1 u_2) ~ \begin{array}{|l} \text{substitute},~ u_1 = \frac{1}{ \sqrt{2}} (i a_1 + b_1) \ \text{substitute},~ u_2 = \frac{1}{ \sqrt{2}} (i a_2 + b_2) \ \text{substitute},~ d_1 = \frac{1}{ \sqrt{2}} (a_1 + i b_1) \ \text{substitute},~ d_2 = \frac{1}{ \sqrt{2}} (a_2 + i b_2) \ \text{simplify} \ \end{array} \rightarrow \Psi = \sqrt{2} \left( \frac{a_1 a_2}{2} + \frac{b_1 b_2}{2} \right) i \nonumber$
Not surprisingly we see bosonic behavior by the photons. They both arrive at the same detector ‐ 50% of the time at detector a and 50% of the time at detector b.
However, recombining the photon beams at a second beam splitter before the detectors (see figure below) appears to invest them with fermionic character.
As was the case above, the entangled two-photon state at BS1 is,
$\Psi = \frac{1}{ \sqrt{2}} (u1 d2 +d1 u2) \nonumber$
In terms of the BS1 output arms A and B, the photons originating in ports u and d evolve as,
$u = \frac{1}{ \sqrt{2}} (iA + B) \nonumber$
$d = \frac{1}{ \sqrt{2}} (A + iB) \nonumber$
While in terms of the detectors a and b, A and B in the presence of BS2 evolve as:
$A = \frac{1}{ \sqrt{2}} (ia + b) \nonumber$
$B = \frac{1}{ \sqrt{2}} (a + ib) \nonumber$
Thus, overall in the basis of the detectors, u and d evolve as follows.
$u = \frac{1}{ \sqrt{2}} \left[ i \left[ \frac{1}{ \sqrt{2}} (ia + b) \right] + \frac{1}{ \sqrt{2}} (a + ib) \right] = ib \nonumber$
$d = \frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2}} (ia + b) + \left[ \frac{1}{ \sqrt{2}} (a + ib) \right] \right] = ia \nonumber$
Attaching photon labels to u and d and substituting into Ψ yields:
$\Psi = - \frac{1}{ \sqrt{2}} (a_1 b_2 + b_1 a_2 ) \nonumber$
Now we see fermionic behavior for the photons. They always arrive at different detectors.
This result can also be achieved as follows:
$\Psi = \frac{i}{ \sqrt{2}} (A_1 A_2 + B_1 B_2) ~ \begin{array}{|l} \text{substitute},~ A_1 = \frac{1}{ \sqrt{2}} (i a_1 + b_1) \ \text{substitute},~ A_2 = \frac{1}{ \sqrt{2}} (i a_2 + b_2) \ \text{substitute},~ B_1 = \frac{1}{ \sqrt{2}} (a_1 + i b_1) \ \text{substitute},~ B_2 = \frac{1}{ \sqrt{2}} (a_2 + i b_2) \ \text{simplify} \ \end{array} \rightarrow \Psi = \frac{ \sqrt{2} a_1 b_2 + a_2 b_1}{2} \nonumber$
The two examples can also be analyzed using matrix algebra. Below we have the vectors for photon direction of propagation, the initial two‐photon entangled state, and the matrix representing a beam splitter.
$u = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$d = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \nonumber$
After the second beam splitter they propagate in opposite directions, behaving like fermions.
$\text{kronecker} (BS,~BS) \text{kronecker} (BS,~BS) \Psi = \begin{pmatrix} 0 \ -0.707 \ -0.707 \ 0 \end{pmatrix} \nonumber$
$\Psi = frac{-1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmarix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.35%3A_Bosonic_and_Fermionic_Photon_Behavior_at_Beam_Splitters.txt
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Photons from separate sources, u and d, arrive simultaneously at the beam splitter shown in the figure below.
Because these photons are indistinguishable they donʹt possess separate identities, and we are forced by quantum mechanical principles to represent their collective state at the beam splitter (BS) by the following entangled wave function. The plus sign in this superposition indicates that photons are bosons; their wave functions are symmetric with respect to the interchange of the photon labels.
$| \Psi _B \rangle = \frac{1}{ \sqrt{2}} \left[ |u \rangle_1 |d \rangle_2 + |d \rangle_1 |u \rangle_2 \right] \nonumber$
The following vector representations are used for the photon states.
$|u \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$|d \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
In tensor format this entangled state becomes,
$\Psi_B \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
Transmission and reflection occur at the 50-50 beam splitters. By convention, reflection is assigned a π/2 phase shift relative to transmission. The matrix representing a 50-50 beam splitter operating on an individual photon is given below, along with its effect on the u and d photon states.
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} i}{2} \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2} i}{2} \ \frac{ \sqrt{2}}{2} \end{pmatrix} \nonumber$
This result clearly shows that the beam splitter converts pure states (u and d) into superpositions of u and d.
In this experiment the beam splitter operates on both photons, and in tensor format the operator becomes the following 4x4 matrix operator.
$\hat{BS} = \hat{BS}_1 \otimes \hat{BS}_2 = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} & i \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \ i \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} & 1 \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \nonumber$
Temporarily thinking of the photon as generic quantum particle (quon to use Nick Herbertʹs phrase), we can identify four possible photon states after the beam splitter, which are shown below in tensor format.
$|uu \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$|ud \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$|du \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$|dd \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The actual result of the interaction of the initial entangled state with the beam splitter is,
$\frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2} i}{2} \ 0 \ 0 \ \frac{ \sqrt{2} i}{2} \end{pmatrix} \nonumber$
By inspection we can see that the photons are always observed in the same output channel, either |uu > or |dd >, with equal probabilities of 0.5. Not surprisingly we see bosonic behavior by the photons.
$\left[ \left| \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow \frac{1}{2} \nonumber$
$\left[ \left| \begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow \frac{1}{2} \nonumber$
Naturally this means they are never observed in different output channels, |ud > or |du>. The photons are not exhibiting fermionic behavior (so far).
$\left[ \left| \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow 0 \nonumber$
$\left[ \left| \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow 0 \nonumber$
However, recombining the photon beams at a second beam splitter appears to invest them with fermionic character.
As is shown below, the addition of a second beam splitter is easily implemented in the tensor format.
$\frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ - \frac{ \sqrt{2}}{2} \ - \frac{ \sqrt{2}}{2} \ 0 \end{pmatrix} \nonumber$
Now we see (by inspection or calculation) fermionic behavior. The photons never appear in the same output channel.
$|uu \rangle$
$\left[ \left| \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow 0 \nonumber$
$|dd \rangle$
$\left[ \left| \begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow 0 \nonumber$
They always appear in different output channels, now behaving like fermions.
$|ud \rangle$
$\left[ \left| \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow \frac{1}{2} \nonumber$
$|du \rangle$
$\left[ \left| \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \right| \right]^2 \rightarrow \frac{1}{2} \nonumber$
7.37: Entangled Photons Can Behave Like Fermions
As shown in the figure, photons from separate sources, a and b, arrive simultaneously at the first beam splitter of a Mach-Zehnder interferometer (MZI).
Because the photons are indistinguishable they don't possess separate identities, and we are forced by quantum mechanical principles to represent their collective state at the beam splitter (BS) by the following entangled wave function. The plus sign in this superposition indicates that photons are bosons; their wave functions are symmetric with respect to the interchange of the photon labels.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |a \rangle_1 |b \rangle_2 + |b \rangle_1 |a \rangle_2 \right] \nonumber$
The following vector representations are used for the photon states, |a> representing presence in the upper arm of the MZI and |b> presence in its lower arm.
$|a \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$|b \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Writing Ψ in tensor format yields,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
There are four detector states after the exit channels of the MZI (|aa>, |ab>, |ba>, |bb>), which are shown below in tensor format.
$| aa \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$aa = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$| ab \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$ab = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$| ba \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$ba = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$| bb \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
$bb = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The matrix operator representing the beam splitter:
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
The following calculations show fermionic behavior for the photons. They never arrive at the same detector. (Kronecker is Mathcad's command for tensor matrix multiplication.)
$\left( \left| aa^T \text{kronecker} (BS,~BS) \text{kronecker} (BS,~BS) \Psi\right| \right)^2 = 0 \nonumber$
$\left( \left| ab^T \text{kronecker} (BS,~BS) \text{kronecker} (BS,~BS) \Psi\right| \right)^2 = 0.5 \nonumber$
$\left( \left| ba^T \text{kronecker} (BS,~BS) \text{kronecker} (BS,~BS) \Psi\right| \right)^2 = 0.5 \nonumber$
$\left( \left| bb^T \text{kronecker} (BS,~BS) \text{kronecker} (BS,~BS) \Psi\right| \right)^2 = 0 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.36%3A_Bosonic_and_Fermionic_Photon_Behavior_at_Beam_Splitters-_A_Tensor_Algebra_Analysis.txt
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Greenberger, Horne and Zeilinger (GHZ) surveyed the then relatively new field of multiparticle interferometry in their August 1993 Physics Today article, "Multiparticle Interferometry and the Superposition Principle." This tutorial will use Mathcad and tensor algebra to analyze the results associated with their Figure 2, shown below.
Richard Feynman identified the superposition principle, as manifested in the double-slit experiment, as the heart of quantum mechanics and its only mystery. Indeed that simple example of a single-photon superposition illustrates a number of quantum fundamentals, but as GHZ point out multi-photon entangled superpositions reveal additional mysteries like quantum correlations that challenge philosophical positions such as local realism.
The operators required to analyze this two-photon interferometer are defined below. α and β are phase shifters.
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Mirror:
$M = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
Phase shift:
$A ( \delta) = \begin{pmatrix} e^{i \delta} & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Beam splitter:
$BS = \sqrt{ \frac{1}{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
The parametric down converter (PDC) creates two photons traveling in opposite directions in the arms of the interferometer, with photon 1 moving to the right and photon 2 to the left. The initial state is the following entangled superposition.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | u_1 \rangle |d_2 \rangle + |d_1 \rangle |u_2 \rangle \right] \nonumber$
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
There are four output states. The appendix shows how the input and output states are constructed in Mathcad.
$| uu \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$| ud \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$| du \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$| dd \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The input and output states are defined in Mathcad syntax.
$u = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$d = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$u1u2 = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$u1d2 = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$d1u2 = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$d1d2 = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The four outcome probabilities are now calculated. The appendix shows how Mathcad is used to carry out the tensor product of two matrices.
$P{u1u2} ( \alpha , \beta ) = \left( \left| u1u2^T \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 \nonumber$
$P{u1d2} ( \alpha , \beta ) = \left( \left| u1d2^T \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 \nonumber$
$P{d1u2} ( \alpha , \beta ) = \left( \left| d1u2^T \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 \nonumber$
$P{d1d2} ( \alpha , \beta ) = \left( \left| d1d2^T \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 \nonumber$
The probability that the individual detectors will fire is given by the following sums.
$P_{u1} = P_{u1u2} + P_{u1d2} \nonumber$
$P_{d1} = P_{d1d2} + P_{d1u2} \nonumber$
$P_{u2} = P_{u1u2} + P_{d1u2} \nonumber$
$P_{d2} = P_{d1d2} + P_{u1d2} \nonumber$
Because only the phase difference between the two arms of the interferometer matters, β is set to zero and the phase difference is expressed in the value of α. The following graphs show the behavior of the detector pairings as a function of the phase difference α.
$\alpha = 0 \text{deg}, 1 \text{deg} .. 360 \text{deg}$
When there is no phase difference in the two branches of the interferometer there is a strong correlation in the detector responses. Fifty percent of the time the photons arrive at the up-detectors and 50% of the time at the down-detectors. There are no coincidences between an up and a down detector. This might be called bosonic behavior - bosons like to do the same thing.
For convenience results will also be presented in numeric format. See the Appendix for a graphical display.
$\alpha = 0 ~ \text{deg}$
$\begin{pmatrix} P_{u1u2} ( \alpha ,0) & P_{u1d2} ( \alpha ,0) \ P_{d1u2} ( \alpha ,0) & P_{d1d2} ( \alpha ,0) \end{pmatrix} = \begin{pmatrix} 0.5 & 0 \ 0 & 0.5 \end{pmatrix} \nonumber$
A 90 degree phase difference results in no correlation. The four detector pairs fire with equal frequency.
$\alpha = 90 ~ \text{deg}$
$\begin{pmatrix} P_{u1u2} ( \alpha ,0) & P_{u1d2} ( \alpha ,0) \ P_{d1u2} ( \alpha ,0) & P_{d1d2} ( \alpha ,0) \end{pmatrix} = \begin{pmatrix} 0.25 & 0.25 \ 0.25 & 0.25 \end{pmatrix} \nonumber$
A 180 degree phase difference results in what might be called fermionic behavior. If one photon is detected at an up-detector, the other is registered at a down detector. They never both arrive at the same type of detector. Fermions don't like to do the same thing at the same time.
$\alpha = 180 ~ \text{deg}$
$\begin{pmatrix} P_{u1u2} ( \alpha ,0) & P_{u1d2} ( \alpha ,0) \ P_{d1u2} ( \alpha ,0) & P_{d1d2} ( \alpha ,0) \end{pmatrix} = \begin{pmatrix} 0 & 0.5 \ 0.5 & 0 \end{pmatrix} \nonumber$
Intermediate correlation is achieved with a 60 degree phase difference.
$\alpha = 60 ~ \text{deg}$
$\begin{pmatrix} P_{u1u2} ( \alpha ,0) & P_{u1d2} ( \alpha ,0) \ P_{d1u2} ( \alpha ,0) & P_{d1d2} ( \alpha ,0) \end{pmatrix} = \begin{pmatrix} 0.375 & 0.125 \ 0.125 & 0.375 \end{pmatrix} \nonumber$
These tabular results can be summarized by displaying the correlation function. Perfect correlation +1; perfect anti-correlation -1; no correlation 0. Another graphical display can be found in the Appendix.
$\alpha = 0, .02 .. 2 \pi$
$\text{Corr} ( \alpha , \beta ) = P_{u1u2} ( \alpha , \beta ) - P_{u1d2} ( \alpha , \beta ) - P_{d1u2} ( \alpha , \beta ) + P_{d1d2} ( \alpha , \beta ) \nonumber$
Note that the sums of the rows and the sums of the columns in the tables above always equal 1/2.
$P_{1u} = P_{2u} = P_{1d} = P_{2d} = \frac{1}{2} \nonumber$
An individual detector fires 50% of the time in spite of the correlations that may occur between two detectors. Looking at individual detectors locally reveals totally random behavior. It is only when coincidences between pairs of detectors on the left and right are examined that correlations are observed.
The random behavior at each of the detectors can be directly shown by calculations in which only one detector is being monitored. This requires the projection operators for up and down motion given below. The calculations are shown for the arbitrary phases shown below. However, all phase relationships give the same results.
Projection operators:
$uu^T = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} ~~~ dd^T = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Phases:
$\alpha = 11~ \text{deg} ~~~ \beta = 58 ~ \text{deg} \nonumber$
$\begin{array}{r} Detector & Probability of Firing \ U1 & \left( \left| \left( uu^T , ~I \right) \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 = 0.5 \ D1 & \left( \left| \left( dd^T , ~I \right) \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 = 0.5 \ U2 & \left( \left| \left( I, ~ uu^T \right) \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 = 0.5 \ D2 & \left( \left| \left( I,~ dd^T \right) \text{kronecker} (BS,~BS) \text{kronecker} (A ( \alpha), A ( \beta )) \text{kronecker} (M,~M) \Psi \right| \right)^2 = 0.5 \ \end{array} \nonumber$
At this point it is important to recall that interference at the individual detectors would be observed if the photons were not entangled. Looking at either half of the interferometer shown above and assuming an unentangled photon in the superposition, $\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix}$, of being in the upper and lower arms of the truncated interferometer results in interference effects as a function of α at the detectors as is illustrated in the following graph.
$\alpha = 0 ~ \text{deg}, 1~ \text{deg} .. 360 ~ \text{deg}$
$Pu ( \alpha ) = \left[ \left| uu^T BS A ( \alpha ) M \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \right| \right]^2 \nonumber$
$Pd ( \alpha ) = \left[ \left| dd^T BS A ( \alpha ) M \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \right| \right]^2 \nonumber$
Recently, Art Hobson posted "Implications of bipartite interferometry for the measurement problem" at arXiv:1301.1673. In this manuscript he deals in depth with the results presented above. Therefore, in concluding this tutorial I will take relevant commentary from his manuscript without further comment. However, some excerpts have been slightly modified, but not in such a way as to change their original meanings.
• When two photons are entangled, the state actually detected by an observer of either photon is the local or reduced state of that photon. The reduced state operator of photon 2 is obtained by averaging (tracing) the total density operator over photon 1 as is shown below. This procedure shows that photon 2's reduced or local state operator is diagonal indicating a classical mixed state - it has been stripped of its off-diagonal interference terms. This result is consistent with the random behavior noted earlier for the measurements on individual photons.
$\hat{ \rho}_{12} = | \Psi \rangle \langle \Psi | = \frac{1}{2} [ |u_1 \rangle |d_2 \rangle \langle d_2 | \langle u_1 | + | u_1 \rangle |d_2 \rangle \langle u_2 | \langle d_1 | + | d_1 \rangle |u_2 \rangle \langle d_2 | \langle u_1 | + | d_1 \rangle | u_2 \rangle \langle u_2 | \rangle d_1 |] \nonumber$
$\hat{ \rho}_2 = \frac{1}{2} [ \langle u_1 | \Psi \rangle \langle \Psi | u_1 \rangle + \langle d_1 | \Psi \rangle \langle \Psi | d_1 \rangle ] = \frac{1}{2} [ |d_2 \rangle \langle d_2 | + | u_2 \rangle \langle u_2 |] \nonumber$
• Therefore, when a bipartite system is in an entangled superposition, its subsystems are not in superpositions but are instead mixed states with each subsystem in a definite, but unknown state.
• The detectors (U1, D1, U2, D2) show no local signs of interference of the type that would occur if the photons were not entangled, because each photon is a "which path" detector for the other photon, decohering the other photon and creating incoherent local mixtures.
• To summarize, an entangled photon is always "in" its local state, the state described by its reduced density operator (see above), because this is the state actually detected by an observer of the photon. The entangled state Ψ is a global superposition of photon correlations, not a superposition of local photon states.
Appendix
The tensor product of two vectors is shown below.
$\begin{pmatrix} a \ b \end{pmatrix} \otimes \begin{pmatrix} c \ d \end{pmatrix} = \begin{pmatrix} ac \ ad \ bc \ bd \end{pmatrix} \nonumber$
Mathcad does not have a command for the vector tensor product, so it is necessary to develop a way of implementing it using kronecker, which requires square matrices. For this reason the spin vector is stored in the left column of a 2x2 matrix by augmenting the spin vector with the null vector. After the matrix tensor products have been carried out using kronecker the final spin vector resides in the left column of the final square matrix. Next the submatrix command is used to save this column, discarding the rest of the matrix.
The Mathcad syntax for the tensor multiplication of two vectors is as follows.
$\Psi ( a,~b) = \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ a, ~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] , \text{augment} \left[ b,~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] \right] , ~1,~4,~1,~1 \right] \nonumber$
The initial photon state:
$\frac{1}{ \sqrt{2}} ( \Psi (u,~d) + \Psi (d,~u)) = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} \nonumber$
Tensor matrix multiplication is also known as Kronecker multiplication. Here it is shown that the Mathcad result using the kronecker command is identical to the hand calculation.
$\text{kronecker} (BS,~BS) = \begin{pmatrix} 0.5 & 0.5i & 0.5i & -0.5i \ 0.5i & 0.5 & -0.5 & 0.5i \ 0.5i & -0.5 & 0.5 & 0.5i \ -0.5 & 0.5i & 0.5i & 0.5 \end{pmatrix} \nonumber$
$\widehat{BS} \otimes \widehat{BS} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} & i \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \ i \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} & 1 \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & i & i & -1 \ i & 1 & -1 & i \ i & -1 & 1 & i \ -1 & i & i & 1 \end{pmatrix} \nonumber$
A graphical display of tabular results presented earlier:
$P ( \alpha , \beta ) = \begin{pmatrix} P_{u1u2} ( \alpha , \beta ) & P_{u1d2} ( \alpha , \beta ) \ P_{d1u2} ( \alpha , \beta ) & P_{d1d2} ( \alpha , \beta ) \end{pmatrix} \nonumber$
The following algebraic analysis of the experiment shows the evolution of the original entangled state. Consistent with the matrix mechanics analysis, it shows that when the phase diference is zero both photons arrive at either the U- or the D-detectors, and when the phase difference is π one photon arrives at U-detector and the other at the D-detector.
$\alpha = 0 ~~~ \beta = 0$
$\frac{1}{ \sqrt{2}} (u_1 d_2 + d_1 u_2)~ \begin{array}{|l} \text{substitute},~ u1 = \frac{e^{i \alpha}}{ \sqrt{2}} (i U1 + D1) \ \text{substitute},~ d2 = \frac{1}{ \sqrt{2}} (U2 + iD2) \ \text{substitute},~ u1 = \frac{1}{ \sqrt{2}} (U1 + D1) \ \text{substitute},~ d2 = \frac{e^{i \beta}}{ \sqrt{2}} (i U2 + D2) \ \end{array} \rightarrow \sqrt{2} \left( \frac{U1 U2 i}{2} + \frac{D1 D2 i}{2} \right) \nonumber$
$\alpha = \pi ~~~ \beta = 0$
$\frac{1}{ \sqrt{2}} (u_1 d_2 + d_1 u_2)~ \begin{array}{|l} \text{substitute},~ u1 = \frac{e^{i \alpha}}{ \sqrt{2}} (i U1 + D1) \ \text{substitute},~ d2 = \frac{1}{ \sqrt{2}} (U2 + iD2) \ \text{substitute},~ u1 = \frac{1}{ \sqrt{2}} (U1 + D1) \ \text{substitute},~ d2 = \frac{e^{i \beta}}{ \sqrt{2}} (i U2 + D2) \ \end{array} \rightarrow \sqrt{2} \left( \frac{D1 U2}{2} - \frac{D2 U2}{2} \right) \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.38%3A_Analyzing_Two-photon_Interferometry_Using_Mathcad_and_Tensor_Algebra.txt
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Greenberger, Horne and Zeilinger (GHZ) surveyed the then relatively new field of multiparticle interferometry in an August 1993 Physics Today article, ʺMultiparticle Interferometry and the Superposition Principle.ʺ This tutorial will use Mathcad and tensor algebra to analyze the results associated with Figure 5, which dealt with a two‐photon quantum eraser. A parametric down converter (PDC) produces two horizontally polarized, entangled photons, one taking the upper path and the other the lower path. The beams are combined at a beam splitter as shown below. Polarizing films oriented at angles of θ and ϕ relative to the horizontal are placed in front of detectors.
As the figure shows both photons, as bosons, arrive at either the U detector or the D detector. This result will now be confirmed using tensor algebra.
The photons emerging from the PDC are entangled and can be moving up or down and be in a particular polarization state. We use the following vectors to represent the motional and polarization states of the photons.
$u = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$d = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\theta = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\phi = \begin{pmatrix} \cos \phi \ \sin \phi \end{pmatrix} \nonumber$
The motional and polarization photon states are combined using vector tensor multiplication. The up motional state is tagged with θ polarization, and the down state with ϕ polarization. The photon polarization states can be the same or different.
$u \theta = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \theta \ \sin \theta \ 0 \ 0 \end{pmatrix} \nonumber$
$d \phi = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} \cos \phi \ \sin \phi \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \cos \phi \ \sin \phi \end{pmatrix} \nonumber$
There are four two‐photon output states at the detectors. These are also represented using tensor algebra. The first two letters refer to photon 1, the second two refer to photon 2. The uθdϕ (|uθ >1 |dϕ>2) state is constructed as an example.
$u \theta d \phi = \begin{pmatrix} \cos \theta \ \sin \theta \ 0 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 0 \ \cos \phi \ \sin \phi \end{pmatrix} \nonumber$
$u \theta d \phi = ud( \theta, ~ \phi ) = \begin{pmatrix} 0 & 0 & \cos \theta \cos \phi & \cos \theta \sin \phi & 0 & 0 & \sin \theta \cos \phi & \sin \theta \sin \phi & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
The four output states are:
$uu ( \theta ) = \begin{pmatrix} \cos \theta ^2 & cos \theta \sin \theta & 0 & 0 & \sin \theta \cos \theta & \sin \theta ^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$ud ( \theta ,~ \phi ) = \begin{pmatrix} 0 & 0 & \cos \theta \cos \phi & \cos \theta \sin \phi & 0 & 0 & \sin \theta \cos \phi & \sin \theta \sin \phi & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$du ( \phi, ~ \theta ) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cos \phi \cos \theta & \cos \phi \sin \theta & 0 & 0 & \sin \phi \cos \theta & \sin \phi \cos \theta & \sin \phi \sin \theta & 0 & 0 \end{pmatrix} ^T \nonumber$
$dd( \phi ) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cos \phi ^2 & \cos \phi \sin \phi & 0 & 0 & \sin \phi \cos \phi & \sin \phi ^2 \end{pmatrix}^T \nonumber$
The matrix operators required for this analysis are,
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Mirror:
$M = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
Beam splitter:
$BS = \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} \nonumber$
The mirrors and the beam splitter operate on the motional degree of freedom. Their operators as configured in the apparatus are constructed using matrix tensor multiplication, implemented with Mathcadʹs kronecker command.
$MI = \text{kronecker} (M,~ \text{kronecker} (I, ~ \text{kronecker} (M,~I))) \nonumber$
$BSI = \text{kronecker} (BS,~ \text{kronecker} (I, ~ \text{kronecker} (BS,~I))) \nonumber$
As mentioned earlier, the PDC produces the following entangled state of horizontally polarized photons:
$\Psi = \frac{1}{ \sqrt{2}} (ud(0,~0) + du(0,~0)) \nonumber$
The output state after the photons interact with the mirrors and the beam splitter is:
$\Psi _{out} = BSI(MI) \Psi \nonumber$
Initially the films in front of the detectors are horizontally polarized and we see that both photons always arrive at the same detector, as is shown graphically in the figure. Fifty percent of the time it is U and fifty percent of the time it is D. No U‐D coincidences are observed (off‐diagonal elements are zero).
$\theta = 0$
$\phi = 0$
$\begin{bmatrix} \left( \left| uu( \theta ) \Psi_{out} \right| \right)^2 & \left( \left| ud( \theta , \phi) \Psi_{out} \right| \right)^2 \ \left( \left| du( \phi , \theta) \Psi_{out} \right| \right)^2 & \left( \left| dd( \phi ) \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{pmatrix} \nonumber$
Now assume that a 90 degree polarization rotator is placed in the lower arm which rotates the horizontal state to the vertical polarization orientation. This provides path information and even though polarization is not measured in this experiment it has a significant affect on the measurement results.
The entangled state is now:
$\Psi = \frac{1}{ \sqrt{2}} \left( ud \left( 0,~ \frac{ \pi}{2} \right) + du \left( \frac{ \pi}{2} , ~ 0 \right) \right) \nonumber$
$\Psi _{out} = BSI(MI) \Psi \nonumber$
No photons are detected if the polarizers remain horizontally oriented or if they are both rotated to the vertical orientation.
$\theta = 0$
$\phi = 0$
$\begin{bmatrix} \left( \left| uu( \theta ) \Psi_{out} \right| \right)^2 & \left( \left| ud( \theta , \phi) \Psi_{out} \right| \right)^2 \ \left( \left| du( \phi , \theta) \Psi_{out} \right| \right)^2 & \left( \left| dd( \phi ) \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
$\theta = \frac{ \pi}{2}$
$\phi = { \pi}{2}$
$\begin{bmatrix} \left( \left| uu( \theta ) \Psi_{out} \right| \right)^2 & \left( \left| ud( \theta , \phi) \Psi_{out} \right| \right)^2 \ \left( \left| du( \phi , \theta) \Psi_{out} \right| \right)^2 & \left( \left| dd( \phi ) \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
However, if one of the polarizers is horizontal and the other vertical, U‐D coincidences are observed (non‐zero off‐diagonal elements).
$\theta = 0$
$\phi = \frac{ \pi}{2}$
$\begin{bmatrix} \left( \left| uu( \theta ) \Psi_{out} \right| \right)^2 & \left( \left| ud( \theta , \phi) \Psi_{out} \right| \right)^2 \ \left( \left| du( \phi , \theta) \Psi_{out} \right| \right)^2 & \left( \left| dd( \phi ) \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & \frac{1}{8} \ \frac{1}{8} & 0 \end{pmatrix} \nonumber$
$\theta = \frac{ \pi}{2}$
$\phi = 0$
$\begin{bmatrix} \left( \left| uu( \theta ) \Psi_{out} \right| \right)^2 & \left( \left| ud( \theta , \phi) \Psi_{out} \right| \right)^2 \ \left( \left| du( \phi , \theta) \Psi_{out} \right| \right)^2 & \left( \left| dd( \phi ) \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & \frac{1}{8} \ \frac{1}{8} & 0 \end{pmatrix} \nonumber$
The photon path information can be erased by a diagonal orientation of the polarizers. Now the coincidences of the previous cases disappears and the original bosonic behavior is restored.
$\theta = \frac{ \pi}{4}$
$\phi = \frac{ \pi}{4}$
$\begin{bmatrix} \left( \left| uu( \theta ) \Psi_{out} \right| \right)^2 & \left( \left| ud( \theta , \phi) \Psi_{out} \right| \right)^2 \ \left( \left| du( \phi , \theta ) \Psi_{out} \right| \right)^2 & \left( \left| dd( \phi ) \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} \frac{1}{8} & 0 \ 0 & \frac{1}{8} \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.39%3A_Analysis_of_a_Two-photon_Quantum_Eraser.txt
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Greenberger, Horne and Zeilinger (GHZ) surveyed the then relatively new field of multiparticle interferometry in an August 1993 Physics Today article, ʺMultiparticle Interferometry and the Superposition Principle.ʺ This tutorial will use Mathcad and tensor algebra to analyze the results associated with Figure 5, which dealt with a two‐photon quantum eraser. A parametric down converter (PDC) produces two horizontally polarized, entangled photons, one taking the upper path and the other the lower path. The beams are combined at a beam splitter as shown below.
As the figure shows both photons arrive at either the U detector or the D detector. This result will now be confirmed using tensor algebra.
The photons emerging from the PDC are entangled and can be moving up or down with horizontal polarization. Later we will consider rotating the polarization in the lower arm to the vertical orientation in order to explore the consequences of providing path information. We use the following vectors to represent the motional and polarization states of the photons.
$u = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~~ d = \begin{pmatrix} 0 \ 1 \end{pmatrix} ~~~ h = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~~ v = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
The four possible photon states are expressed using vector tensor multiplication.
$uh = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$uh = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
There are sixteen photon measurement (output) states at the detectors. These are also represented using tensor algebra. The first two letters refer to photon 1, the second two refer to photon 2. The uhdv (|uh>1|dv>2) state is constructed as an example.
$uhdv = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uhuh = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uhuv = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uhdh = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uhdv = \begin{pmatrix} = 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uvuh = \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uvuv = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uvdh = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$uvdv = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dhuh = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dhuv = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dhdh = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dhdv = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dvuh = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dvuv = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} ^T \nonumber$
$dvdh = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} ^T \nonumber$
$dvdv = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} ^T \nonumber$
Matrix operators:
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Mirror:
$M = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
Beam splitter:
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
The mirrors and the beam splitter operate on the motional degree of freedom. Their operators as configured in the apparatus are constructed using matrix tensor multiplication, implemented with Mathcadʹs kronecker command.
$MI = \text{kronecker} (M,~ \text{kronecker} (I,~ \text{kronecker} (M,~I))) \nonumber$
$BSI = \text{kronecker} (BS,~ \text{kronecker} (I,~ \text{kronecker} (BS,~I))) \nonumber$
The entangled state produced by the PDC is:
$\Psi_{boson} = \frac{1}{ \sqrt{2}} (uhdh + dhuh) \nonumber$
The output state after the photons interact with the mirrors and the beam splitter is:
$\Psi _{out} = BSI (MI) \Psi_{boson} \nonumber$
An equivalent algebraic analysis clearly shows the constructive and destructive interference between the probability amplitudes for the measurement states.
$\Psi _{out} = \frac{1}{ 2 \sqrt{2}} (i (uh)uh - uh (dh) + dh(uh) + i(dh)dh + i (uh)uh + uh (dh) - dh(uh) + i (dh) dh = \frac{i}{ \sqrt{2}} (uh (uh) - dh (dh)) \nonumber$
We now calculate a matrix of all possible experimental outcomes, recognizing at this point that because the photons are horizontally polarized we could have just calculated a 2x2 matrix, eliminating columns 2 and 4, and rows 2 and 4.
$P_{boson} = \begin{bmatrix} \left[ \left| ( \overline{uhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{uvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdv} )^T \Psi _{out} \right| \right]^2 \ \end{bmatrix} = \begin{pmatrix} \frac{1}{2} & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & \frac{1}{2} & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
$P_{boson} = P_{uhuh} = P_{dhdh} = \frac{1}{2} \nonumber$
We see that this calculation is in agreement with the experimental results represented in the figure above: both photons always arrive at the same detector. 50 % of the time it is U and 50% of the time it is D.
Now assume that a 90 degree polarization rotator is placed in the lower arm which rotates the horizontal state to the vertical polarization orientation. This provides path information and even though polarization is not measured in this experiment it has a significant affect on the measurement results.
The entangled photon state after the PDC now is:
$\Psi _{boson} = \frac{1}{ \sqrt{2}} (uhdv + dvuh) \nonumber$
This leads to the following output state after interaction with the mirrors and the beam splitter:
$\Psi _{out} = BSI(MI) \Psi _{boson} \nonumber$
The measurement outcome matrix now shows that the photons arrive at different detectors 50% of the time and the same detector 50% of the time. Remember that polarization is not being measured in this experiment. But the fact that polarization information exists changes the experimental results. The following algebraic expression for the output states shows that the interence effects that were seen previous do not occur because of the h/v polarization markers on the motional states.
$\Psi _{out} = \frac{1}{ \sqrt{2}} (i (uh) uv - uh (dv) + dh (uv) + i (dh) dv + i (uv) uh - uv (dh) - dv (uh) + i (dv) dh) \nonumber$
$P_{boson} = P_{uhuv} = P_{uhdv} = P_{dhdv} = P_{uvuh} = P_{uvdh} = P_{dvuh} = P_{dvdh} = \frac{1}{8} \nonumber$
$P_{boson} = \begin{bmatrix} \left[ \left| ( \overline{uhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{uvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdv} )^T \Psi _{out} \right| \right]^2 \ \end{bmatrix} = \begin{pmatrix} 0 & \frac{1}{8} & 0 & \frac{1}{8} \ \frac{1}{8} & 0 & \frac{1}{8} & 0 \ 0 & \frac{1}{8} & 0 & \frac{1}{8} \ \frac{1}{8} & 0 & \frac{1}{8} & 0 \end{pmatrix} \nonumber$
The path information provided by the h/v polarization states of the photons can be ʺerasedʺ by placing diagonally (45 degrees to the veritcal and labelled s for slant) oriented polarizers after the beam splitter and before the detectors. Polarizers are projection operators, consequently only diagonally polarized photons reach the detectors. There are only four possible final measurement states,
$usus = \frac{1}{2} \begin{pmatrix} 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
$usds = \frac{1}{2} \begin{pmatrix} 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
$dsus = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \end{pmatrix} \nonumber$
$dsds = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{pmatrix} \nonumber$
where for example dsus is calculated as follows. Tensor multiplication is implied between the vector states.
$dsus = \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \right]^T \nonumber$
where
$s = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \nonumber$
The recalculated experimental outcome matrix shows that the photons again always arrive at the same detector. The accompanying algebraic analysis reveals the revived interference effects that lead to the final result.
$\begin{bmatrix} \left( \left| usus \Psi_{out} \right| \right)^2 & \left( \left| usds \Psi_{out} \right| \right)^2 \ \left( \left| dsus \Psi_{out} \right| \right)^2 & \left( \left| dsds \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} \frac{1}{8} & 0 \ 0 & \frac{1}{8} \end{pmatrix} \nonumber$
$\Psi_s = \frac{1}{ 4 \sqrt{2}} = (i (us) us - us (ds) + ds (us) + i (ds) ds + i (us) us +us (ds) - ds (us) + i (ds) ds) = \frac{1}{2 \sqrt{2}} (us (us) + ds (ds)) \nonumber$
Photons are bosons and therefore have symmetric wave functions. This is why in the initial experiment they always arrive at the same detector. We will now assume that they are fermions, which have anti‐symmetric wave functions, and repeat the calculations and observe the consequences.
The fermionic entangled photon wave function:
$\Psi_{fermion} = \frac{1}{ \sqrt{2}} (uhuh - dhdh) \nonumber$
The output state after the photons interact with the mirrors and the beam splitter is:
$\Psi _{out} = BMI (MI) \Psi_{fermion} \nonumber$
The measurement outcome matrix shows that the fermions, as expected, always arrive at different detectors:
$P_{fermion} = \begin{bmatrix} \left[ \left| ( \overline{uhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{uvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdv} )^T \Psi _{out} \right| \right]^2 \ \end{bmatrix} = \begin{pmatrix} 0 & 0 & \frac{1}{2} & 0 \ 0 & 0 & 0 & 0 \ \frac{1}{2} & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
Naturally an algebraic analysis yields the same result.
$\Psi _{out} = \frac{1}{2 \sqrt{2}} (i (uh) uh - uh (dh) + dh (uh) + i (dh) dh - i (uh) uh - uh (dh) + dh (uh) - i (dh) dh) = \frac{1}{ sqrt{2}} (dh (uh) - uh(dh)) \nonumber$
$P_{fermion} = P_{uhdh} = P_{dhuh} = \frac{1}{2} \nonumber$
As algebraic and matrix calculations show, introduction of path information for fermions yields the same result as for bosons, the photons sometimes arrive at the same detector and sometimes at different detectors.
$\Psi _{fermion} = \frac{1}{ \sqrt{2}} = \frac{1}{ sqrt{2}} (uhdv = dvuh) \nonumber$
$\Psi _{out} = BSI(MI) \Psi_{fermion} \nonumber$
$\Psi_{out} = \frac{1}{2 \sqrt{2}} (i (uh) uv - uh (dv) + dh (uv) + i (dh) dv- i (uv) uh - uv (dh) + dv (uh) - i (dv)dh) \nonumber$
$P_{fermion} = P_{uhuv} = P_{uhdv} = P_{dhuv} = P_{dhdv} = P_{uvuh} = P_{uvdh} = P_{dvuh} = P_{dvdh} = \frac{1}{8} \nonumber$
$P_{fermion} = \begin{bmatrix} \left[ \left| ( \overline{uhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{uvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{uvdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dhuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dhdv} )^T \Psi _{out} \right| \right]^2 \ \left[ \left| ( \overline{dvuh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvuv} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdh} )^T \Psi _{out} \right| \right]^2 & \left[ \left| ( \overline{dvdv} )^T \Psi _{out} \right| \right]^2 \ \end{bmatrix} = \begin{pmatrix} 0 & \frac{1}{8} & 0 & \frac{1}{8} \ \frac{1}{8} & 0 & \frac{1}{8} & 0 \ 0 & \frac{1}{8} & 0 & \frac{1}{8} \ \frac{1}{8} & 0 & \frac{1}{8} & 0 \end{pmatrix} \nonumber$
With erasure of path information the fermionic photons again always arrive at different detectors.
$\begin{bmatrix} \left( \left| usus \Psi_{out} \right| \right)^2 & \left( \left| usds \Psi_{out} \right| \right)^2 \ \left( \left| dsus \Psi_{out} \right| \right)^2 & \left( \left| dsds \Psi_{out} \right| \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & \frac{1}{8} \ \frac{1}{8} & 0 \end{pmatrix} \nonumber$
$\Psi _s = \frac{1}{4 \sqrt{2}} (i (us) us - us (ds) + ds (us) + i (ds) ds - i (us) us - us (ds) + ds(us) - i (ds) ds = \frac{1}{2 \sqrt{2}} (ds (us) - us (ds)) \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.40%3A_Another_Example_of_a_Two-photon_Quantum_Eraser.txt
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This tutorial works through the first half of the Science report (Nov. 2, 2012, pp . 634‐637) by Peruzzo et al. with the title given above. The Wheeler‐type delayed‐choice experiment dealt with in this paper is shown schematically in the figure below. An observer has the option of inserting or removing a second beam splitter (BS) in a Mach‐Zehnder interferometer (MZI) after a photon has passed the first BS.
This apparatus is implemented using Hadamard and controlled‐Hadamard gates as shown below. A phase shift is also possible in the lower arm of the interferometer. The operation of the second Hadamard gate is controlled by an ancillary photon which can be prepared as a superposition. For this reason the second beam splitter is called a quantum beam splitter (QBS).
The Hadamard gate is a single‐photon gate as is the phase shifter, while the controlled‐Hadamard gate is a two‐photon gate. Their matrix representations are as follows.
$H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
$A ( \phi ) = \begin{pmatrix} 1 & 0 \ 0 & e^{ i \phi} \end{pmatrix} \nonumber$
$A ( \phi ) H \begin{pmatrix} 1 \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} e^{ \phi i}}{2} \end{pmatrix} \nonumber$
$CH = \begin{pmatrix} 1 & 0 & & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \ 0 & 0 & \frac{1}{ \sqrt{2}} & \frac{-1}{ \sqrt{2}} \end{pmatrix} \nonumber$
The H gate and the phase‐shifter create the superposition at the point of the arrow (see highlighted area above), just before the CH gate. This is the point at which this analysis begins. Using vector tensor multiplication, the composite state (a = ancilla; s = system) is shown below.
$\begin{pmatrix} \cos \alpha \ \sin \alpha \end{pmatrix}_a \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ e^{i \phi} \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} \cos \alpha \ \cos \alpha e^{i \phi} \ \sin \alpha \ \sin \alpha e^{i \phi} \end{pmatrix} \nonumber$
The four output states expressed in the |0 > ‐ |1 > basis are as follows.
$| 0 \rangle_a |0 \rangle_s = \begin{pmatrix} 1 \ 0 \end{pmatrix}_a \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_s = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$| 1 \rangle_a |0 \rangle_s = \begin{pmatrix} 0 \ 1 \end{pmatrix}_a \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_s = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \nonumber$
$| 0 \rangle_a |1 \rangle_s = \begin{pmatrix} 1 \ 0 \end{pmatrix}_a \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_s = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
$| 1 \rangle_a |1 \rangle_s = \begin{pmatrix} 0 \ 1 \end{pmatrix}_a \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_s = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The probability that the photon is detected at Dʹʹ (|0 >s) is:
$D" ( \alpha , \phi ) = \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} ^T CH \frac{1}{ \sqrt{2}} \begin{pmatrix} \cos \alpha \ \cos \alpha e^{i \phi} \ \sin \alpha \ \sin \alpha e^{i \phi} \end{pmatrix} \right| \right]^2 + \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} ^T CH \frac{1}{ \sqrt{2}} \begin{pmatrix} \cos \alpha \ \cos \alpha e^{i \phi} \ \sin \alpha \ \sin \alpha e^{i \phi} \end{pmatrix} \right| \right]^2 \nonumber$
The probability that the photon is detected at Dʹ (|1 >s) is:
$D' ( \alpha , \phi ) = \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} ^T CH \frac{1}{ \sqrt{2}} \begin{pmatrix} \cos \alpha \ \cos \alpha e^{i \phi} \ \sin \alpha \ \sin \alpha e^{i \phi} \end{pmatrix} \right| \right]^2 + \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} ^T CH \frac{1}{ \sqrt{2}} \begin{pmatrix} \cos \alpha \ \cos \alpha e^{i \phi} \ \sin \alpha \ \sin \alpha e^{i \phi} \end{pmatrix} \right| \right]^2 \nonumber$
The following graph shows that if α = 0 (the ancillary photon is in the |0 > state), the interferometer is open and the detectors each fire 50% of the time no matter what phase relationship exists between the arms of the interferometer.
However, if α = π/2 (the ancillary photon is in the |1 > state) the interferometer is closed and interference is observed. The firing of the detectors depends upon the phase relation between the two arms of the interferometer. If ϕ = 0 only Dʹʹ fires, while if ϕ = π only Dʹ fires. As shown below, at other ϕ values both detectors fire.
If α = π/4 (the ancilla photon is in a 50‐50 superposition of |0 > and |1 >), both detectors fire at all ϕ values.
$N = 50 ~~~ i = 0 .. N ~~~ j = 0 .. N ~~~ \alpha_1 = \frac{ \frac{ \pi}{2} i}{N} ~~~ \phi_j = \frac{ - \pi}{2} + \frac{2 \pi j}{N} ~~~ I_{i,~j} = D'( \alpha_i , \phi_j)$
In summary, according to Peruzzo et al. this is a delayed‐choice experiment for 0 < α < π/2 because the system photon is in the interferometer before it becomes entangled with the ancillary photon.
While this work represents an exquisite experimental tour de force, in my opinion it doesnʹt reveal anything more about wave‐particle duality than any other quantum optics experiment. We always measure particles (detectors click, photographic film is darkened, etc.), but we interpret what happened or predict what will happen by assuming wavelike behavior. In other words, quantum particles exhibit both wave and particle properties in every experiment. To paraphrase Nick Herbert (Quantum Reality), particles are always detected, but the experimental results observed are the result of wavelike behavior. Richard Feynman put it this way (The Character of Physical Law), ʺI will summarize, then, by saying that electrons arrive in lumps, like particles, but the probability of arrival of these lumps is determined as the intensity of waves would be. It is in this sense that the electron behaves sometimes like a particle and sometimes like a wave. It behaves in two different ways at the same time.ʺ Bragg said, ʺEverything in the future is a wave, everything in the past is a particle.ʺ
In 1951 in his treatise Quantum Theory, David Bohm described wave‐particle duality as follows: ʺOne of the most characteristic features of the quantum theory is the wave‐particle duality, i.e. the ability of matter or light quanta to demonstrate the wave‐like property of interference, and yet to appear subsequently in the form of localiziable particles, even after such interference has taken place.ʺ In other words, to explain interference phenomena wave properties must be assigned to matter and light quanta prior to detection as particles.
Equation (1) in this paper is not a ʺparticleʺ wave function, it is a superposition of the photon being in both arms of the interferometer. It is a wave function (period) and it accurately predicts the future detection results in both the absence and presence of the second beam splitter. See the Appendix for further comment. John Wheeler, designer of several delayed‐choice experiments (both terrestrial and cosmological), had the following to say about the interpretation of such experiments.
... in a loose way of speaking, we decide what the photon shall have done after it has already done it. In actuality it is wrong to talk of the ʹrouteʹ of the photon. For a proper way of speaking we recall once more that it makes no sense to talk of a phenomenon until it has been brought to a close by an irreversible act of amplification. ʹNo elementary phenomenon is a phenomenon until it is a registered (observed) phenomenon.ʹ
This statement was given a convincing and amusing graphical representation by Field Gilbertʹs sketch of Wheelerʹs ʹGreat Smokey Dragonʹ which can be found on page 154 of Jim Baggotʹs The Meaning of Quantum Theory.
Heisenberg said pretty much the same thing as Wheeler when he wrote in 1958: ʺIf we want to describe what happens in an atomic event, we have to realize that the word ʹhappensʹ can only apply to the observation, not to the state of affairs between two observations.ʺ
I would also like to point out that equation (4) is mislabeled. The authors claim it is the detection probability at Dʹ as a function of α and ϕ. However, the plot below shows that designation is not consistent with Figure 3B. Equation (4) is actually the detection probability at detector Dʹʹ.
$\ln ( \alpha , \phi ) = \frac{1}{2} \cos \alpha ^2 + \cos \left( \frac{ \phi }{2} \right)^2 \sin \alpha ^2 ~~~ I_{i,~j} = \ln ( \alpha_i , \phi_j ) \nonumber$
Appendix
Removing the phase shifter makes it easier to follow the evolution of the photons through the interferometer and to evaluate the authorsʹ claim that they have demonstrated wave‐particle duality as indicated in their equations 1, 2 and 3. We begin with the composite state prior to the controlled Hadamard gate.
$\begin{pmatrix} \cos \alpha \ \sin \alpha \end{pmatrix} _a \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} _s = \frac{ \cos \alpha |0 \rangle_a (0| \rangle_s + |1 \rangle_s ) + \sin \alpha |1 \rangle_a (|0 \rangle_s + |1 \rangle_s )}{ \sqrt{2}} \nonumber$
The operation of the controlled Hadamard (CH) gate is written algebraically.
$CH | 0 \rangle_a | 0 \rangle_s = |0 \rangle_a |0 \rangle_s \nonumber$
$CH|0 \rangle_a |1 \rangle_s = |0 \rangle_a |1 \rangle_s \nonumber$
$CH |1 \rangle_a |0 \rangle_s = |1 \rangle_a \frac{1}{ \sqrt{2}} (|0 \rangle_s + |1 \rangle_s) \nonumber$
$CH | 1 \rangle_a |1 \rangle_s = |1 \rangle_a \frac{1}{ \sqrt{2}} (|0 \rangle_s - |1 \rangle_s) \nonumber$
The action of the controlled Hadamard gate leads to the following state which shows a superposition in blue and interference in red, depending on the state of the control qubit.
$\frac{ \cos \alpha |) \rangle_a (|0 \rangle_s + |1 \rangle_s) + \sin \alpha |1 \rangle_a \left( \frac{|0 \rangle_s + |1 \rangle_s}{ \sqrt{2}} + \frac{|0 \rangle_s - |1 \rangle_s}{ \sqrt{2}} \right)}{ \sqrt{2}} = \cos \alpha |0 \rangle_a \frac{|0 \rangle_s + |1 \rangle_s}{ \sqrt{2}} + \sin \alpha |1 \rangle_a |0 \rangle_s \nonumber$
According to Peruzzo et al. this leads to the following interpretive equation expressing their view of wave‐particle duality.
$\cos \alpha | 0 \rangle_a \Psi ( \text{particle}) \rangle_s + \sin \alpha |1 \rangle_a| \Psi ( \text{wave}) \rangle_s \nonumber$
My objection is that a superposition, even if it dosenʹt lead to interference, is an example of wavelike (delocalized) behavior. The fact that the superposition doesnʹt lead to interference doesnʹt mean the photon took a particular path to a particular detector, it simply means on observation the superposition collapsed with appropriate probability at that detector.
Quantum mechanical entities (quons) do not, as Peruzzo et al. assert, exhibit particle‐ or wavelike behavior depending on the design of the experimental apparatus they encounter. As Bohm, Feynman and Herbert have said quons exhibit wave and particle characteristics in every experiment.
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This note presents a critique of "Entanglement-Enabled Delayed-Choice Experiment," F. Kaiser, et al. Science 338, 637 (2012). This experiment was also summarized in section 6.3 of Quantum Weirdness, by William Mullin, Oxford University Press, 2017.
A source, S, emits two photons in opposite directions on the x-axis in the following polarization state, where v and h represent vertical and horizontal polarization, respectively.
$| \Psi \rangle_{AB} \frac{1}{ \sqrt{2}} \left[ |xv \rangle_A |xv \rangle_B + |xh \rangle_A |xh \rangle_B \right] \nonumber$
Photon B travels to the left to a polarizing beam splitter, PBS. Photon A travels to the right entering an interferometer whose elements are a beam splitter (BS), two mirrors (M), and a polarization-dependent beam splitter (PDBS).
The implementation of the PDBS is shown here.
The operation of the optical elements are as follows. A mirror simply reflects the photon's direction of motion.
$M = |y \rangle \langle x | + |x \rangle \langle y | \nonumber$
A 50-50 BS splits the photon beam into a superposition of motion in the x- and y-directions. By convention the reflected beam collects a pi/2 (i) phase shift relative to the transmitted beam.
$BS = \frac{|x \rangle + i | y \rangle}{ \sqrt{2}} \langle x | + \frac{i | x \rangle + | y \rangle}{ \sqrt{2}} \langle y | = \frac{1}{ \sqrt{2}} (|x \rangle \langle x | + i | y \rangle \langle x | + i | x \rangle \langle y | + |y \rangle \langle y |) \nonumber$
A PBS transmits vertically polarized photons and reflects horizontally polarized photons.
$M = |y \rangle \langle x | + |x \rangle \langle y | \nonumber$
A 50-50 BS splits the photon beam into a superposition of motion in the x- and y-directions. By convention the reflected beam collects a pi/2 (i) phase shift relative to the transmitted beam.
$BS = \frac{|x \rangle + i |y \rangle}{ \sqrt{2}} \rangle x| + \frac{i|x \rangle + |y \rangle}{ \sqrt{2}} \langle y | = \frac{1}{ \sqrt{2}} (|x \rangle \langle x | + i |y \rangle \langle x | + i | x \rangle \langle y | + |y \rangle \langle y| ) \nonumber$
A PBS transmits vertically polarized photons and reflects horizontally polarized photons.
$PBS = |xv \rangle \langle xv | + |yh \rangle \langle xh | + |yv \rangle \langle yv | + |xh \rangle \langle yh | \nonumber$
The PDBS uses an initial PBS to reflect horizontally polarized photons to a second PBS which reflects them to the detectors. Vertically polarized photons are transmitted by the first PBS to a BS which has the action shown above, after which they are transmitted to the detectors by the second PBS. PBS/PDBS blue/red color coding highlights the action of the central BS.
$PDPS = \frac{|xv \rangle + i | yv \rangle}{ \sqrt{2}} \langle xv | + |yh \rangle \langle xh + \frac{i|xv \rangle + |yv \rangle}{ \sqrt{2}} \langle yv| + |xh \rangle \langle yh| \nonumber$
Given the state produced by the source, half the time a horizontal photon will enter the interferometer and half the time a vertical photon will enter. The following algebraic analysis shows the progress of the h- and v-polarized photons entering the interferometer. It is clear from this analysis that D1 will fire 25% of the time and D2 75% of the time.
$\begin{array}{c} |xh \rangle & |xv \rangle \ BS & BS \ \frac{ |xh \rangle + i |yh \rangle}{ \sqrt{2}} & \frac{ |xv \rangle + i |yv \rangle}{ \sqrt{2}} \ M & M \ \frac{ |yh \rangle + i |xh \rangle}{ \sqrt{2}} & \frac{ |yv \rangle + i |xv \rangle}{ \sqrt{2}} \ PDBS & PDBS \ \frac{|D2 \rangle |h \rangle + i|D1 \rangle |h \rangle}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \left[ \frac{|D1 \rangle |v \rangle + i| D2 \rangle |v \rangle}{ \sqrt{2}} + \frac{i (i|D1 \rangle |v \rangle + D2 \rangle |v \rangle}{ \sqrt{2}} \right] \ ~ & \downarrow \ ~ & i|D2 \rangle |v \rangle \ \end{array} \nonumber$
This analysis suggests to some that inside the interferometer h-photons behave like particles and v-photons behave like waves. The argument for this view is that interference occurs at the PDBS for v-photons, but not for h-photons. However, in both cases the state illuminating the PDBS, highlighted in blue, is a superposition of the photon being in both arms of the interferometer. In my opinion this superposition implies delocalization which implies wavelike behavior. At the PDBS the h-photon superposition is reflected away from the central BS to D1 and D2, leading to the final superposition in the left-hand column above, which collapses on observation to either D1 or D2. The v-photon superposition is transmitted at the PDBS to the central BS allowing for destructive interference at D1 and constructive interference at D2 as is shown at the bottom of the right-hand column.
Those who interpret this experiment in terms of particle or wave behavior also invoke the concept of delayed-choice, claiming that if D2 fires we don't know for sure which behavior has occurred because both h and v photons can arrive there. They argue that until photon B has been observed at Dv or Dh, which by design can be long after photon A has exited the interferometer, the polarization of the photon detected at D2 is unknown and therefore so is whether particle or wave behavior has occurred. These analysts write the final two photon wavefunction as the following entangled superposition, where particle behavior is highlighted in red and wave behavior in blue.
$| \Psi \rangle_{final} = \frac{1}{ \sqrt{2}} \left( \left( \frac{|D2,h \rangle + i|D1,h \rangle}{ \sqrt{2}} \right)_A |Dh \rangle_B + i|D2,v \rangle_A |Dv \rangle_B \right) = \frac{1}{ \sqrt{2}} ( | Particle \rangle_A |Dh \rangle_B + | Wave \rangle_A |Dv \rangle_B ) \nonumber$
For the reasons expressed above I do not find this interpretation convincing. We always observe particles (detectors click, photographic film is darkened, etc.), but we interpret what happened or predict what will happen by assuming wavelike behavior. In other words, objects governed by quantum mechanical principles (quons) exhibit both wave and particle properties in every experiment. To paraphrase Nick Herbert (Quantum Reality), particles are always detected, but the experimental results observed are the result of wavelike behavior. Bragg summarized wave-particle duality saying, "Everything in the future is a wave, everything in the past is a particle."
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As shown in the graphic below (Nature, December 11, 1997, page 576), quantum teleportation is a form of information transfer that requires pre-existing entanglement and a classical communication channel to send information from one location to another. Alice has the photon to be teleported and a photon of an entangled pair (β00) that she shares with Bob. She performs a measurement on her photons that projects them into one of the four Bell states and Bob's photon, via the entangled quantum channel, into a state that has a unique relationship to the state of the teleportee. Bob carries out one of four unitary operations on his photon depending on the results of Alice's measurement, which she sends him through a classical communication channel.
The teleportee and the Bell states indexed in binary notation:
Teleportee:
$\begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
Bell states:
$\beta_{00} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} ~~~ \beta_{01} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} ~~~ \beta_{10} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} ~~~ \beta_{11} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The three-cubit initial state is rewritten as a linear superposition of the four possible Bell states that Alice can find on measurement.
$| \Psi \rangle = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \otimes \beta_{00} = \frac{1}{2} \left[ \beta_{00} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{01} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + \beta_{10} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{11} \otimes \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \nonumber$
Alice's Bell state measurement result (β00, β01, β10 or β11) determines the operation (I, X, Z or ZX) that Bob performs on his photon. The matrices for these operations are as follows.
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} ~~~ X = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} ~~~ Z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} ~~~ ZX = \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix} \nonumber$
Tabular summary of teleportation experiment:
$\begin{pmatrix} \text{Alice}~ \text{Measurement} ~ \text{Result} & \beta_{00} & \beta_{01} & \beta_{10} & \beta_{11} \ \text{Bob's} ~ \text{Action} & I & X & Z & XZ \end{pmatrix} \nonumber$
Summary of the quantum teleportation protocol: "Quantum teleportation provides a 'disembodied' way to transfer quantum states from one object to another at a distant location, assisted by previously shared entangled states and a classical communication channel." Nature 518, 516 (2015)
8.02: Quantum Teleportation - A Brief Introduction
Alice wishes to teleport a photon in the polarization state, αh1 + βv1, to Bob, where α and β are complex coefficients such that the sum of the square of their absolute magnitudes is unity. The letters h and v refer to the vertical and horizontal polarization states. In preparation for the teleportation event, Alice and Bob first prepare an entangled state involving photons 2 and 3, $\frac{ h_2 v_3 - v_2 h_3}{ \sqrt{2}}$, as shown in the figure below [Nature 390, 576 (1997)].
Alice has photon 2 and Bob photon 3, but because they are in an entangled state the photons do not have well‐defined individual polarization states.
Alice arranges for photons 1 and 2 to arrive at opposite sides of a beam splitter at the same time. This gives rise to the following state,
$( \alpha h_1 \beta v_1 ) \frac{h_2 v_3 - v_2 h_3}{ \sqrt{2}} \nonumber$
which upon expansion yields,
$\frac{1}{2} \sqrt{2} \alpha h_1 h_2 v_3 - \frac{1}{2} \alpha h_1 v_2 h_3 + \frac{1}{2} \beta v_1 h_2 v_3 - \frac{1}{2} \beta v_1 v_2 h_3 \nonumber$
Alice now makes a Bell‐state measurement (vide infra) at the detectors (a and b) to the left and right of her beam splitter. Bell states are the four maximally entangled h‐v polarization states of photons 1 and 2. They are as follows:
$\Phi_p = \frac{h_1 h_2 + v_1 v_2}{ \sqrt{2}} ~~~ \Phi_m = \frac{h_1 h_2 - v_1 v_2}{ \sqrt{2}} ~~~ \Psi_p = \frac{h_1 v_2 + v_1 h_2}{ \sqrt{2}} ~~~ \Psi_m = \frac{h_1 v_2 - v_1 h_2}{ \sqrt{2}} \nonumber$
The products of the polarization states of photons 1 and 2 in equation (3) can be expressed as linear superpositions of the Bell states.
$h_1 h_2 = \frac{ \sqrt{2}}{2} ( \Phi_p + \Phi_m) ~~~ h_1 v_2 = \frac{ \sqrt{2}}{2} ( \Psi_p + \Psi_m) ~~~ v_1 h_2 = \frac{ \sqrt{2}}{2} ( \Psi_p - \Psi_m) ~~~ v_1 v_2 = \frac{ \sqrt{2}}{2} ( \Phi_p - \Phi_m) \nonumber$
We can let Mathcad do the heavy lifting by having it expand (1), substitute equations (4) and collect on the Bell states.
$( \alpha h_1 \beta v_1) \frac{h_2v_3 - v_2 h_3}{ \sqrt{2}} \begin{array}{|l} \text{expand} \ \text{substitute}, h_1 h_2 = \frac{ \sqrt{2}}{2} ( \Phi_p + \Phi_m) \ \text{substitute}, h_1 v_2 = \frac{ \sqrt{2}}{2} ( \Psi_p + \Psi_m) \rightarrow \left( \frac{ \beta h_3}{2} + \frac{ \alpha v_3}{2} \right) \Phi_m + \left( \frac{ \alpha v_3}{2} - \frac{ \beta h_3}{2} \right) \Phi_p + \left( - \frac{ \alpha h_3}{2} - \frac{ \beta v_3}{2} \right) \Psi_m + \left( \frac{ \beta v_3}{2} - \frac{ \alpha h_3}{2} \right) \Psi_p \ \text{substitute}, v_1 h_2 = \frac{ \sqrt{2}}{2} ( \Psi_p - \Psi_m) \ \text{substitute}, v_1 v_2 = \frac{ \sqrt{2}}{2} ( \Phi_p - \Phi_m) \ \text{collect},~ \Phi_m,~ \Phi_p,~ \Psi_m,~ \Psi_p \ \end{array} \nonumber$
Thus Aliceʹs Bell‐state measurement has four equally likely (0.52 = 0.25) outcomes. We will restrict our attention to the third term which says that if Alice measures Ψm, then Bob receives photon 1ʹs polarization state without further action by him.
$- ( \alpha h_3 + \beta v_3 ) \Psi_m \nonumber$
This will occur 25% of the time. The other three possible measurement outcomes are more complicated to analyze and will not be discussed further here. So we assume Alice measures Ψm and communicates this to Bob by classical means so that he knows that his photon (#3) now has the polarization state of the original photon 1. But, how does Alice know that the results she observes at detectors a and b mean that photons 1 and 2 are in Bell state Ψm?
First the short, qualitative answer. Of the four Bell states, Ψm is the only one that is antisymmetric with respect to the interchange of the labels of the photons. Thus, in spite of the fact that photons individually are bosons, this entangled state is fermionic ‐ collectively the photons are behaving as fermions. This means that they canʹt be in the same (measurement) state at the same time. If photon 1 is detected at a, then photon 2 will be detected at b. Therefore, if Alice observes a‐b coincidences it means that photons 1 and 2 are in the Ψm Bell state and photon 1ʹs polarization state has been teleported to Bobʹs photon (#3).
Bob confirms that he has received the polarization state of photon 1 using a polarizing beam splitter as shown in the figure above. Suppose Alice encodes photon 1 with a 45o polarization, then Bob sets his polarizing beam splitter to detect +45o/‐45o polarized photons. A three‐fold coincidence between detectors a, b and Bobʹs +45o detector confirms teleportation. This was procedure employed in the original experiment published in Nature on December 11, 1997.
It should be pointed out that the initial entangled state used by Alice and Bob involving photons 2 and 3 is the Bell state Ψm. As was just seen, if Alice observes Ψm in her Bell‐state measurement Bob receives photon 1ʹs polarization state without further action by him. As is shown below the analogous thing occurs if the initial entangled state is Ψp, Φp or Φm.
$\Psi_p > \left( \frac{ \alpha h_3}{2} + \frac{ \beta v_3}{2} \right) \Psi_p \nonumber$
$( \alpha h_1 \beta v_1) \frac{h_2v_3 + v_2 h_3}{ \sqrt{2}} \begin{array}{|l} \text{expand} \ \text{substitute}, h_1 h_2 = \frac{ \sqrt{2}}{2} ( \Phi_p + \Phi_m) \ \text{substitute}, h_1 v_2 = \frac{ \sqrt{2}}{2} ( \Psi_p + \Psi_m) \rightarrow \left( \frac{ \alpha v_3}{2} - \frac{ \beta h_3}{2} \right) \Phi_m + \left( \frac{ \beta h_3}{2} + \frac{ \alpha v_3}{2} \right) \Phi_p + \left( \frac{ \alpha h_3}{2} - \frac{ \beta v_3}{2} \right) \Psi_m + \left( \frac{ \alpha h_3}{2} + \frac{ \beta v_3}{2} \right) \Psi_p \ \text{substitute}, v_1 h_2 = \frac{ \sqrt{2}}{2} ( \Psi_p - \Psi_m) \ \text{substitute}, v_1 v_2 = \frac{ \sqrt{2}}{2} ( \Phi_p - \Phi_m) \ \text{collect},~ \Phi_m,~ \Phi_p,~ \Psi_m,~ \Psi_p \ \end{array} \nonumber$
$\Phi_p > \left( \frac{ \alpha h_3}{2} + \frac{ \beta v_3}{2} \right) \Phi_p \nonumber$
$( \alpha h_1 \beta v_1) \frac{h_2 h_3 + v_2 v_3}{ \sqrt{2}} \begin{array}{|l} \text{expand} \ \text{substitute}, h_1 h_2 = \frac{ \sqrt{2}}{2} ( \Phi_p + \Phi_m) \ \text{substitute}, h_1 v_2 = \frac{ \sqrt{2}}{2} ( \Psi_p + \Psi_m) \rightarrow \left( \frac{ \alpha h_3}{2} - \frac{ \beta v_3}{2} \right) \Phi_m + \left( \frac{ \alpha h_3}{2} + \frac{ \beta v_3}{2} \right) \Phi_p + \left( \frac{ \alpha v_3}{2} - \frac{ \beta h_3}{2} \right) \Psi_m + \left( \frac{ \beta h_3}{2} + \frac{ \alpha v_3}{2} \right) \Psi_p \ \text{substitute}, v_1 h_2 = \frac{ \sqrt{2}}{2} ( \Psi_p - \Psi_m) \ \text{substitute}, v_1 v_2 = \frac{ \sqrt{2}}{2} ( \Phi_p - \Phi_m) \ \text{collect},~ \Phi_m,~ \Phi_p,~ \Psi_m,~ \Psi_p \ \end{array} \nonumber$
$\Phi_m > \left( \frac{ \alpha h_3}{2} + \frac{ \beta v_3}{2} \right) \Phi_m \nonumber$
$( \alpha h_1 \beta v_1) \frac{h_2 h_3 + v_2 v_3}{ \sqrt{2}} \begin{array}{|l} \text{expand} \ \text{substitute}, h_1 h_2 = \frac{ \sqrt{2}}{2} ( \Phi_p + \Phi_m) \ \text{substitute}, h_1 v_2 = \frac{ \sqrt{2}}{2} ( \Psi_p + \Psi_m) \rightarrow \left( \frac{ \alpha h_3}{2} + \frac{ \beta v_3}{2} \right) \Phi_m + \left( \frac{ \alpha h_3}{2} - \frac{ \beta v_3}{2} \right) \Phi_p + \left( - \frac{ \beta h_3}{2} - \frac{ \alpha v_3}{2} \right) \Psi_m + \left( \frac{ \beta h_3}{2} - \frac{ \alpha v_3}{2} \right) \Psi_p \ \text{substitute}, v_1 h_2 = \frac{ \sqrt{2}}{2} ( \Psi_p - \Psi_m) \ \text{substitute}, v_1 v_2 = \frac{ \sqrt{2}}{2} ( \Phi_p - \Phi_m) \ \text{collect},~ \Phi_m,~ \Phi_p,~ \Psi_m,~ \Psi_p \ \end{array} \nonumber$
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The purpose of this tutorial is to provide a brief mathematical outline of the basic elements of quantum teleportation, as illustrated in the figure below, using matrix and tensor algebra.
Alice wishes to teleport the following state (X in the figure) to Bob,
$| \Phi \rangle = a | 0 \rangle + b | 1 \rangle = \begin{pmatrix} a \ b \end{pmatrix} ~~~ |a|^2 + |b|^2 = 1 \nonumber$
where,
$| 0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} ~~~ | 1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
They prepare the following entangled two-particle state, involving A and B, in which Alice has particle A and Bob has B.
$| \Psi_{ab} = \frac{1}{ \sqrt{2}} [|00 \rangle + |11 \rangle] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
Alice arranges for the particle to be teleported, |Φ >, and her entangled particle to meet simultaneously on opposite sides of a beam splitter, creating the following the three-particle state.
$| \Phi \rangle | \Psi_{AB} \rangle = \begin{pmatrix} a \ b \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} a \ 0 \ 0 \ a \ b \ 0 \ 0 \ b \end{pmatrix} \nonumber$
This state can be written as a superposition of the following 4-vectors, which are the well-known Bell states. Please see the Appendix for definitions of the Bell states.
$\frac{1}{ \sqrt{2}} \begin{pmatrix} a \ 0 \ 0 \ a \ b \ 0 \ 0 \ b \end{pmatrix} = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} a \ b \end{pmatrix} + \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \otimes \begin{pmatrix} a \ -b \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} b \ a\ \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} -b \ a \end{pmatrix} \right] \nonumber$
We now write this three-particle state in terms of the Bell basis labels.
$| \Phi \rangle | \Psi_{AB} \rangle = \frac{1}{2} \left[ | \Phi ^+ \begin{pmatrix} a \ b \end{pmatrix} + | \Phi^- \rangle \begin{pmatrix} a \ -b \end{pmatrix} + | \Psi^+ \begin{pmatrix} b \ a \end{pmatrix} + | \Psi^- \begin{pmatrix} -b \ a \end{pmatrix} \right] \nonumber$
Next, Alice makes a Bell-state measurement on her two particles, getting any of the four possible outcomes (Φ+ , Φ- , Ψ+ , or Ψ- ) with equal probability, 25%. Her measurement collapses the state of Bob’s particle into the companion of the result of her Bell-state measurement. Alice then sends the result of her measurement through a classical channel to Bob. Depending on her report, he carries out one of the following operations on his particle to complete the teleportation process.
$| \Phi ^+ \rangle = \hat{I} \begin{pmatrix} a \ b \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \begin{pmatrix} a \ b \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix} \nonumber$
$| \Phi ^- \rangle = \widehat{ \sigma_z} \begin{pmatrix} a \ -b \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \begin{pmatrix} a \ -b \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix} \nonumber$
$| \Psi ^+ \rangle = \widehat{ \sigma_x} \begin{pmatrix} b \ a \end{pmatrix} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} b \ a \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix} \nonumber$
$| \Psi ^- \rangle = \widehat{ \sigma_z} \widehat{ \sigma_x} \begin{pmatrix} -b \ a \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} -b \ a \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix} \nonumber$
Appendix
The Bell basis is the following collection of maximally entangled two-qubit states.
$| \Phi ^+ \rangle = \frac{1}{ \sqrt{2}} [| 00 \rangle + | 11 \rangle] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
$| \Phi ^- \rangle = \frac{1}{ \sqrt{2}} [| 00 \rangle - | 11 \rangle] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
$| \Psi ^+ \rangle = \frac{1}{ \sqrt{2}} [| 01 \rangle + | 10 \rangle] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$| \Psi ^- \rangle = \frac{1}{ \sqrt{2}} [| 01 \rangle - | 10 \rangle] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The initial entangled state Alice and Bob prepare is the Bell state |Φ+ >.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.03%3A_Quantum_Teleportation_at_a_Glance.txt
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The purpose of this tutorial is to provide a brief outline of the basic elements of quantum teleportation, as illustrated in the figure below, using matrix and tensor algebra in the Mathcad programming environment. The methods used here are closely related to those presented in the preceding tutorial on teleportation.
As is generally the case, the analysis requires the use of the maximally entangled two‐quibit Bell states, the identity matrix and two of the unitary Pauli matrices.
Bell states:
$\Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
$\Phi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
$\Psi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$\Psi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
Identity and x- and z-Pauli matrices:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
$\sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \nonumber$
Alice wishes to teleport this state to Bob.
$\begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \nonumber$
$\left[ \left| \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \right| \right]^2 = 1 \nonumber$
They prepare the first Bell state (Φp) in which Alice has the first qubit and Bob the second. Alice arranges for the state to be teleported and her entangled qubit to meet creating the following three qubit state.
$\begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0.577 \ 0 \ 0 \ 0.577 \ 0.816 \ 0 \ 0 \ 0.816 \end{pmatrix} \nonumber$
$\Psi^T \Psi = 1 \nonumber$
Next Alice performs a Bell‐state measurement, in other words she projects Ψ onto the Bell‐state basis given above. The four equally likely outcomes are calculated below. But first letʹs use Φp as an example. Its projector is |Φp >< Φp|.
$\Phi _p \Phi _p ^T = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0.5 & 0 & 0 & 0.5 \end{pmatrix} \nonumber$
However, Ψ is a three‐qubit state in which Alice has the first two qubits and Bob the third. In other words, the projector does not operate directly on Bobʹs qubit. Thus the appropriate matrix for this operation uses the identity matrix to leave Bobʹs qubit alone. Kronecker is Mathcadʹs command for tensor multiplication of matrices.
$\text{kronecker} ( \Phi_p, \Phi_p,^T, I) = \begin{pmatrix} 0.5 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0 \ 0 & 0.5 & 0 & 0 & 0 & 0 & 0 & 0.5 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0.5 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0 \ 0 & 0.5 & 0 & 0 & 0 & 0 & 0 & 0.5 \end{pmatrix} \nonumber$
Next we show that indeed there are four equally likely outcomes for the Bell-state measurement.
$\left( \left| \text{kronecker} \left( \Phi_p, \Phi_p^T, I \right) \Psi \right| \right)^2 = 0.25 \nonumber$
$\left( \left| \text{kronecker} \left( \Phi_m, \Phi_m^T, I \right) \Psi \right| \right)^2 = 0.25 \nonumber$
$\left( \left| \text{kronecker} \left( \Psi_p, \Psi_p^T, I \right) \Psi \right| \right)^2 = 0.25 \nonumber$
$\left( \left| \text{kronecker} \left( \Psi_m, \Psi_m^T, I \right) \Psi \right| \right)^2 = 0.25 \nonumber$
Aliceʹs Bell‐state measurement yields one of these results, which she communicates to Bob through a classical channel. As shown below, depending on Aliceʹs report Bob carries out the following unitary operations on his qubit to receive the teleported state: I, σz, σx, and σzσx. The multiplicative factor of 2 normalizes the result of the Bell state measurement.
$2 \text{kronecker} ( \Phi_p, \Phi_p^T, I) \Psi = \begin{pmatrix} 0.408 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0.408 \ 0.577 \end{pmatrix} \nonumber$
Factors to:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \nonumber$
So Bob does nothing:
$I \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \nonumber$
$2 \text{kronecker} ( \Phi_m, \Phi_m^T, I) \Psi = \begin{pmatrix} 0.408 \ -0.577 \ 0 \ 0 \ 0 \ 0 \ -0.408 \ 0.577 \end{pmatrix} \nonumber$
Factors to:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \begin{pmatrix} 0.577 \ -0.816 \end{pmatrix} \nonumber$
So Bob does this:
$\sigma_z \begin{pmatrix} 0.577 \ -0.816 \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \nonumber$
$2 \text{kronecker} ( \Psi_p, \Psi_p^T, I) \Psi = \begin{pmatrix} 0 \ 0 \ 0.577 \ 0.408 \ 0.577 \ 0.408 \ 0 \ 0 \end{pmatrix} \nonumber$
Factors to:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \nonumber$
So Bob does this:
$\sigma_x \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \nonumber$
$2 \text{kronecker} ( \Psi_m, \Psi_m^T, I) \Psi = \begin{pmatrix} 0 \ 0 \ -0.577 \ 0.408 \ 0.577 \ 0.408 \ 0 \ 0 \end{pmatrix} \nonumber$
Factors to:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \nonumber$
So Bob does this:
$\sigma_x \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \nonumber$
Of course, only 25% of the time is the state of Aliceʹs photon teleported to Bob without further action on his part. Note that we began with Alice and Bob sharing the p Bell state. Consequently when Aliceʹs Bell state measurement yields Φp, Bob has Aliceʹs photon. We now show that teleportation can be presented quite simply using the density matrix formulation under this circumstance.
Bell state measurement result (normalized):
$2 \text{kronecker} ( \Phi_p, \Phi_p^T, I) \Psi = \begin{pmatrix} 0.408 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0.408 \ 0.577 \end{pmatrix} \nonumber$
Density matrix of measurement state:
$\begin{pmatrix} 0.408 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0.408 \ 0.577 \end{pmatrix} \begin{pmatrix} 0.408 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0.408 \ 0.577 \end{pmatrix}^T = \begin{pmatrix} 0.166 & 0.235 & 0 & 0 & 0 & 0 & 0.166 & 0.235 \ 0.235 & 0.333 & 0 & 0 & 0 & 0 & 0.235 & 0.333 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0.166 & 0.235 & 0 & 0 & 0 & 0 & 0.166 & 0.235 \ 0.235 & 0.333 & 0 & 0 & 0 & 0 & 0.235 & 0.333 \end{pmatrix} \nonumber$
Next the density matrix is calculated assuming Bob has the teleported state. The density matrices are identical confirming that Aliceʹs initial state has been successfully teleported to Bob.
$\text{kronecker} \left[ \Phi_p, \Phi_p^T, \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} ^T \right] \begin{pmatrix} 0.166 & 0.235 & 0 & 0 & 0 & 0 & 0.166 & 0.235 \ 0.235 & 0.333 & 0 & 0 & 0 & 0 & 0.235 & 0.333 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0.166 & 0.235 & 0 & 0 & 0 & 0 & 0.166 & 0.235 \ 0.235 & 0.333 & 0 & 0 & 0 & 0 & 0.235 & 0.333 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.04%3A_Another_Look_at_Quantum_Teleportation.txt
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Implementation of the following 8-step circuit using quantum gates teleports cubit |Ψ > from the top wire to the bottom wire. The circuit can be found on page 226 of Julian Brown's The Quest for the Quantum Computer. A final Bell state measurement on the top wires teleports |Ψ > to the bottom wire.
$\begin{matrix} \text{Initial} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & ~ & \text{Final} \ | \Psi \rangle & \cdots & \cdots & \cdot & \fbox{H} & \cdots & \cdots & \cdot & \cdots & \triangleright & \text{Measure,}~0~ \text{or}~1 \ ~ & ~ & ~ & | & ~ & ~ & ~ & | & ~ & ~ & \text{Bell state measurement} \ | 0 \rangle & \fbox{H} & \cdot & \oplus & \cdots & \cdot & \cdots & | & \cdots & \triangleright & \text{Measure,}~0~ \text{or}~1 \ ~ & ~ | ~ & ~ & | & ~ & | \ | 0 \rangle & \cdots & \oplus & \cdots & \cdots & \oplus & \fbox{H} & \oplus & \fbox{H} & \triangleright & | \Psi \rangle \end{matrix} \nonumber$
In the matrix version of quantum mechanics, vectors represent states and matrices represent operators or, in this application, quantum gates. Quantum gates are required to be unitary matrices.
The necessary quantum bits or qubit states are:
Base states:
$0 = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$1 = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
A superposition of base states:
$\begin{pmatrix} \alpha \ \beta \end{pmatrix} = \alpha \begin{pmatrix} 1 \ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
where
$(|a|)^2 + (|b|)^2 = 1 \nonumber$
The identity operator and the following quantum gates are required to calculate the result of the teleportation circuit displayed above.
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Hadamard gate:
$H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
Controlled-NOT gate:
$CNOT = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \nonumber$
Step-7 Controlled-NOT gate:
$\text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \nonumber$
Using $\begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix}$ as the input qubit, the three-qubit initial state is,
$\begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{2}{3}} \ 0 \ 0 \ 0 \ \sqrt{ \frac{1}{3}} \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$\Psi = \begin{pmatrix} \sqrt{ \frac{2}{3}} \ 0 \ 0 \ 0 \ \sqrt{ \frac{1}{3}} \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
The matrix operators required for the steps of the teleportation circuit are now constructed. In the Mathcad programming environment, kronecker is the command for matrix tensor multiplication.
$\text{Step1} = \text{kronecker} (I, \text{kronecker} (H,~I)) \nonumber$
$\text{Step2} = \text{kronecker} (I, CNOT) \nonumber$
$\text{Step3} = \text{kronecker} (CNOT, I) \nonumber$
$\text{Step4} = \text{kronecker} (H, \text{kronecker} (I,~I)) \nonumber$
$\text{Step5} = \text{kronecker} (I, CNOT) \nonumber$
$\text{Step6} = \text{kronecker} (I, \text{kronecker} (I,~H)) \nonumber$
$\text{Step7} = CnNOT \nonumber$
$\text{Step1} = \text{kronecker} (I, \text{kronecker} (I,~H)) \nonumber$
$\text{QuantumCircuit} = \text{Step}8 \text{Step}7 \text{Step}6 \text{Step}5 \text{Step}4 \text{Step}3 \text{Step}2 \text{Step}1 \nonumber$
$\Psi = \text{QuantumCircuit} \Psi \nonumber$
$\Psi^T = \begin{pmatrix} 0.408 & 0.289 & 0.408 & 0.289 & 0.408 & 0.289 & 0.408 & 0.289 \end{pmatrix} \nonumber$
A Bell state measurement on the top wires in the computational basis collapses the wave function and achieves the desired teleportation no matter what the actual measurement results are. There are, of course, four possibilites |0>|0>, |0>|1>, |1>|0> and |1>|1>. All result in |Ψ > on the bottom wire without further action as is now shown. (The calculations are multiplied by a factor 2 to normalize the result.)
Measurement operator for |0 >:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Measurement operator for |1 >:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} ) 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\begin{bmatrix} a' & b' & c' \ \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} , \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} ,~I \right] \right] \Psi' = \begin{pmatrix} 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$\begin{bmatrix} a' & b' & c' \ \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} , \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} ,~I \right] \right] \Psi' = \begin{pmatrix} 0 \ 0 \ 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$\begin{bmatrix} a' & b' & c' \ \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} , \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} ,~I \right] \right] \Psi' = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.816 \ 0.577 \ 0 \ 0 \end{pmatrix} \nonumber$
$\begin{bmatrix} a' & b' & c' \ \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} , \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} ,~I \right] \right] \Psi' = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0.816 \ 0.577 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.05%3A_Teleportation_Using_Quantum_Gates.txt
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In this example of teleportation using quantum gates we will dispense with Alice, Bob and Carol, and talk instead about transferring the cubit |Ψ > in the quantum circuit below from the first wire to the third wire.
$\begin{matrix} \text{Initial} & 1 & 2 & 3 & 4 & ~ & \text{Final} \ | \Psi \rangle & \cdots & \cdots & \cdot & \fbox{H} & \triangleright & \text{Measure}~ |a \rangle ~ 0~ \text{or} ~ 1 \ ~ & ~ & ~ & | & ~ & ~ & \text{Bell}~ \text{state}~ \text{measurement} \ |0 \rangle & \fbox{H} & \cdot & \oplus & \cdots & \triangleright & \text{Measure}~ |b \rangle ~ 0~ \text{or}~ 1 \ ~ & ~ & | \ |0 \rangle & \cdots & \oplus & \cdots & \cdots & \triangleright & X^bZ^a \rightarrow | \Psi \rangle \end{matrix} \nonumber$
The quantum teleportation circuit is adapted from the one shown on page 226 of The Quest for the Quantum Computer by Julian Brown. This tutorial also draws on Brad Rubin's "Quantum Teleportation" at the Wolfram Demonstration Project: http:demonstrations.wolfram.com/QuantumTeleportation/.
In the matrix version of quantum mechanics, vectors represent states and matrices represent operators or, in this application, quantum gates. Quantum gates are required to be unitary matrices.
The necessary quantum bits or qubit states are:
Base states:
$0 = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$1 = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
A superposition of base states:
$\begin{pmatrix} \alpha \ \beta \end{pmatrix} = \alpha \begin{pmatrix} 1 \ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
where
$(|a|)^2 + (|b|)^2 = 1 \nonumber$
The identity operator and the following quantum gates are required to calculate the result of the teleportation circuit displayed above up to the point of the measurements on qubits 1 and 2.
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Hadamard gate:
$H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
Controlled-NOT gate:
$\text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \nonumber$
We begin by using $\begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix}$ as the input qubit on wire 1. This choice means that the three-qubit initial state is,
$\begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{2}{3}} \ 0 \ 0 \ 0 \ \sqrt{ \frac{1}{3}} \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
$\Psi = \begin{pmatrix} \sqrt{ \frac{2}{3}} \ 0 \ 0 \ 0 \ \sqrt{ \frac{1}{3}} \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
The following operations on the initial state yield the state (Ψ') prior to the measurements on qubits 1 and 2. In the Mathcad programming environment, kronecker is the command for matrix tensor multiplication.
$\text{Step1} = \text{kronecker} (I,~ \text{kronecker} (H,~I)) \nonumber$
$\text{Step2} = \text{kronecker} (I,~ CNOT) \nonumber$
$\text{Step13} = \text{kronecker} (CNOT, ~ I) \nonumber$
$\text{Step1} = \text{kronecker} (H,~ \text{kronecker} (I,~I)) \nonumber$
$\Psi = \text{Step}4 \text{Step}3 \text{Step}2 \text{Step}1 \nonumber$
$\begin{pmatrix} 0.408 & 0.289 & 0.289 & 0.408 & 0.408 & -0.289 & -0.289 & 0.408 \end{pmatrix} \nonumber$
There are four possible measurement outcomes on qubits 1 and 2 in the z-basis: |00>, |01>, |10 > and |11 >. The projection operators for |0 > and |1 > are given below.
Projection operator for I0>:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Projection operator for I1>:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
As shown below, depending on the measurement results the following unitary operations are required to complete the transfer of $\begin{pmatrix} 0.816 \ 0.577 \end{pmatrix}$ to the third wire: I, σx, σz, and σzσx. The needed Pauli operators are:
$\sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
$\sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \nonumber$
$\begin{matrix} \text{Measurement result for qubits 1 and 2} & \text{Final 3-qubit state} & \text{Required operation} \ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~ I \right] \right] \Psi' = \begin{pmatrix} 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} & I \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~ I \right] \right] \Psi' = \begin{pmatrix} 0 \ 0 \ 0.577 \ 0.816 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} & \sigma_x \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~ I \right] \right] \Psi' = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.816 \ -0.577 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0.816 \ -0.577 \end{pmatrix} & \sigma_z \begin{pmatrix} 0.816 \ -0.577 \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~ I \right] \right] \Psi' = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.577 \ 0.816 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} -0.577 \ 0.816 \end{pmatrix} & \sigma_z \sigma_x \begin{pmatrix} -0.577 \ 0.816 \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{matrix} \nonumber$
In the interest of relating this example to other examples of quantum teleportation, it is pointed out that the first two steps in the diagram above create an entangled Bell state involving qubits 2 and 3.
$\text{CNOT kronecker}(H,~I) \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.06%3A_Another_Example_of_Teleportation_Using_Quantum_Gates.txt
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This quantum teleportation circuit transfers Ψ from the top wire on the left to the lower wire on the right without the requirement of measurements on the top two wires, which are occupied by identical superposition states.
$\begin{matrix} \text{Initial} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \text{Final} \ | \Psi \rangle & \cdots & \cdots & \cdot & \text{H} & \cdots & \cdots & x & \cdots & \cdots & \frac{1}{ \sqrt{2}} (| 0 \rangle + |1 \rangle ) \ ~ & ~ & ~ & | & ~ & ~ & ~ & | \ |0 \rangle & H & \cdot & \oplus & \cdots & \cdot & \cdots & x & \cdot & \cdot & \frac{1}{ \sqrt{2}} (| 0 \rangle + |1 \rangle ) \ ~ & ~ & | & ~ & ~ & | & ~ & ~ & | \ |0 \rangle & \cdots & \oplus & \cdots & \cdots & \oplus & \text{H} & \cdots & \oplus & \text{H} & | \Psi \rangle \end{matrix} \nonumber$
The required matrix operators:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
$\text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \nonumber$
$\text{Swap} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \nonumber$
The initial state:
$\Psi = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\Psi = \begin{pmatrix} \sqrt{ \frac{1}{3}} & 0 & 0 & 0 & \sqrt{ \frac{2}{3}} & 0 & 0 & 0 \end{pmatrix}^T \nonumber$
Because of its length, the quantum circuit is written in steps.
$\text{Step1} = \text{kronecker}(I, \text{kronecker}( \text{H, I})) \nonumber$
$\text{Step2} = \text{kronecker}( \text{I, CNOT}) \nonumber$
$\text{Step3} = \text{kronecker}( \text{CNOT, I}) \nonumber$
$\text{Step4} = \text{kronecker}(H, \text{kronecker}( \text{I, I})) \nonumber$
$\text{Step5} = \text{kronecker}( \text{I, CNOT}) \nonumber$
$\text{Step6} = \text{kronecker}(I, \text{kronecker}( \text{I, H})) \nonumber$
$\text{Step7} = \text{kronecker}( \text{Swap, I}) \nonumber$
$\text{Step8} = \text{kronecker}( \text{I, CNOT}) \nonumber$
$\text{Step9} = \text{kronecker}(I, \text{kronecker}( \text{I, H})) \nonumber$
$QC = \text{Step9Step8Step7Step6Step5Step4Step3Step2Step1} \nonumber$
The quantum circuit operates on the initial state, transfering Ψ from the top wire to the bottom wire.
$QC \Psi = \begin{pmatrix} 0.289 \ 0.408 \ 0.289 \ 0.408 \ 0.289 \ 0.408 \ 0.289 \ 0.408 \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 0.289 \ 0.408 \ 0.289 \ 0.408 \ 0.289 \ 0.408 \ 0.289 \ 0.408 \end{pmatrix} \nonumber$
$\frac{1}{2} \sqrt{ \frac{1}{3}} = 0.289 \nonumber$
$\frac{1}{2} \sqrt{ \frac{2}{3}} = 0.408 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.07%3A_Yet_Another_Quantum_Teleportation_Circuit.txt
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A graphic representation of the teleportation procedure is accompanied by a quantum circuit for its implementation.
$\begin{matrix} ~ & \text{Initial} & ~ & ~ & ~ & ~ & ~ & \text{Final} \ ~ & | \Psi \rangle & \cdots & \cdot & \cdots & \fbox{H} & \triangleright & \text{Measure}~ | \text{a} \rangle ~ 0~ \text{or}~ 1 \ \text{Alice} & ~ & ~ & | & ~ & ~ & ~ & \text{Bell state measurement} \ ~ & \cdot & \cdots & \oplus & \cdots & \cdots & \triangleright & \text{Measure}~ | \text{b} \rangle ~ 0~ \text{or}~ 1 \ ~ & \beta_{00} \ \text{Bob} & \cdot & \cdots & \cdots & \cdots & \cdots & \triangleright & X^b Z^a \rightarrow | \Psi \rangle \end{matrix} \nonumber$
Alice has the photon to be teleported (teleportee) and another photon belonging to an entangled pair that she shares with Bob.
Teleportee photon state:
$\Psi = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
Entangled Bell state photon pair:
$\beta_{00} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
The shared Bell state and the remaining Bell states shown below are indexed in binary notation. The Appendix contains an explanation of the indexing scheme.
$\beta_{01} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$\beta_{10} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
$\beta_{11} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
In the teleportation circuit, the teleportee is initially on the top wire and the Bell state photons are on the two lower wires. Alice controls the top two wires, while Bob has the bottom wire. The initial product state is formed by vector tensor multiplication.
$\Psi_{in} = \Psi \beta_{00} \nonumber$
$\Psi _{in} = \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} & 0 & 0 & \sqrt{ \frac{1}{3}} & \sqrt{ \frac{2}{3}} & 0 & 0 & \sqrt{ \frac{2}{3}} \end{pmatrix}^T \nonumber$
The matrix operators required to run the circuit and carry out the subsequent measurements and transformations are now identified.
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & X = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & Z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Measurement operator for |0 >:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Measurement operator for |1 >:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
The foward slashes shown on the top two wires of the circuit represent measurements done by Alice. The matrix operator representing the quantum circuit prior to Alice's measurements is formed using tensor multiplication.
$QC = \text{kronecker(H, kronecker (I, I)) kronecker(CNOT, I)} \nonumber$
After the controlled-not and Hadamard gates, but before measurement by Alice, the system is in a superposition state involving the Bell state indices on the top two registers. The third register contains a state that can be easily transformed into the teleportee once Alice tells Bob which Bell state she observed.
$QC \Psi_{in} = \begin{pmatrix} 0.289 \ 0.408 \ 0.408 \ 0.289 \ 0.289 \ -0.408 \ -0.408 \ 0.289 \end{pmatrix} \nonumber$
which can be written
$\frac{1}{2} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \nonumber$
If Alice observes β00 Bob does nothing (the identity operation) because he has Ψ on his register. If Alice observes β01 Bob applies the X operator, if she finds β10 he uses the Z operator, and finally if Alice observes β11 Bob applies the X operator followed by the Z operator. Further mathematical detail is provided by showing explicitly the four equally probable measurement outcomes that Alice observes, and Bob's subsequent action on his register.
$\begin{matrix} 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~I \right] \right] QC \Psi_{in} = \begin{pmatrix} 0.577 \ 0.816 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} & I \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~I \right] \right] QC \Psi_{in} = \begin{pmatrix} 0 \ 0 \ 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} & X \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~I \right] \right] QC \Psi_{in} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.816 \ 0.577 \end{pmatrix} & \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} & Z \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~I \right] \right] QC \Psi_{in} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.816 \ 0.577 \end{pmatrix} & \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} & ZX \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} 0.577 \ 0.816 \end{pmatrix} \end{matrix} \nonumber$
The teleportation circuit can also be analyzed algebraically.
$\left( \sqrt{ \frac{1}{3}} \left| 0 \left\rangle + \sqrt{ \frac{2}{3}} \right| 1 \right\rangle \right) \frac{1}{ \sqrt{2}} \left( \left| 00 \left\rangle + \right| 11 \right\rangle \right) = \frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} \left( \left| 000 \left\rangle + \right| 011 \right\rangle \right) + \sqrt{ \frac{2}{3}} \left( \left| 100 \left\rangle + \right| 111 \right\rangle \right) \right] \nonumber$
$\text{CNOT} \otimes \text{I}$
$\frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} | 0 \rangle \left( \left| 00 \left\rangle + \right| 11 \right\rangle \right) + \sqrt{ \frac{2}{3}} | 1 \rangle \left( \left| 10 \left\rangle + \right| 01 \right\rangle \right) \right] \nonumber$
$\text{H} \otimes \text{I} \otimes \text{I}$
$\frac{1}{ \sqrt{2}} \left[ ( \sqrt{ \frac{1}{3}} | 0 \rangle ( | 1 \rangle ) + ( | 00 \rangle + |11 \rangle ) + \sqrt{ \frac{2}{3}} (0 \rangle - |1 \rangle ) ( | 10 \rangle + | 01 \rangle ) \right] \nonumber$
$\downarrow$
$\frac{1}{ \sqrt{2}} \left[ | 00 \rangle \left( \sqrt{ \frac{1}{3}} | 0 \rangle + \sqrt{ \frac{2}{3}} | 1 \rangle \right) + | 01 \rangle \left( \sqrt{ \frac{2}{3}} | 0 \rangle + \sqrt{ \frac{1}{3}} | 1 \rangle \right) + |10 \rangle \left( \sqrt{ \frac{1}{3}} | 0 \rangle - \sqrt{ \frac{2}{3}} | 1 \rangle \right) + |11 \rangle \left( - \sqrt{ \frac{2}{3}} | 0 \rangle + \sqrt{ \frac{1}{3}} | 1 \rangle \right) \right] \nonumber$
$\downarrow$
$\frac{1}{2} \left[ |00 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + |01 \rangle \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + |10 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + | 11 \rangle \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \overset{Action}{\rightarrow} \frac{1}{2} \left[ I \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + X \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + Z \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + ZX \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.08%3A_Quantum_Teleportation_-_Another_Look.txt
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This Mathcad document examines the math involved in a teleportation experiment for undergraduates using IBM's 5-qubit quantum processor (IBM Quantum Experience) posted by S. Fedortchenko at arXiv:1607.02398v1. Except for state preparation it is identical to the following teleportation circuit, which can be found in my "Teleportation: Another Look."
$\begin{matrix} ~ & \text{Initial} & ~ & ~ & ~ & ~ & ~ & \text{Final} \ ~ & | \Psi \rangle & \cdots & \cdot & \cdots & \fbox{H} & \triangleright & \text{Measure}~ | \text{a} \rangle~ \text{0 or 1} \ \text{Alice} & ~ & ~ & | & ~ & ~ & ~ & \text{Bell state measurement} \ ~ & \cdot & \cdots & \oplus & \cdots & \cdots & \triangleright & \text{Measure}~ | \text{b} \rangle~ \text{0 or 1} \ ~ & \beta_{00} \ \text{Bob} & \cdot & \cdots & \cdots & \cdots & \cdots & \triangleright & X^b Z^a \rightarrow | \Psi \rangle \end{matrix} \nonumber$
Fedortchenko's teleportation circuit is shown below.
$\begin{matrix} \text{Initial} & ~ & 1 & 2 & 3 & 4 & 6 & \text{Final} \ |0 \rangle & \triangleright & \fbox{H} & \fbox{T} & \fbox{H} & \fbox{S} & \cdot & \fbox{H} & \triangleright & \text{Measure}~ | \text{a} \rangle~ \text{0 or 1} \ ~&~&~&~&~&~ & | & ~ & ~ & \text{Bell state measurement} \ |0 \rangle & \triangleright & \cdots & \cdots & \cdots & \oplus & \oplus & \cdots & \triangleright & \text{Measure}~ | \text{b} \rangle~ \text{0 or 1} \ ~&~&~&~&~ & | \ |0 \rangle & \triangleright & \cdots & \cdots & \fbox{H} & \cdot & \cdots & \cdots & \triangleright & X^b Z^a \rightarrow | \Psi \rangle \end{matrix} \nonumber$
Single qubit operators:
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & S = \begin{pmatrix} 1 & 0 \ 0 & i \end{pmatrix} \ T = \begin{pmatrix} 1& 0 \ 0 & e^{ i \frac{ \pi}{4}} \end{pmatrix} & X = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & Z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
Two qubit operators:
$\begin{matrix} \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \text{ICNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
Demonstrate the generation of the teleportee state held by Alice:
$\text{SHTH} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
$e^{i \frac{ \pi}{8}} \left[ \cos \left( \frac{ \pi}{8} \right) \begin{pmatrix} 1 \ 0 \end{pmatrix} + \sin \left( \frac{ \pi}{8} \right) \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
Demonstrate the creation of the entangled Bell state shared by Alice and Bob:
$\begin{matrix} \text{ICNOT (kronecker(I, H))} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} & \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \end{matrix} \nonumber$
Creation of the teleportee state and the entangled Bell state occurs in the first four steps.
$\text{StatePrep = kronecker (S, ICNOT) kronecker (H, kronecker(I, H)) kronecker (T, kronecker(I, I)) kronecker (H kronecker (I, I))} \nonumber$
Teleportation occurs in steps 5 and 6.
$text{TC = kronecker (H, kronecker (I, I)) kronecker (CNOT, I)} \nonumber$
$\text{TC StatePrep} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.427 + 0.177i \ 0.177 + 0.073i \ 0.177 + 0.073i \ 0.427 + 0.177i \ 0.427 + 0.177i \ -0.177 - 0.073i \ -0.177 - 0.073i \ 0.427 + 0.177i \end{pmatrix} \nonumber$
Measurement occurs in the final step on the top two wires with possible outcomes |00>, |01>, |10> and |11>. In other words, Alice makes a Bell state measurement (see first figure above) on the two qubits in her possession and informs Bob of the result through a classical channel. He then performs an operation on his qubit to recover the teleported state.
If Alice observes |00> Bob does nothing (the identity operation) because he has the teleportee on his register. If Alice observes |01> Bob applies the X operator, if she finds |10> he uses the Z operator, and finally if Alice observes |11> Bob applies the X operator followed by the Z operator. Further mathematical detail is provided by showing explicitly the four equally probable measurement outcomes that Alice observes, and Bob's subsequent action on his register.
Computational Details
Measurement operator for |0>:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Measurement operator for |1>:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
Alice measures |00>:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
Bob's action:
$I \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{I} \right] \right] \text{TC StatePrep} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
Alice measures |01>:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0.354 + 0.146i \ 0.854 + 0.354i \end{pmatrix} \nonumber$
Bob's action:
$X \begin{pmatrix} 0.354 + 0.146i \ 0.854 + 0.354i \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{I} \right] \right] \text{TC StatePrep} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0.354 + 0.146i \ 0.854 + 0.354i \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
Alice measures |10>:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ -0.354 - 0.146i \end{pmatrix} \nonumber$
Bob's action:
$Z \begin{pmatrix} 0.854 + 0.354i \ -0.354 - 0.146i \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{I} \right] \right] \text{TC StatePrep} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.854 + 0.354i \ -0.354 - 0.146i \ 0 \ 0 \end{pmatrix} \nonumber$
Alice measures |11>:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} -0.354 - 0.146i \ 0.854 + 0.354i \end{pmatrix} \nonumber$
Bob's action:
$Z X \begin{pmatrix} 0.854 + 0.354i \ -0.354 - 0.146i \end{pmatrix} = \begin{pmatrix} 0.854 + 0.354i \ 0.354 + 0.146i \end{pmatrix} \nonumber$
$2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{I} \right] \right] \text{TC StatePrep} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.354 - 0.146i \ 0.854 + 0.354i \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.09%3A_A_Quantum_Teleportation_Experiment_for_Undergraduates.txt
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This circuit teleports |Ψ> from the top wire to the bottom wire. Measurement on the top wire yields |0> or |1>, the power to which the Z matrix on the bottom wire is raised. As shown below Z0 is the identity matrix, in other words do nothing because |Ψ> is already on the bottom wire.
$\begin{matrix} | \Psi \rangle & \triangleright & \cdot & H & \triangleright & \text{Measure m = 0 or 1} \ |0 \rangle & \triangleright & \oplus & ~ & \triangleright & Z^m \rightarrow | \Psi \rangle \end{matrix} \nonumber$
$\begin{matrix} |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} & | \Psi \rangle = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \end{matrix} \nonumber$
The required quantum gates in matrix format:
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & Z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & Z^0 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Form the quantum circuit for teleportation:
$\begin{matrix} \text{QC = kronecker(H, I) CNOT)} & \text{QC} = \begin{pmatrix} 0.707 & 0 & 0 & 0.707 \ 0 & 0.707 & 0.707 & 0 \ 0.707 & 0 & 0 & -0.707 \ 0 & 0.707 & -0.707 & 0 \end{pmatrix} & \text{QC} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \ 1 & 0 & 0 & -1 \ 0 & 1 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Write the initial state as a 4-vector using tensor multiplication.
$\begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ 0 \ \sqrt{ \frac{2}{3}} \ 0 \end{pmatrix} \nonumber$
Calculate the output state of the circuit and write it as a superposition of |0> and |1> on the top wire.
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \ 1 & 0 & 0 & -1 \ 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ 0 \ \sqrt{ \frac{2}{3}} \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{2}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{2}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber$
It is clear that if |0> is measured on the top wire, |Ψ> is on the bottom wire without further action. If |1> is measured on the top wire the qubit on the bottom wire is converted to |Ψ> by multiplication by Z.
$\begin{matrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} & ~ & ~ & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \end{matrix} \nonumber$
Matrix summary:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \ 1 & 0 & 0 & -1 \ 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ 0 \ \sqrt{ \frac{2}{3}} \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber$
Algebraic summary:
$\begin{matrix} Z = \begin{pmatrix} 0 & \text{to} & 0 \ 1 & \text{to} & -1 \end{pmatrix} & H = \begin{bmatrix} 0 & \text{to} & \frac{(0 + 1)}{ \sqrt{2}} \ 1 & \text{to} & \frac{(0-1)}{ \sqrt{2}} \end{bmatrix} & \text{CNOT} = \begin{pmatrix} 00 & \text{to} & 00 \ 01 & \text{to} & 01 \ 10 & \text{to} & 11 \ 11 & \text{to} & 10 \end{pmatrix} \end{matrix} \nonumber$
$\left( \sqrt{ \frac{1}{3}} \left| 0 \left\rangle + \sqrt{ \frac{2}{3}} \right|1 \right\rangle \right) |0 \rangle = \sqrt{ \frac{1}{3}} \left| 00 \right\rangle + \sqrt{ \frac{2}{3}} \left|10 \right\rangle \nonumber$
$\text{CNOT}$
$\sqrt{ \frac{1}{3}} \left| 00 \right\rangle + \sqrt{ \frac{2}{3}} \left|11 \right\rangle \nonumber$
$\text{H} \otimes \text{I}$
$\frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} \left( |0 \rangle + |1 \rangle \right) |0 \rangle + \sqrt{ \frac{2}{3}} \left( | 0 \rangle - |1 \rangle \right) |1 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ |0 \rangle \left( \sqrt{ \frac{1}{3}} | 0 \rangle + \sqrt{ \frac{2}{3}} |1 \rangle \right) | 1 \rangle + \left( \sqrt{ \frac{1}{3}} |0 \rangle - \sqrt{ \frac{2}{3}} |1 \rangle \right) \right] \nonumber$
$\downarrow$
$\frac{1}{ \sqrt{2}} \left[ |0 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + |1 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \xrightarrow{Action} \frac{1}{ \sqrt{2}} \left[ I \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + Z \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber$
Total matrix operator approach:
m = 0
$\text{kronecker}(I,~ Z^0 ) \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~I \right] \text{kronecker} (H,~I) CNOT \begin{pmatrix} \sqrt{ \frac{1}{3}} \ 0 \ \sqrt{ \frac{2}{3}} \ 0 \end{pmatrix} = \begin{pmatrix} 0.408 \ 0.577 \ 0 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
m = 1
$\text{kronecker}(I,~ Z^1 ) \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~I \right] \text{kronecker} (H,~I) CNOT \begin{pmatrix} \sqrt{ \frac{1}{3}} \ 0 \ \sqrt{ \frac{2}{3}} \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0.408 \ 0.577 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
The measurement projection operators used above in the next to the last step are:
m = 0
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
m = 1
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.10%3A_A_Simple_Teleportation_Exercise.txt
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This tutorial works through the following teleportation circuit provided by Gilles Brassard in "Teleportation as a Quantum Computation" (arXiv:quant-ph/9605035v1). The computational methodology employed here is similar to that used in the other teleportation examples given in this series of tutorials.
$\begin{matrix} \text{Initial} & ~ & 1 & 2 & 3 & 4 & ~ & 5 & ~ & 6 & 7 & ~ & \text{Final} \ | \Psi \rangle & \triangleright & \cdots & \cdots & \cdot & \fbox{R} & \cdots & \fbox{S} & \oplus & \fbox{S} & \oplus & \triangleright & \frac{1}{ \sqrt{2}} ( | 0 \rangle + |1 \rangle ) \ ~ & ~ & ~ & ~ & | ~ & ~ & ~ & ~ & | & ~ & | \ | 0 \rangle & \triangleright & \fbox{L} & \cdot & \oplus & \cdots & \cdots & \cdots & | & \cdots & | & \triangleright & \frac{1}{ \sqrt{2}} (|0 \rangle + |1 \rangle ) \ ~ & ~ & ~ & | & ~ & ~ & ~ & | & | & ~ & | \ | 0 \rangle & \triangleright & \cdots & \oplus & \cdots & \cdots & \cdots & \oplus & \cdot & \fbox{T} & \cdot & \triangleright & | \Psi \rangle \end{matrix} \nonumber$
The necessary quantum bits or qubit states are:
Base states:
$\begin{matrix} 0 = \begin{pmatrix} 1 \ 0 \end{pmatrix} & & 1 = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Superposition of base states:
$\begin{pmatrix} \alpha \ \beta \end{pmatrix} = \alpha \begin{pmatrix} 1 \ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
where
$(| \alpha |)^2 + (| \beta |)^2 = 1 \nonumber$
Using $\begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix}$ as the teleportee (|Ψ>), Brassard's circuit yields the final states shown in the circuit above. In other words |Ψ > is teleported from the first wire on the left to the third wire on the right.
Initial state:
$\begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{2}{3}} \ 0 \ 0 \ 0 \ \sqrt{ \frac{1}{3}} \ 0 \ 0 \ 0 \end{pmatrix} \nonumber$
Final state:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} = \begin{pmatrix} \sqrt{ \frac{1}{6}} \ \sqrt{ \frac{1}{12}} \ \sqrt{ \frac{1}{6}} \ \sqrt{ \frac{1}{12}} \ \sqrt{ \frac{1}{6}} \ \sqrt{ \frac{1}{12}} \ \sqrt{ \frac{1}{6}} \ \sqrt{ \frac{1}{12}} \end{pmatrix} \nonumber$
The identity operator and the following quantum gates are required to calculate the result of the teleportation circuit. L and R are single qubit rotations, S and T are single qubit phase shifts. The other gates (CNOT, CnNOT, and ICnNOT) are well-known in quantum circuitry.
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & L = \begin{pmatrix} 1 & -1 \ 1 & 1 \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ -1 & 1 \end{pmatrix} & S = \begin{pmatrix}i & 0 \ 0 & 1 \end{pmatrix} & T = \begin{pmatrix} -1 & 0 \ 0 & -i \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} & \text{ICnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{Step1 = kronecker(I, kronecker(L, I))} & \text{Step2 = kronecker(I, CNOT)} & \text{Step3 = kronecker(CNOT, I)} \ \text{Step4 = kronecker(R, kronecker(I, I))} & \text{Step5 = kronecker(S, CNOT)} & \text{Step6 = ICnNOT} \ \text{Step7 = kronecker(S, kronecker(I, T))} &\text{Step8 = ICnNOT} \end{matrix} \nonumber$
According to this result |Ψ > has indeed been teleported to the bottom wire on the right, so the goal has been achieved. However, Brassard suggests that measurements on the top wires in the |0 >-|1 > basis are also instructive. There are four possible measurement outcomes: |00>, |01>, |10 > and |11 >. The projection operators for |0 > and |1 > are as follows.
Projection operator for |0>:
$\begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
Projection operator for |1>:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
These calculations also show successful teleportation.
$\begin{matrix} \text{Measurement result for qubits x and y} & \text{Final 3-qubit state} \ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{bmatrix} x & y & z \ \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0 \ 0 \ 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{bmatrix} x & y & z \ \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.816 \ 0.577 \ 0 \ 0 \end{pmatrix} & \begin{bmatrix} x & y & z \ \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \ 2 \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix}, I \right] \right] \Psi_{final} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0.816 \ 0.577 \end{pmatrix} & \begin{bmatrix} x & y & z \ \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0.816 \ 0.577 \end{pmatrix} \end{bmatrix} \end{matrix} \nonumber$
Appendix
In the inverse controlled-NOT gate of steps 6 and 8 of Brassard's teleportation circuit, c is the control, a is the target and b is unchanged.
$\begin{pmatrix} a & b & c & ' & a' & b' & c' \ 0 & 0 & 0 & ' & 0 & 0 & 0 \ 0 & 0 & 1 & ' & 1 & 0 & 1 \ 0 & 1 & 0 & ' & 0 & 1 & 0 \ 0 & 1 & 1 & ' & 1 & 1 & 1 \ 1 & 0 & 0 & ' & 1 & 0 & 0 \ 1 & 0 & 1 & ' & 0 & 0 & 1 \ 1 & 1 & 0 & ' & 1 & 1 & 0 \ 1 & 1 & 1 & ' & 0 & 1 & 1 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.11%3A_Teleportation_as_a_Quantum_Computation.txt
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The science fiction dream of "beaming" objects from place to place is now a reality - at least for particles of light. Anton Zeilinger, Scientific American, April 2000, page 50.
Quantum teleportation is a way of transferring the state of one particle to a second, effectively teleporting the initial particle. (Tony Sudbery, Nature, December 11, 1997, page 551.)
As shown in the graphic below, quantum teleportation is a form of information transfer that requires pre-existing entanglement and a classical communication channel to send information from one location to another.
Alice has the photon to be teleported and a photon of an entangled pair that she shares with Bob. She performs a measurement on her photons that projects them into one of the four Bell states and Bob's photon, via the entangled quantum channel, into a state that has a unique relationship to the state of the teleportee. Bob carries out one of four unitary operations on his photon depending on the results of Alice's measurement, which she sends him through a classical communication channel.
The figure (Nature, December 11, 1997, page 576) on the left provides a graphic summary of the first successful teleportation experiment. The quantum circuit on the right shows a method of implementation.
$\begin{matrix} ~ & \text{Initial} ~ & ~ & ~ & ~ \text{Final} \ ~ & | \Psi \rangle & \cdot & \cdots & \fbox{H} & \triangleright & \text{Measure}~ | \text{a} \rangle ~ \text{0 or 1} \ \text{Alice} & ~ & | ~ & ~ & ~ & ~ & \text{Bell state measurement} \ ~ & \cdot & \oplus & \cdots & \cdots & \triangleright & \text{Measure}~ | \text{b} \rangle ~ \text{0 or 1} \ ~ & \beta_{00} \ \text{Bob} & \cdot & \cdots & \cdots & \cdots & \triangleright & X^b Z^a \rightarrow | \Psi \rangle \end{matrix} \nonumber$
The teleportee and the Bell states indexed in binary notation:
Teleportee:
$\begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
Bell states:
$\begin{matrix} \beta_{00} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} & \beta_{01} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} & \beta_{10} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & \beta_{11} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
Truth tables and matrices representing the various teleportation circuit operations:
$\begin{matrix} |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} & X = \begin{pmatrix} 0~ \text{to} ~1 \ 1~ \text{to} ~ -1 \end{pmatrix} & H = \begin{bmatrix} 0~ \text{to}~ \frac{(0 + 1)}{ \sqrt{2}} \ 1~ \text{to}~ \frac{(0 - 1)}{ \sqrt{2}} \end{bmatrix} & \text{CNOT} = \begin{pmatrix} 00~ \text{to} ~00 \ 01~ \text{to} ~01 \ 10~ \text{to} ~11 \ 11~ \text{to} ~10 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & X = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & Z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Perspective I. Using the truth tables, the operation of the teleportation circuit is expressed in Dirac notation.
$\left( \sqrt{ \frac{1}{3}} |0 \rangle + \sqrt{ \frac{2}{3}} |1 \rangle \right) \frac{1}{ \sqrt{2}} \left( |00 \rangle + |11 \rangle \right) = \frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} \left( |00 \rangle |0 \rangle + |01 \rangle |1 \rangle \right) + \sqrt{ \frac{2}{3}} \left( |10 \rangle |0 \rangle + |11 \rangle |1 \rangle \right) \right] \nonumber$
$\text{CNOT} \otimes I$
$\frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} (|00 \rangle |0 \rangle + |01 \rangle |1 \rangle ) + \sqrt{ \frac{2}{3}} (|11 \rangle |0 \rangle + |10 \rangle |1 \rangle ) \right] = \frac{1}{ \sqrt{2}} \left[ \sqrt{ \frac{1}{3}} |0 \rangle (|00 \rangle + |11 \rangle ) + \sqrt{ \frac{2}{3}} |1 \rangle ( |10 \rangle + |01 \rangle ) \right] \nonumber$
$H \otimes I \otimes I$
$\frac{1}{2} \left[ \sqrt{ \frac{1}{3}} (|0 \rangle + |1 \rangle ) (|00 \rangle + | 11 \rangle ) + \sqrt{ \frac{2}{3}} (|0 \rangle - |1 \rangle) (|10 \rangle + |01 \rangle ) \right] \nonumber$
$\downarrow$
$\frac{1}{2} \left[ |00 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + |01 \rangle \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + |10 \rangle \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + |11 \rangle \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \xrightarrow{Action} \frac{1}{2} \left[ I \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + X \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + Z \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + Z X \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \nonumber$
Alice's Bell state measurement result (|00>, |01>. |10> or |11>, see indexed Bell states above) determines the operation (I, X, Z or ZX) that Bob performs on his photon.
Perspective II. The three-cubit initial state is re-written as a linear superposition of the four possible Bell states that Alice can find on measurement. Note that this is a equivalent to the expression on the left side of the equation immediately above if the Bell states are replaced by their binary indices, as they would be after the Bell state measurement.
$\Psi \rangle = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \otimes \sqrt{ \frac{1}{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} = \sqrt{ \frac{1}{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ 0 \ 0 \ \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \ 0 \ 0 \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
$= \frac{1}{2} \left[ \sqrt{ \frac{1}{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \sqrt{ \frac{1}{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + \sqrt{ \frac{1}{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + \sqrt{ \frac{1}{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \nonumber$
Condensed version of the equation:
$| \Psi \rangle = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \otimes \beta_{00} = \frac{1}{2} \left[ \beta_{00} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{01} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + \beta_{10} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{11} \otimes \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \nonumber$
Another way to write the equation:
$| \Psi \rangle = \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \otimes \beta_{00} = \frac{1}{2} \left[ \beta_{00} \otimes I \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{01} \otimes X \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{10} \otimes Z \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \beta_{11} \otimes XZ \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \right] \nonumber$
Perspective III. The teleportation circuit (TC) is written as a composite matrix operator which then operates on the initial three-qubit state.
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} & 0 & 0 & \sqrt{ \frac{1}{3}} \sqrt{ \frac{2}{3}} & 0 & 0 & \sqrt{ \frac{2}{3}} \end{pmatrix}^2 & & \text{TC = kronecker(H, kronecker(I, I)) kronecker(CNOT, I)} \end{matrix} \nonumber$
After operation of the circuit the system is in a superposition state involving the Bell state indices on the top two registers. The third register contains a state that can easily be transformed into the teleportee once Alice tells Bob which Bell state she observed.
$\text{TC} \Psi = \begin{pmatrix} 0.289 \ 0.408 \ 0.408 \ 0.289 \ 0.289 \ -0.408 \ -0.408 \ 0.289 \end{pmatrix} \begin{array}{l} = \frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \ = \frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sqrt{ \frac{1}{3}} \ -\sqrt{ \frac{2}{3}} \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix} \right] \end{array} \nonumber$
Tabular summary of teleportation experiment:
$\begin{pmatrix} \text{Alice Measurement Result} & \beta_{00} & \beta_{01} & \beta{10} & \beta_{11} \ \text{Bob's Action} & I & X & Z & ZX \end{pmatrix} \nonumber$
Bell state indices:
$\begin{matrix} \text{kronecker (H, I) CNOT} \beta_{00} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{kronecker (H, I) CNOT} \beta_{01} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \ \text{kronecker (H, I) CNOT} \beta_{10} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{kronecker (H, I) CNOT} \beta_{11} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
A similar approach is to use projection operators on the top two qubits to simulate the four measurement outcomes. See the Appendix for more on this method.
$\begin{matrix} ~& ~ & \text{Bob's Action} \ \text{Measure} |00 \rangle & 2 \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0.577 \ 0.816 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{No action required} \ \text{Measure} |01 \rangle & 2 \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0 \ 0 \ 0.816 \ 0.577 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{Operate with X} \ \text{Measure} |10 \rangle & 2 \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.577 \ -0.816 \ 0 \ 0 \end{pmatrix} & \text{Operate with Z} \ \text{Measure} |11 \rangle & 2 \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.816 \ 0.577 \end{pmatrix} & \text{Operate with ZX} \end{matrix} \nonumber$
Perspective IV. Projecting the teleportee photon 1 (green) onto the result of Alice's Bell state measurement (blue) yields the state of photon 2 which was initially entangled with Bob's photon 3. Projection of this state onto the original 2-3 entangled state (red) transforms Bob's photon to the teleportee state 25% of the time. As is now shown this happens when Alice's Bell state measurment yields β00.
$_{12} \langle \beta_{00} | \Psi \rangle_1 | \beta_{00} \rangle_{23} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_1 ^T \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} _2 ^T + \begin{pmatrix} 0 \ 1 \end{pmatrix} _1 ^T \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} _2 ^T\right] \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix}_1 \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_2 \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_3 + \begin{pmatrix} 0 \ 1 \end{pmatrix}_2 \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_3 \right]= \frac{1}{2} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix}_3 \nonumber$
An algebraic analysis of this example is as follows.
$\text{Teleportee}$
$\frac{1}{ \sqrt{2}} \left[ _1 \langle 0|_2 \langle 0| +~ _1 \langle 1|_2 \langle 1| \right] \left[ \alpha |0 \rangle_1 + \beta |1 \rangle_1 \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle_2 |0 \rangle_3 + |1 \rangle_2 |1 \rangle_3 \right] \nonumber$
$\downarrow$
$\frac{1}{ \sqrt{2}} \left[ \alpha _2 \langle 0| + \beta _2 \langle 1| \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle_2 |0 \rangle_3 + |1 \rangle_2 |1 \rangle_3 \right] \nonumber$
$\downarrow$
$\frac{1}{2} \left[ \alpha |0 \rangle_3 + \beta |1 \rangle_3 \right] \nonumber$
Naturally this approach yields the same results as the previous perspectives when Alice's Bell state measurement is β01, β10 and β11. As demonstrated previously for these results Bob's action is X, Z and ZX, respectively.
$_{12} \langle \beta_{01} | \Psi \rangle_1 | \beta_{00} \rangle_{23} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_1 ^T \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} _2 ^T + \begin{pmatrix} 0 \ 1 \end{pmatrix} _1 ^T \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} _2 ^T\right] \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix}_1 \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_2 \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_3 + \begin{pmatrix} 0 \ 1 \end{pmatrix}_2 \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_3 \right]= \frac{1}{2} \begin{pmatrix} \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix}_3 \nonumber$
$_{12} \langle \beta_{10} | \Psi \rangle_1 | \beta_{00} \rangle_{23} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_1 ^T \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} _2 ^T - \begin{pmatrix} 0 \ 1 \end{pmatrix} _1 ^T \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} _2 ^T\right] \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix}_1 \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_2 \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_3 + \begin{pmatrix} 0 \ 1 \end{pmatrix}_2 \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_3 \right]= \frac{1}{2} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ - \sqrt{ \frac{2}{3}} \end{pmatrix}_3 \nonumber$
$_{12} \langle \beta_{11} | \Psi \rangle_1 | \beta_{00} \rangle_{23} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_1 ^T \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} _2 ^T - \begin{pmatrix} 0 \ 1 \end{pmatrix} _1 ^T \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} _2 ^T\right] \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix}_1 \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_2 \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_3 + \begin{pmatrix} 0 \ 1 \end{pmatrix}_2 \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_3 \right]= \frac{1}{2} \begin{pmatrix} - \sqrt{ \frac{2}{3}} \ \sqrt{ \frac{1}{3}} \end{pmatrix}_3 \nonumber$
Summary of the quantum teleportation protocol: "Quantum teleportation provides a 'disembodied' way to tranfer quantum states from one object to another at a distant location, assisted by previously shared entangled states and a classical communication channel." Nature 518, 516 (2015)
The paper cited above reported the first successful teleportation of two degrees of freedom of a single photon. The analysis is somewhat more complicated than that provided in this tutorial, but the general principle is the same. The quantum channel is a hyper-entangled state shared by Alice and Bob, rather than one of the simple entangled Bell states.
Appendix
Addendum to Perspective III. In these calculations the required operations by Bob are actually carried out on the third qubit.
$\begin{matrix} \text{Measure} |00 \rangle ~ \text{do nothing} & \text{kronecker(I, kronecker(I, I)) 2 kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0.577 \ 0.816 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \ \text{Measure} |01 \rangle ~ \text{operate with X} & \text{kronecker(I, kronecker(I, X)) 2 kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0 \ 0 \ 0.577 \ 0.816 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \ \text{Measure} |10 \rangle ~ \text{operate with Z} & \text{kronecker(I, kronecker(I, Z)) 2 kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.577 \ 0.816 \ 0 \ 0 \end{pmatrix} \ \text{Measure} |11 \rangle ~ \text{operate with Z} & \text{kronecker(I, kronecker(I, Z, X)) 2 kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} ^T, I \right] \right] \text{TC} \Psi = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0.577 \ 0.816 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.12%3A_Quantum_Teleportation_-_Four_Perspectives.txt
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The following quantum circuit teleports |a>|b> from the top two wires on the left to the bottom two wires on the right.
$\begin{matrix} \text{Input} & 1 & 2 & 3 & 4 & 5 & 6 & ~ & \text{Output} \ | a \rangle & \cdots & \cdots & \cdots & \cdot & \fbox{H} & \cdots & \triangleright & \text{Measure, 0 or 1} \ | b \rangle & \cdots & \cdots & \cdots & | & \cdot & \fbox{H} & \triangleright & \text{Measure, 0 or 1} \ |0 \rangle & \fbox{H} & \cdot & \cdots & \oplus & | & \cdots & \triangleright & \text{Measure, 0 or 1} \ |0 \rangle & \fbox{H} & | & \cdot & \cdots & \oplus & \cdots & \triangleright & \text{Measure, 0 or 1} \ |0 \rangle & \cdots & \oplus & | & \cdots & \cdots & \cdots & ~ & |a \rangle \ |0 \rangle & \cdots & \cdots & \oplus & \cdots & \cdots & \cdots & ~ & |b \rangle \end{matrix} \nonumber$
There are 16 possible measurement results on the top four wires on the right ranging from |0> to |15> (|000000> to |111111> in binary notation), and these results determine the action required to transform the bottom two wires to |a>|b>.
As an initial example, if the teleportee in decimal notation is |3> the total input state is |48>.
Teleportee:
$\begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} = |3 \rangle \nonumber$
Total input state:
$\Psi = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = | 48 \rangle \nonumber$
Ψ48 is the only non-zero element in the input state vector.
$\begin{matrix} i = 0 .. 63 & \Psi_i = 0 & \Psi_48 = 1 \end{matrix} \nonumber$
$\begin{array}{| c | c | c | c | c | c | c | c | c | c |} ~ & 44 & 45 & 46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 \ \hline 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots \end{array} \nonumber$
With the assistance of truth tables provided in the Appendix and a bit of tedious algebra the operation of the teleportation circuit on this input state can be summarized as follows.
Given $| \Psi \rangle_{input} = |1 \rangle |1 \rangle |0 \rangle |0 \rangle |0 \rangle |0 \rangle = |48 \rangle$ the teleportation circuit yields
$| \Psi \rangle_{output} = \frac{1}{4} \begin{bmatrix} (|3 \rangle - |7 \rangle - |11 \rangle + |15 \rangle ) |00 \rangle + (|2 \rangle - |6 \rangle - |10 \rangle + |14 \rangle ) |01 \rangle \ + ( |1 \rangle - |5 \rangle - |9 \rangle + |13 \rangle ) |10 \rangle + (|0 \rangle - |4 \rangle - |8 \rangle + |12 \rangle ) |11 \rangle \end{bmatrix} \nonumber$
The output state shows that if the measurement result is 3, 7, 11 or 15, the NOT operator (|0> to |1> and |1> to |0>) should be applied to wires 5 and 6. If the result is 2, 6, 10 or 14, the NOT operator is applied to wire 5; if the result 1, 5, 9 or 13, the NOT operator is applied to wire 6; if the result is 0, 4, 8 or 12, no action is required.
Now an example teleportation calculation will be carried out on the teleportee state given above. The following matrix operators are required for the teleportation circuit. They are the identity, the Hadamard gate, the measurement operators for |0> and |1>, and the CnNOT gate.
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & M_0 = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} & M_1 = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} & \text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
The matrix operators for the six steps in the circuit are formed using tensor (Kronecker) multiplication.
$\begin{matrix} \text{Step1} = \text{kronecker(I, kronecker(I, kronecker(H, kronecker(H, kronecker(I, I))))} \ \text{Step2} = \text{kronecker(I, kronecker(I, kronecker(CnNOT, I)))} & \text{Step3} = \text{kronecker(I, kronecker(I, kronecker(I, CnNOT)))} \ \text{Step4} = \text{kronecker(CnNOT, kronecker(I, kronecker(I, I)))} & \text{Step5} = \text{kronecker(H, kronecker(CnNOT, kronecker(I, I)))} \ \text{Step6} = \text{kronecker(I, kronecker(H, kronecker(I, kronecker(I, kronecker(I, I)))))} \ \text{Step7} = \text{kronecker} (M_0, \text{kronecker}(M_0, \text{kronecker}(M_0, \text{kronecker}(M_0, \text{kronecker}(I,~I))))) \end{matrix} \nonumber$
The output vector is calculated for the simple case of measuring |0>|0>|0>|0> on the top four wires. The factor of 4 takes into account that measuring |0>|0>|0>|0> has a probability amplitude of 1/4. It therefore normalizes the calculation.
$\begin{matrix} \Psi_{out} = 4 \text{Step7 Step6 Step 5 Step4 Step3 Step2 Step1} \Psi \ \Psi_{out}^T = \begin{array}{| c | c | c | c | c | c | c | c | c | c |} ~ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \ \hline 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \end{array} \ \Psi_{out} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = |3 \rangle \end{matrix} \nonumber$
In a second example a superposition of |0 > + |1 > + |2 > + |3> is teleported.
Teleportee:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} = \frac{|0 \rangle + |1 \rangle + |2 \rangle + 3 \rangle}{2} \nonumber$
Total input vector:
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Non-zero vector elements:
$\begin{matrix} \Psi_0 = \frac{1}{2} & \Psi_{16} = \frac{1}{2} & \Psi_{32} = \frac{1}{2} & \Psi_{48} = \frac{1}{2} \end{matrix} \nonumber$
The output vector is calculated for the case of measuring |0>|0>|0>|0> on the top four wires.
$\begin{matrix} \Psi_{out} = 4 \text{Step7 Step6 Step 5 Step4 Step3 Step2 Step1} \Psi \ \Psi_{out}^T = \begin{array}{| c | c | c | c | c | c | c | c | c | c |} ~ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \ \hline 0 & 0.5 & 0.5 & 0.5 & 1 & 0.5 & 0 & 0 & 0 & 0 & \cdots \end{array} \ \Psi_{out} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{2} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} = \frac{|0 \rangle + |1 \rangle + |2 \rangle + 3 \rangle}{2} \end{matrix} \nonumber$
A final example:
Teleportee:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \nonumber$
Total input vector:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Non-zero vector elements:
$\begin{matrix} \Psi_0 = \sqrt{ \frac{1}{6}} & \Psi_{16} = \sqrt{ \frac{1}{3}} & \Psi_{32} = \sqrt{ \frac{1}{6}} & \Psi_{48} = \sqrt{ \frac{1}{3}} \end{matrix} \nonumber$
$\Psi_{out} = 4 \text{Step7Step6Step5Step4Step3Step2Step1} \Psi \nonumber$
$\Psi_{out}^T = \begin{array}{| c | c | c | c | c | c | c | c | c | c |} ~ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \ \hline 0 & 0.408 & 0.577 & 0.408 & 0.577 & 0 & 0 & 0 & 0 & \cdots \end{array} \nonumber$
$\Psi_{out} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} \sqrt{ \frac{1}{3}} \ \sqrt{ \frac{2}{3}} \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0.408 \ 0.577 \ 0.408 \ 0.577 \end{pmatrix} \nonumber$
Appendix
$\begin{matrix} \text{NOT} = \begin{pmatrix} \text{0 to 1} \ \text{1 to 0} \end{pmatrix} & H = \begin{bmatrix} \text{ 0 to}~ \frac{1}{ \sqrt{2}} (0 + 1) \ \text{1 to}~ \frac{1}{ \sqrt{2}} (0-1) \end{bmatrix} & \text{CnNOT} = \begin{pmatrix} \text{Decimal} & \text{Binary to Binary} & \text{Decimal} \ 0 & 000 ~ \text{to}~ 000 & 0 \ 1 & 001 ~ \text{to}~ 001 & 1 \ 2 & 010 ~ \text{to}~ 010 & 2 \ 3 & 011 ~ \text{to}~ 011 & 3 \ 4 & 011 ~ \text{to}~ 011 & 5 \ 5 & 101 ~ \text{to}~ 100 & 4 \ 6 & 110 ~ \text{to}~ 111 & 7 \ 7 & 111 ~ \text{to}~ 110 & 6 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.13%3A_Teleportation_of_Two_Qubits.txt
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This tutorial summarizes experimental results on GHZ entanglement reported by Anton Zeilinger and collaborators in the 3 February 2000 issue of Nature (pp. 515‐519). The GHZ experiment employs three‐photon entanglement to provide a stunning attack on local realism.
Definitions
First some definitions:
• Realism ‐ experiments yield values for properties that exist independent of experimental observation
• Locality ‐ the experimental results obtained at location A at time t, do not depend on the results at some other location B at time t.
• H/V = horizontal/vertical linear polarization. R/L = right/left circular polarization. Hʹ/Vʹ rotated by 45o with respect to H/V.
Next some relationships between the various photon polarization states: See the appendix for vector definitions of |H>, |V>, |Hʹ>, |Vʹ>, |R> and |L>.
$\begin{matrix} H' = \frac{1}{ \sqrt{2}} (H + V) & V' = \frac{1}{ \sqrt{2}} (H - V) & R = \frac{1}{ \sqrt{2}} (H + iV) & L = \frac{1}{ \sqrt{2}} (H - iV) \end{matrix} \nonumber$
$\begin{matrix} H = \frac{1}{ \sqrt{2}} (H' + V') & V = \frac{1}{ \sqrt{2}} (H' - V') & H = \frac{1}{ \sqrt{2}} (R + L) & V = \frac{i}{ \sqrt{2}} (L - R) \end{matrix} \nonumber$
The initial GHZ three-photon entangled state:
$\Psi = \frac{1}{ \sqrt{2}} (H_1 H_2 H_3 + V_1 V_2 V_3) \nonumber$
After preparation of the initial GHZ state (see figure 1 in the reference cited above), polarization measurements are performed on the three photons. Zeilinger and collaborators use y to stand for a circular polarization measurement and x for a linear polarization measurement. Initially they perform circular polarization measurements on two of the photons and a linear polarization measurement on the other photon. The quantum mechanically predicted results and actual experimental measurements are given below. The quantum predictions are obtained by substituting equations (2) into equation (3).
yyx - experiment
$\Psi_{yyx} = \frac{1}{ \sqrt{2}} \left[ \frac{R_1 + L_1}{ \sqrt{2}} \frac{R_2 + L_2}{ \sqrt{2}} \frac{ H'_3 + V'_3}{ \sqrt{2}} + \frac{ i(L_1 - R_1)}{ \sqrt{2}} \frac{i (L_2 - R_2)}{ \sqrt{2}} \frac{H'_3 - V'_3}{ \sqrt{2}} \right] \nonumber$
on expansion yields
$\Psi_{yyx} = \frac{1}{2} R_1 R_2 V'_3 + \frac{1}{2} R_1 L_2 H'_3 + \frac{1}{2} L_1 R_2 H'_3 + \frac{1}{2} L_1 R_2 H'_3 + \frac{1}{2} L_1 L_2 V'_3 \nonumber$
Each measurement (R/L or Hʹ/Vʹ) has two possible outcomes so, in principle, there could be 8 possible results. However, quantum mechanics predicts that only four equally probable, $\left( \frac{1}{2} \right)^2 = 0.25$, outcomes are possible. The quantum mechanical prediction is in agreement with the experimental results shown in Figure 1 to within experimental error.
For the two remaining experiments in this class (yxy and xyy), the agreement between theoretical prediction and experimental results is basically the same. While these agreements between quantum mechanics and experiment are impressive they do not directly challenge the local realist position. As will be shown later that will be accomplished by a fourth experiment involving the measurement of the linear polarization on all three photons ‐ the xxx experiment.
yxy ‐ experiment
$\Psi_{yxy} = \frac{1}{ \sqrt{2}} \left[ \frac{R_1 + L_1}{ \sqrt{2}} \frac{H'_2 + V'_2}{ \sqrt{2}} \frac{ R_3 + L_3}{ \sqrt{2}} + \frac{ i(L_1 - R_1)}{ \sqrt{2}} \frac{i (H'_2 - V'_2)}{ \sqrt{2}} \frac{L_3 - R_3}{ \sqrt{2}} \right] \nonumber$
on expansion yields
$\Psi_{yxy} = \frac{1}{2} R_1 H'_2 L_3 + \frac{1}{2} R_1 V'_2 R_3 + \frac{1}{2} L_1 H'_2 R_3 + \frac{1}{2} L_1 V'_2 L_3 \nonumber$
xyy - experiment
$\Psi_{xyy} = \frac{1}{ \sqrt{2}} \left[ \frac{H'_1 + V'_1}{ \sqrt{2}} \frac{R_2 + L_2}{ \sqrt{2}} \frac{ R_3 + L_3}{ \sqrt{2}} + \frac{ H'_1 - V'_1)}{ \sqrt{2}} \frac{i (L_2 - R_2)}{ \sqrt{2}} \frac{i(L_3 - R_3)}{ \sqrt{2}} \right] \nonumber$
on expansion yields
$\Psi_{xyy} = \frac{1}{2} H'_1 R_2 L_3 + \frac{1}{2} H'_1 L_2 R_3 + \frac{1}{2} V'_1 R_2 R_3 + \frac{1}{2} V'_1 L_2 L_3 \nonumber$
Now for the critical experiment.
xxx - experiment
$\Psi_{xyx} = \frac{1}{ \sqrt{2}} \left[ \frac{H'_1 + V'_1}{ \sqrt{2}} \frac{H'_2 + V'_2}{ \sqrt{2}} \frac{ H'_3 + V'_3}{ \sqrt{2}} + \frac{ H'_1 - V'_1)}{ \sqrt{2}} \frac{i (H'_2 - V'_2)}{ \sqrt{2}} \frac{H'_3 - V'_3)}{ \sqrt{2}} \right] \nonumber$
on expansion yields
$\Psi_{xxx} = \frac{1}{2} H'_1 H'_2 H'_3 + \frac{1}{2} H'_1 V'_2 V'_3 + \frac{1}{2} V'_1 H'_2 V'_3 + \frac{1}{2} V'_1 V'_2 H'_3 \nonumber$
This quantum mechanical prediction is displayed graphically in Figure 4.
According to local realism the experimental outcome should be as shown in Figure 5. The origin of this prediction will be outlined shortley.
The quantum mechanical prediction for the xxx experiment agrees with experiment, the local realist prediction doesnʹt.
To explain how the local realist prediction shown in Figure 5 is derived, we recall that this position assumes that physical properties exist independent of measurement. We associate with polarization measurements the following eigenvalues: Hʹ = +1, Vʹ = ‐1, R = +1 and L = ‐1. Substituting these measurement eigenvalues into the first three experimental results (yyx, yxy, xyy) yields,
$Y_1 Y_2 Y_3 = Y_1 X_2 Y_3 = X_1 Y_2 Y_3 = -1 \nonumber$
Therefore,
$(Y_1 Y_2 Y_3) (Y_1 X_2 Y_3) (X_1 Y_2 Y_3) = -1 \nonumber$
However, this means that $X_1 X_2 X_3 = -1$ because $Y_1 Y_1 = Y_2 Y_2 = Y_3 Y_3 = 1$. According to the local realist view there are four ways to achieve this result V1V2V3, H1H2V3, H1V2H3 and V1H2H3, none of which are observed at a statistically meaningful level in the GHZ experiment.
For the xxx experiment, the mathematical predictions of quantum mechanics (Figure 4) and the local realist view (Figure 5) when compared with the actual experimental results (Figure 6) present a convincing refutation of local realism.
Appendix
$\begin{matrix} H = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V = \begin{pmatrix} 0 \ 1 \end{pmatrix} & H' = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & V' = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} & L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \end{matrix} \nonumber$
8.15: GHZ Math Appendix
This appendix shows another way of "doing the math" in the GHZ experiment.
$\Psi_{yyx} = \frac{1}{ \sqrt{2}} (H_1 H_2 H_3 + V_1 V_2 V_3)~ \begin{array}{|l} \text{substitute,}~ H_1 = \frac{1}{ \sqrt{2}} (R_1 + L_1) \ \text{substitute,}~ H_2 = \frac{1}{ \sqrt{2}} (R_2 + L_2) \ \text{substitute,}~ H_3 = \frac{1}{ \sqrt{2}} (H'_3 + V'_3) \ \text{substitute,}~ V_1 = \frac{i}{ \sqrt{2}} (L_1 - R_1) \ \text{substitute,}~ V_2 = \frac{i}{ \sqrt{2}} (L_2 - R_2) \ \text{substitute,}~ V_3 = \frac{1}{ \sqrt{2}} (H'_3 - V'_3) \ \text{simplify} \end{array} \rightarrow \Psi_{yyx} = \frac{1}{2} R_1 R_2 V'_3 + \frac{1}{2} R_1 L_2 H'_3 + \frac{1}{2} L_1 R_2 H'_3 + \frac{1}{2} L_1 L_2 V'_3 \nonumber$
$\Psi_{yxy} = \frac{1}{ \sqrt{2}} (H_1 H_2 H_3 + V_1 V_2 V_3)~ \begin{array}{|l} \text{substitute,}~ H_1 = \frac{R_1 + L_1}{ \sqrt{2}} \ \text{substitute,}~ H_2 = \frac{(H'_2 + V'_2)}{ \sqrt{2}} \ \text{substitute,}~ H_3 = \frac{R_3 + L_3}{ \sqrt{2}} \ \text{substitute,}~ V_1 = \frac{i}{ \sqrt{2}} (L_1 - R_1) \ \text{substitute,}~ V_2 = \frac{H'_2 - V'_2}{ \sqrt{2}} \ \text{substitute,}~ V_3 = \frac{i (L_3 - R_3)}{ \sqrt{2}} \ \text{simplify} \end{array} \rightarrow \Psi_{yxy} = \frac{1}{2} R_1 H'_2 L_3 + \frac{1}{2} R_1 V'_2 R_3 + \frac{1}{2} L_1 H'_2 R_3 + \frac{1}{2} L_1 V'_2 L_3 \nonumber$
$\Psi_{xyy} = \frac{1}{ \sqrt{2}} (H_1 H_2 H_3 + V_1 V_2 V_3)~ \begin{array}{|l} \text{substitute,}~ H_1 = \frac{H'_1 + V'_1}{ \sqrt{2}} \ \text{substitute,}~ H_2 = \frac{R_2 + L_2}{ \sqrt{2}} \ \text{substitute,}~ H_3 = \frac{R_3 + L_3}{ \sqrt{2}} \ \text{substitute,}~ V_1 = \frac{H'_1 - V'_1}{ \sqrt{2}} \ \text{substitute,}~ V_2 = \frac{i (L_2 - R_2)}{ \sqrt{2}} \ \text{substitute,}~ V_3 = \frac{i (L_3 - R_3}{ \sqrt{2}} \ \text{simplify} \end{array} \rightarrow \Psi_{xyy} = \frac{1}{2} H'_1 R_2 L_3 + \frac{1}{2} H'_1 L_2 R_3 + \frac{1}{2} V'_1 R_2 R_3 + \frac{1}{2} V'_1 L_2 L_3 \nonumber$
$\Psi_{xxx} = \frac{1}{ \sqrt{2}} (H_1 H_2 H_3 + V_1 V_2 V_3)~ \begin{array}{|l} \text{substitute,}~ H_1 = \frac{H'_1 + V'_1}{ \sqrt{2}} \ \text{substitute,}~ H_2 = \frac{H'_2 + V'_2}{ \sqrt{2}} \ \text{substitute,}~ H_3 = \frac{H'_3 + V'_3}{ \sqrt{2}} \ \text{substitute,}~ V_1 = \frac{H'_1 - V'_1}{ \sqrt{2}} \ \text{substitute,}~ V_2 = \frac{H'_2 - V'_2)}{ \sqrt{2}} \ \text{substitute,}~ V_3 = \frac{H'_3 - V'_3}{ \sqrt{2}} \ \text{simplify} \end{array} \rightarrow \Psi_{xxx} = \frac{1}{2} H'_1 H'_2 H'_3 + \frac{1}{2} H'_1 V'_2 V'_3 + \frac{1}{2} V'_1 H'_2 V'_3 + \frac{1}{2} V'_1 V'_2 H'_3 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.14%3A_Greenaberger-Hrne-Zeilinger_%28GHZ%29_Entanglement_and_Local_Realism.txt
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This tutorial analyses experimental results on a GHZ entanglement reported by Anton Zeilinger and collaborators in the 3 February 2000 issue of Nature (pp. 515‐519) using tensor algebra. The GHZ experiment employs three‐photon entanglement to provide a stunning attack on local realism.
First some definitions:
• Realism ‐ experiments yield values for properties that exist independent of experimental observation
• Locality ‐ the experimental results obtained at location A at time t, do not depend on the results at some other location B at time t.
• H/V = horizontal/vertical linear polarization.
• R/L = right/left circular polarization. Hʹ/Vʹ rotated by 45o with respect to H/V.
Next the vector representations of the various photon polarization states:
$\begin{matrix} H = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V = \begin{pmatrix} 0 \ 1 \end{pmatrix} & H' = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & V' = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} & L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} & N = \begin{pmatrix} 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
Some additional matrices needed to form state vectors in Mathcad:
$\begin{matrix} \text{HN = augment (H, N)} & \text{VN = augment (V, N)} & \text{RN = augment (R, N)} \ \text{LN = augment (L, N)} & \text{H'N = augment (H', N)} & \text{V'N = augment (V', N)} \end{matrix} \nonumber$
The operators associated with linear and cicular polarization:
$\begin{matrix} H'V = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & RL = \begin{pmatrix} 0 & -1 \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
The initial GHZ three-phon entangled state:
$\Psi = \frac{1}{ \sqrt{2}} (H_1 H_2 H_3 + V_1 V_2 V_3) \nonumber$
Initial state expressed in tensor format:
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
Constructing the initial state using Mathcad:
$\Psi = \frac{1}{ \sqrt{2}} \text{submatrix}(\text{kronecker (HN, HN)}) + \text{kronecker( VN, kronecker (VN, VN))}, 1,~8,~1,~1) \nonumber$
$\Psi^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0.707 \end{pmatrix} \nonumber$
After preparation of the initial GHZ state (see figure 1 in the reference cited above), polarization measurements are performed on the three photons. Zeilinger and collaborators use y to stand for a circular polarization measurement and x for a linear polarization measurement. Initially they perform circular polarization measurements on two of the photons and a linear polarization measurement on the other photon. The quantum mechanically predicted results and actual experimental measurements are given below.
yyx - experiment
The yyx operator is calculated and it is shown that Ψ is an eigenstate of the operator with eigenvalue ‐1.
$\begin{matrix} \text{yyx} = \text{kronecker(RL, kronecker(RL, H'V'))} & \Psi^T \text{yyx} \Psi = -1 \end{matrix} \nonumber$
Each measurement (R/L or Hʹ/Vʹ) has two possible outcomes so, in principle, there could be 8 possible results. However, quantum mechanics predicts that only four equally probable outcomes are possible. This is because the eigenvalues of the individual results must be consistent with the eigenvalue the total operator. As shown below, the eigenvalues for Hʹ and R are +1, and for Vʹ and L they are ‐1.
$\begin{matrix} H'V'H' = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & H'V'V' = \begin{pmatrix} -0.707 \ 0.707 \end{pmatrix} & RLR = \begin{pmatrix} 0.707 \ 0.707i \end{pmatrix} & RLL = \begin{pmatrix} -0.707 \ 0.707i \end{pmatrix} \end{matrix} \nonumber$
Thus quantum mechanics predicts that RRVʹ (++‐), LRHʹ (‐++), RLHʹ (+‐+), and LLVʹ (‐‐‐) should be observed, but RRHʹ (+++), LRVʹ (‐+‐), RLVʹ (+‐‐) and LLHʹ (‐‐+) should not be observed. This prediction is in agreement with the experimental results shown in Figure 1 to within experimental error.
Confirm the first two results in figure 1.
$\begin{matrix} RRV' = \text{submatrix(kronecker(RN, kronecker(RN, V'N))},~ 1,~8,~1,~1) (|RRV' \Psi |)^2 = 0.25 \ RRH' = \text{submatrix(kronecker(RN, kronecker(RN, H'N))},~ 1,~8,~1,~1) (|RRH' \Psi |)^2 = 0 \end{matrix} \nonumber$
For the two remaining experiments in this class (yxy and xyy), the agreement between theoretical prediction and experimental results is basically the same.
While these agreements between quantum mechanics and experiment are impressive they do not directly challenge the local realist position. As will be shown later that will be accomplished by a fourth experiment involving the measurement of the linear polarization on all three photons ‐ the xxx experiment.
yxy - experiment
The yxy operator is calculated and it is shown that Ψ is an eigenstate of the operator with eigenvalue -1.
$\begin{matrix} \text{yxy} = \text{kronecker(RL, kronecker(H'V', RL))} & \Psi^T yxy \Psi = -1 \end{matrix} \nonumber$
Using reasoning identical to the yyx experiment, quantum mechanics predicts that RVʹR (+‐+), LHʹR (‐++), RHʹL (++‐), and LVʹL (‐‐‐) should be observed, but RHʹR (+++), LVʹR (‐‐+), RVʹL (+‐‐) and LHʹL (‐+‐) should not be observed. This prediction is in agreement with the experimental results shown in Figure 2 to within experimental error.
Confirm the first two results in Figure 2.
$\begin{matrix} RV'R = \text{submatrix(kronecker(RN, kronecker(V'N, RN))},~ 1,~8,~1,~1) \ (|RV'R \Psi |)^2 = 0.25 \ RH'R = \text{submatrix(kronecker(RN, kronecker(H'N, RN))},~ 1,~8,~1,~1) \ (|RH'R \Psi |)^2 = 0 \end{matrix} \nonumber$
xyy - experiment
The xyy operator is calculated and it is shown that Ψ is an eigenstate of the operator with eigenvalue ‐1.
$\begin{matrix} \text{xyy} = \text{kronecker(H'V', kronecker(RL, RL))} & \Psi^T \text{xyy} \Psi = -1 \end{matrix} \nonumber$
Using reasoning identical to the previous experiments, quantum mechanics predicts that VʹRR (‐++), HʹLR (+‐+), HʹRL (++‐), and VʹLL (‐‐‐) should be observed, but HʹRR (+++), VʹLR (‐‐+), VʹRL (‐+‐) and HʹLL (+‐‐) should not be observed. This prediction is in agreement with the experimental results shown in Figure 3 to within experimental error.
Confirm the first two results in Figure 3.
$\begin{matrix} \text{V'RR} = \text{submatrix(kronecker(V'N, kronecker(RN RN))},~ 1,~ 8,~ 1,~ 1) (|V'RR \Psi |)^2 = 0.25 \ \text{H'RR} = \text{submatrix(kronecker(H'N, kronecker(RN, RN))},~ 1,~ 8,~ 1,~ 1) (| H'RR \Psi |)^2 = 0 \end{matrix} \nonumber$
Now for the critical experiment.
xxx - experiment
The xxx operator is calculated and it is shown that Ψ is an eigenstate of the operator with eigenvalue +1.
$\begin{matrix} \text{xxx} = \text{kronecker(H'V', kronecker(H'V', H'V'))} & \Psi^T \text{xxx} \Psi = 1 \end{matrix} \nonumber$
Using reasoning identical to the previous experiments, quantum mechanics predicts that only HʹVʹVʹ (+‐‐), VʹHʹVʹ (‐+‐), VʹVʹHʹ (‐‐+), and HʹHʹHʹ (+++) should be observed. This prediction is displayed graphically in Figure 4.
QM calculation for xxx experiment.
$\begin{matrix} V'V'V' = \text{submatrix(kronecker(V'N, kronecker(V'N, V'N))},~ 1,~ 8,~ 1,~ 1) (|V'V'V' \Psi |)^2 = 0 \ H'V'V' = \text{submatrix(kronecker(H'N, kronecker(V'N, V'N))},~ 1,~ 8,~ 1,~ 1) (|H'V'V' \Psi |)^2 = 0.25 \end{matrix} \nonumber$
According to local realism the experimental outcome should be as shown in Figure 5. To explain how the local realist prediction is derived, we recall that this position assumes that physical properties exist independent of measurement. Previously it has been shown that eigenvalues of the three‐photon operators according to quantum mechanics are,
$\begin{matrix} y_1 y_2 y_3 = y_1 x_2 y_3 = x_1 y_2 y_3 = -1 & x_1 x_2 x_3 = 1 \end{matrix} \nonumber$
It follows that,
$(y_1 y_2 x_3 )(y_1 x_2 y_3)(x_1 y_2 y_3) = -1 \nonumber$
However, this means that x1x2x3 = -1, because y1y1 = y2y2 = y3y3 = 1. (All of the three-photon operators commute, and therefore can have simultaneous eigenvalues. The commutation relations are shown in the Appendix.) According to the local realist view there are four ways to achieve this result (x1x2x3 = -1) V'V'V', H'H'V', H'V'H' and V'H'H'.
LR results are contradicted by QM calculation.
$\begin{matrix} V'V'V' = \text{submatrix(kronecker(V'N, kronecker(V'N, V'N))},~ 1,~ 8,~ 1,~ 1) (| V'V'V' \Psi |)^2 = 0 \ H'H'V' = \text{submatrix(kronecker(H'N, kronecker(H'N, V'N))},~ 1,~ 8,~ 1,~ 1) (| H'H'V' \Psi |)^2 = 0 \end{matrix} \nonumber$
Figure 6 shows that none local realistic predictions are observed at a statistically meaningful level in th GHZ experiment. The quantum mechanical prediction for the xxx experiment agrees with experiment, the local realist prediction doesnʹt.
QM calculations confirmed by experimental results.
$\begin{matrix} V'V'V' = \text{submatrix(kronecker(V'N, kronecker(V'N, V'N))},~ 1,~ 8,~1,~1) (| V'V'V' \Psi |)^2 = 0 \ H'V'V' = \text{submatrix(kronecker(H'N, kronecker(V'N, V'N))},~ 1,~ 8,~1,~1) (| H'V'V' \Psi |)^2 = 0.25 \end{matrix} \nonumber$
For the xxx experiment, the mathematical predictions of quantum mechanics (Figure 4) and the local realist view (Figure 5) when compared with the actual experimental results (Figure 6) present a convincing refutation of local realism.
Appendix
Demonstration that the three‐photon operators commute, allowing for simultaneous eigenvalues.
$\begin{matrix} \text{xyy} \text{yxy} - \text{yxy} \text{xyy} \rightarrow 0 & \text{xyy} \text{yyx} - \text{yyx} \text{xyy} \rightarrow 0 & \text{xyy} \text{xxx} - \text{xxx} \text{xyy} \rightarrow 0 \ \text{yxy} \text{yyx} - \text{yyx} \text{yxy} \rightarrow 0 & \text{yxy} \text{xxx} - \text{xxx} \text{yxy} \rightarrow 0 & \text{yyx} \text{xxx} - \text{xxx} \text{yyx} \rightarrow 0 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.16%3A_GHZ_Entanglement_-_A_Tensor_Algebra_Analysis.txt
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This tutorial analyzes a GHZ (Greenberger-Horne-Zeilinger) thought experiment involving three spin-1/2 particles that illustrates the clash between local realism and the quantum view of reality. The analysis consists of three parts: a traditional theoretical analysis, a quantum computer simulation, and an analysis based local realism. In the 1990s N. David Mermin published two articles in the general physics literature (Physics Today, June 1990; American Journal of Physics, August 1990) on the GHZ gedanken experiment. I drew heavily on these articles in developing this tutorial.
Theoretical Analysis
We begin with a quantum mechanical analysis of a version of the GHZ thought experiment suggested by Mermin. The initial state, Ψ, and Mermin's operator are as follows.
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & i \end{pmatrix}^T \nonumber$
$\hat{M} = \sigma_y \sigma_x \sigma_x + \sigma_x \sigma_y \sigma_x + \sigma_x \sigma_x \sigma_y - \sigma_y \sigma_y \sigma_y \nonumber$
The Pauli σx and σy individual spin operators and their composites required for Mermin's operator:
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \ \sigma_{xxy} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_x, \sigma_y )) & \sigma_{xyx} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_y, \sigma_x )) \ \sigma_{yxx} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_x, \sigma_s )) & \sigma_{yyy} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_y, \sigma_y )) \end{matrix} \nonumber$
The composite operators commute (σx and σy don't) and therefore can have simultaneous eigenvalues.
$\begin{matrix} \sigma_{xxy} \sigma_{xyx} - \sigma_{xyx} \sigma_{xxy} \rightarrow 0 & \sigma_{xxy} \sigma_{yxx} - \sigma_{yxx} \sigma_{xxy} \rightarrow 0 & \sigma_{xxy} \sigma_{yyy} - \sigma_{yyy} \sigma_{xxy} \rightarrow 0 \ \sigma_{xyx} \sigma_{yxx} - \sigma_{yxx} \sigma_{xyx} \rightarrow 0 & \sigma_{xyx} \sigma_{yyy} - \sigma_{yyy} \sigma_{xyx} \rightarrow 0 & \sigma_{yxx} \sigma_{yyy} - \sigma_{yyy} \sigma_{yxx} \rightarrow 0 \end{matrix} \nonumber$
The calculations of various expectation values bases on the proposed initial state and operator:
$M = \sigma_{yxx} + \sigma_{xyx} + \sigma_{xxy} - \sigma_{yyy} \nonumber$
$\begin{matrix} ( \overline{ \Psi})^T \sigma_{yxx} \Psi = 1 & ( \overline{ \Psi})^T \sigma_{xyx} \Psi = 1 & ( \overline{ \Psi})^T \sigma_{xxy} \Psi = 1 & ( \overline{ \Psi})^T \sigma_{yyy} \Psi = -1 & ( \overline{ \Psi})^T M \Psi = 4 \end{matrix} \nonumber$
The key result is that the expectation value for M is 4. Subsequently it will be shown that a quantum simulator circuit is in agreement with this result but that local realism is not.
Quantum Computer Simulation
"Quantum simulation is a process in which a quantum computer simulates another quantum system. Because of the various types of quantum weirdness, classical computers can simulate quantum systems only in a clunky, inefficient way. But because a quantum computer is itself a quantum system, capable of exhibiting the full repertoire of quantum weirdness, it can efficiently simulate other quantum systems. The resulting simulation can be so accurate that the behavior the computer will be indistinguishable from the behavior of the simulated system itself." (Seth Lloyd, Programming the Universe, page 149.)
This is the most important part of the tutorial - a demonstration that the thought experiment can be simulated using the quantum circuit shown below which can be found at: arXiv:1712.06542v2; "Five Experimental Tests on the 5-qubit IBM Quantum Computer," Diego Garcia-Martin and German Sierra.
$\begin{matrix} |0 \rangle & \triangleright & H & \cdot & \cdot & S & S^{ \dagger} & H & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & | & | \ | 0 \rangle & \triangleright & \cdots & \oplus & | & \cdots & \cdots & H & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & ~ & | \ | 0 \rangle & \triangleright & \cdots & \cdots & \oplus & \cdots & \cdots & H & \triangleright & \text{Meaure, 0 or 1} \end{matrix} \nonumber$
The required matrix operators and the build-up of the quantum circuit:
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & S = \begin{pmatrix} 1 & 0 \ 0 & i \end{pmatrix} & S' = \begin{pmatrix} 1 & 0 \ 0 & -i \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{HII} = \text{kronecker(H, kronecker(I, i))} & \text{CNOTI} = \text{kronecker(CNOT, I)} & \text{SII} = \text{kronecker(S, kronecker,(I, I))} \ \text{S'S'S'} = \text{kronecker(S', kronecker(S', S'))} & \text{HHH} = \text{kronecker(H, kronecker(H, H))} & \text{S'II} = \text{kronecker(S', kronecker(I, I))} \ \text{IS'I} = \text{kronecker(I, kronecker(S', I))} & \text{IIS'} = \text{kronecker(I, kronecker(I, S'))} \end{matrix} \nonumber$
First it is demonstrated that the first four steps of the circuit create the initial state.
$\begin{bmatrix} \text{SII (CnNOT) CNOT (HII)} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix}^T = \begin{pmatrix} 0.707 & 0 & 0 & 0 & 0 & 0 & 0 & 0.707i \end{pmatrix} \nonumber$
The complete circuit shows the simulation for the expectation value of σyσxσx. The presence of S' on a line before the final H gates indicates the measurement of the σy. The subsequent simulations show the presence of S' on the middle and last line, and finally on all three lines for the simulation of the expectation value for σyσyσy.
Eigenvalue of |0> = +1; eigenvalue of |1> = -1
$\begin{matrix} \langle \sigma_y \sigma_x \sigma_x \rangle = 1 & \text{HHH S'II SII CnNOT CNOTI HII} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.5 \ 0 \ 0 \ 0.5 \ 0 \ 0.5 \ 0.5 \ 0 \end{pmatrix} = \frac{1}{2} \left[|000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right] \end{matrix} \nonumber$
Given the eigenvalue assignments above the expectation value associated with this outcome is 1/4[(1)(1)(1)+(1)(-1)(-1)+(-1)(1)(-1)+(-1)(-1)(1)] = 1. Note that 1/2 is the probability amplitude for the product state. Therefore the probability of each member of the superposition being observed is 1/4. Similar reasoning is used for the remaining simulations.
$\begin{matrix} \langle \sigma_x \sigma_y \sigma_x \rangle = 1 & \text{HHH IS'I SII CnNOT CNOTI HII} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.5 \ 0 \ 0 \ 0.5 \ 0 \ 0.5 \ 0.5 \ 0 \end{pmatrix} = \frac{1}{2} \left[|000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right] \end{matrix} \nonumber$
$\begin{matrix} \langle \sigma_x \sigma_x \sigma_y \rangle = 1 & \text{HHH IIS' SII CnNOT CNOTI HII} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.5 \ 0 \ 0 \ 0.5 \ 0 \ 0.5 \ 0.5 \ 0 \end{pmatrix} = \frac{1}{2} \left[|000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right] \end{matrix} \nonumber$
$\begin{matrix} \langle \sigma_y \sigma_y \sigma_y \rangle = 1 & \text{HHH S'S'S' SII CnNOT CNOTI HII} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.5 \ 0.5 \ 0 \ 0.5 \ 0 \ 0 \ 0.5 \end{pmatrix} = \frac{1}{2} \left[|001 \rangle + |010 \rangle + |100 \rangle + |111 \rangle \right] \end{matrix} \nonumber$
$\langle M \rangle = \langle \sigma_y \sigma_x \sigma_x \rangle + \langle \sigma_x \sigma_y \sigma_x \rangle + \langle \sigma_x \sigma_x \sigma_y \rangle - \langle \sigma_y \sigma_y \sigma_y \rangle = 4 \nonumber$
The simulation is in exact agreement with the initial theoretical analysis.
EPR Local Realistic Analysis
Local realism asserts that objects have definite properties independent of measurement. In this experiment it assumes that the x- and y-components of spin have definite values prior to measurement. This position leads to a contradiction with the quantum mechanical calculation and the simulation. There is no way to assign consistent eigenvalues (+/-1) to the results of the individual spin measurements that is consistent with the quantum mechanical result. Using a variety of possible x- and y-spin values shows that local realism predicts that $M \leq 2$, in sharp disagreement with the quantum mechanical result of M = 4.
$\begin{matrix} \text{Sx1} = 1 & \text{Sx2} = 1 & \text{Sx3} = 1 & \text{Sy1} = 1 & \text{Sy2} = 1 & \text{Sy3} = 1 \end{matrix} \nonumber$
$\text{M} = \text{Sy1 Sx2 Sx3} + \text{Sx1 Sy2 Sx3} + \text{Sx1 Sx2 Sy3} - \text{Sy1 Sy2 Sy3} = 2 \nonumber$
$\begin{matrix} \text{Sx1} = -1 & \text{Sx2} = 1 & \text{Sx3} = 1 & \text{Sy1} = 1 & \text{Sy2} = -1 & \text{Sy3} = -1 \end{matrix} \nonumber$
$\text{M} = \text{Sy1 Sx2 Sx3} + \text{Sx1 Sy2 Sx3} + \text{Sx1 Sx2 Sy3} - \text{Sy1 Sy2 Sy3} = 2 \nonumber$
$\begin{matrix} \text{Sx1} = 1 & \text{Sx2} = -1 & \text{Sx3} = 1 & \text{Sy1} = -1 & \text{Sy2} = 1 & \text{Sy3} = -1 \end{matrix} \nonumber$
$\text{M} = \text{Sy1 Sx2 Sx3} + \text{Sx1 Sy2 Sx3} + \text{Sx1 Sx2 Sy3} - \text{Sy1 Sy2 Sy3} = 2 \nonumber$
$\begin{matrix} \text{Sx1} = -1 & \text{Sx2} = 1 & \text{Sx3} = -1 & \text{Sy1} = -1 & \text{Sy2} = 1 & \text{Sy3} = -1 \end{matrix} \nonumber$
$\text{M} = \text{Sy1 Sx2 Sx3} + \text{Sx1 Sy2 Sx3} + \text{Sx1 Sx2 Sy3} - \text{Sy1 Sy2 Sy3} = 2 \nonumber$
$\begin{matrix} \text{Sx1} = -1 & \text{Sx2} = -1 & \text{Sx3} = -1 & \text{Sy1} = -1 & \text{Sy2} = 1 & \text{Sy3} = -1 \end{matrix} \nonumber$
$\text{M} = \text{Sy1 Sx2 Sx3} + \text{Sx1 Sy2 Sx3} + \text{Sx1 Sx2 Sy3} - \text{Sy1 Sy2 Sy3} = -2 \nonumber$
$\begin{matrix} \text{Sx1} = 1 & \text{Sx2} = 1 & \text{Sx3} = 1 & \text{Sy1} = -1 & \text{Sy2} = -1 & \text{Sy3} = -1 \end{matrix} \nonumber$
$\text{M} = \text{Sy1 Sx2 Sx3} + \text{Sx1 Sy2 Sx3} + \text{Sx1 Sx2 Sy3} - \text{Sy1 Sy2 Sy3} = -2 \nonumber$
Alsina and Latorre ("Experimental test of Mermin inequalities on a five qubit quantum computer" available at arXiv:1605.04220V2) used the following alternative quantum circuit for the GHZ simulation. Here it is shown that it provides the same results for the σyσyσy simulation.
$\begin{matrix} |0 \rangle & \triangleright & \cdots & \cdots & \text{H} & \cdot & \text{H} & \cdots & \text{S}^\dagger & \text{H} & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & ~ & ~ & | \ |0 \rangle & \triangleright & \text{H} & \cdot & \text{H} & | & \cdots & \cdots & \text{S}^\dagger & \text{H} & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & | & ~ & | \ |0 \rangle & \triangleright & \cdots & \oplus & \cdots & \oplus & \text{H} & \text{S} & \text{S}^\dagger & \text{H} & \triangleright & \text{Measure, 0 or 1} \end{matrix} \nonumber$
$\begin{matrix} \text{IIS} = \text{kronecker(I, kronecker(I, S))} & \text{HIH} = \text{kronecker(H, kronecker(I, H))} & \text{HHI} = \text{kronecker(H, kronecker(H, I))} \ \text{IHI} = \text{kronecker(I, kronecker(H, I))} & \text{ICNOT} = \text{kronecker(I, CNOT)} \end{matrix} \nonumber$
$\text{HHH S'S'S' IIS HIH CnNOT HHI ICNOT IHI} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.5 \ 0.5 \ 0 \ 0.5 \ 0 \ 0 \ 0.5 \end{pmatrix} = \frac{1}{2} \left[|001 \rangle + |010 \rangle + |100 \rangle + |111 \rangle \right] \nonumber$
$\langle \sigma_y \sigma_y \sigma_y \rangle = -1 \nonumber$
The first six steps of this circuit generate the initial state.
$\begin{bmatrix} \text{IIS HIH CnNOT HHI ICNOT IHI} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix}^T = \begin{pmatrix} 0.707 & 0 & 0 & 0 & 0 & 0 & 0 & 0.707i \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.17%3A_Simulation_of_a_GHZ_Gedanken_Experiment.txt
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Many years ago N. David Mermin published two articles (Physics Today, June 1990; American Journal of Physics, August 1990) in the general physics literature on a Greenberger-Horne-Zeilinger (American Journal of Physics, December 1990; Nature, 3 February 2000) thought experiment involving spins that sharply revealed the clash between local realism and the quantum view of reality.
Three spin-1/2 particles are created in a single event and move apart in the horizontal y-z plane. Subsequent spin measurements will be carried out in units of h/4π with spin operators in the x- and y-directions. The z-basis eigenfunctions are:
$\begin{matrix} Sz_{up} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & Sz_{down} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The x- and y-direction spin operators:
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{eigenvals} ( \sigma_x ) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & \text{eigenvals} ( \sigma_y) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix}\ The initial entangeld spin state for the three spin-1/2 particles in tensor notation is: \[ \begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -1 \end{pmatrix} & \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The following operators represent the measurements to be carried out on spins 1, 2 and 3, in that order.
$\begin{matrix} \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 & \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \end{matrix} \nonumber$
The matrix tensor product is also known as the Kronecker product, which is available in Mathcad. The four operators in tensor format are formed as follows.
$\begin{matrix} \sigma_{xyy} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_y,~ \sigma_y )) & \sigma_{yxy} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_x,~ \sigma_y )) \ \sigma_{yyx} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_y,~ \sigma_x )) & \sigma_{xxx} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_x,~ \sigma_x )) \end{matrix} \nonumber$
These composite operators are Hermitian and mutually commute which means they can have simultaneous eigenvalues.
$\begin{matrix} \sigma_{xyy} \sigma_{yxy} - \sigma_{yxy} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{yyx} - \sigma_{yyx} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{xxx} - \sigma_{xxx} \sigma_{xyy} \rightarrow 0 \ \sigma_{yxy} \sigma_{yyx} - \sigma_{yyx} \sigma_{yxy} \rightarrow 0 & \sigma_{yxy} \sigma_{xxx} - \sigma_{xxx} \sigma_{yxy} \rightarrow 0 & \sigma_{yyx} \sigma_{xxx} - \sigma_{xxx} \sigma_{yyx} \rightarrow 0 \end{matrix} \nonumber$
The expectation values of the operators are now calculated.
$\begin{matrix} \Psi^T \sigma_{xyy} \Psi = 1 & \Psi^T \sigma_{yxy} \Psi = 1 & \Psi^T \sigma_{yyx} \Psi = 1 & \Psi^T \sigma_{xxx} \Psi = 1 \end{matrix} \nonumber$
Consequently the product of the four operators has the expectation value of -1.
$\Psi^T \sigma_{xyy} \sigma_{yxy} \sigma_{yyx} \sigma_{xxx} \Psi = -1 \nonumber$
Local realism assumes that objects have definite properties independent of measurement. In this example it assumes that the x- and y-components of the spin have definite values prior to measurement. This position leads to a contradiction with the above result as demonstrated by Mermin (Physics Today, June 1990). Looking again at the measurement operators, notice that there is a σx measurement on the first spin in the first and fourth experiment. If the spin state is well-defined before measurement those results have to be the same, either both +1 or both -1, so that the product of the two measurements is +1.
$\left( \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 \right) \left( \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 \right) \left( \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 \right) \left( \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \right) \nonumber$
Likewise there is a σy measurement on the second spin in experiments one and three. By similar arguments those results will lead to a product of +1 also. Continuing with all pairs in the total operator using local realistic reasoning unambiguously shows that its expectation value should be +1, in sharp disagreement with the quantum mechanical result of -1. This result should cause all mathematically literate local realists to renounce and recant their heresy. However, they may resist saying this is just a thought experiment. It hasn't actually been performed. However, if you believe in quantum simulation it has been performed.
Quantum Simulation
"Quantum simulation is a process in which a quantum computer simulates another quantum system. Because of the various types of quantum weirdness, classical computers can simulate quantum systems only in a clunky, inefficient way. But because a quantum computer is itself a quantum system, capable of exhibiting the full repertoire of quantum weirdness, it can efficiently simulate other quantum systems. The resulting simulation can be so accurate that the behavior the computer will be indistinguishable from the behavior of the simulated system itself." (Seth Lloyd, Programming the Universe, page 149.) The thought experiment can be simulated using the quantum circuit shown below which is an adaptation of one that can be found at: arXiv:1712.06542v2.
$\begin{matrix} |1 \rangle & \triangleright & \text{H} & \cdot & \cdots & \cdots & \text{H} & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & | \ |0 \rangle & \triangleright & \cdots & \oplus & \cdot & \text{S} & \text{H} & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & ~ & | \ |0 \rangle & \triangleright & \cdots & \cdots & \oplus & \text{S} & \text{H} & \triangleright & \text{Measure, 0 or 1} \end{matrix} \nonumber$
The matrix operators required for the implementation of the quantum circuit:
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & S = \begin{pmatrix} 1 & 0 \ 0 & -i \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{HII} = \text{kronecker(H, kronecker(I, I))} & \text{CNOTI} = \text{kronecker(CNOT, I)} & \text{ICNOT} = \text{kronecker(I, CNOT)} \ \text{ISS} = \text{kronecker(I, kronecker(S, S))} & \text{SIS} = \text{kronecker(S, kronecker(I, S))} & \text{SSI} = \text{kronecker(S, kronecker(S, I))} \ \text{HHH} = \text{kronecker(H, kronecker(H, H))} \end{matrix} \nonumber$
First it is demonstrated that the first three steps of the circuit create the initial state
$\begin{bmatrix} \text{ICNOT CNOTI HII} \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix}^T = \begin{pmatrix} 0.707 & 0 & 0 & 0 & 0 & 0 & 0 & -0.707 \end{pmatrix} \nonumber$
The complete circuit shown above simulates the expectation value of the σxσyσy operator. The presence of S on a line before the final H gates indicates the measurement of the σy, its absence a measurement of σx. The subsequent simulations show the absence of S on the middle and last line, and finally on all three lines for the simulation of the expectation value for σxσxσx.
Eigenvalue |0 > = +1; eigenvalue |1 > = -1
$\begin{bmatrix} \text{ HHH ISS ICNOT CNOT HII} \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix} ^T = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 & 0 & 0.5 & 0.5 & 0 \end{pmatrix} \nonumber$
$\frac{1}{2} \left( |000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right) \Rightarrow \langle \sigma_x \sigma_y \sigma_y \rangle =1 \nonumber$
Given the eigenvalue assignments above the expectation value associated with this measurement outcome is 1/4[(1)(1)(1)+(1)(-1)(-1)+(-1)(1)(-1)+(-1)(-1)(1)] = 1. Note that 1/2 is the probability amplitude for the product state. Therefore the probability of each member of the superposition being observed is 1/4. The same reasoning is used for the remaining simulations.
$\begin{bmatrix} \text{ HHH ISS ICNOT CNOT HII} \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix} ^T = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 & 0 & 0.5 & 0.5 & 0 \end{pmatrix} \nonumber$
$\frac{1}{2} \left( |000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right) \Rightarrow \langle \sigma_y \sigma_x \sigma_y \rangle =1 \nonumber$
$\begin{bmatrix} \text{ HHH ISS ICNOT CNOT HII} \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix} ^T = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 & 0 & 0.5 & 0.5 & 0 \end{pmatrix} \nonumber$
$\frac{1}{2} \left( |000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right) \Rightarrow \langle \sigma_y \sigma_y \sigma_x \rangle =1 \nonumber$
$\begin{bmatrix} \text{ HHH ISS ICNOT CNOT HII} \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}^T \end{bmatrix} ^T = \begin{pmatrix} 0 & 0.5 & 0.5 & 0 & 0.5 & 0 & 0 & 0.5 \end{pmatrix} \nonumber$
$\frac{1}{2} \left( |001 \rangle + |010 \rangle + |100 \rangle + |111 \rangle \right) \Rightarrow \langle \sigma_x \sigma_x \sigma_x \rangle = -1 \nonumber$
Individually and in product form the simulated results are in agreement with the previous quantum mechanical calculations.
$\langle \sigma_x \sigma_x \sigma_x \rangle \langle \sigma_x \sigma_y \sigma_y \rangle \langle \sigma_y \sigma_x \sigma_y \rangle \langle \sigma_y \sigma_y \sigma_x \rangle = -1 \nonumber$
The appendix provides algebraic calculations of $\langle \sigma_x \sigma_y \sigma_y \rangle$ and $\langle \sigma_x \sigma_x \sigma_x \rangle$.
Appendix
Truth tables for the operation of the circuit elements:
$\begin{matrix} I = \begin{pmatrix} \text{0 to 0} \ \text{1 to 1} \end{pmatrix} & H = \begin{bmatrix} \text{0 to}~ \frac{(0 + 1)}{ \sqrt{2}} \ \text{1 to}~ \frac{(0 - 1)}{ \sqrt{2}} \end{bmatrix} & \text{CNOT} = \begin{pmatrix} \text{00 to 00} \ \text{01 to 01} \ \text{10 to 11} \ \text{11 to 10} \end{pmatrix} & S = \begin{pmatrix} \text{0 to 0} \ \text{1 to -i} \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} |100 \rangle & |100 \rangle \ H \otimes I \otimes I & H \otimes I \otimes I \ \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |100 \rangle \right] & \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |100 \rangle \right] \ CNOT \otimes I & CNOT \otimes I \ \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |110 \rangle \right] & \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |110 \rangle \right] \ I \otimes CNOT & I \otimes CNOT \ \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |111 \rangle \right] & \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |111 \rangle \right] \ I \otimes S \otimes S & H \otimes H \otimes H \ \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |1 - i - i \rangle \right] & \frac{1}{ \sqrt{2}} \left[ |001 \rangle + |010 \rangle + |100 \rangle + |111 \rangle \right] \ H \otimes H \otimes H & \langle \sigma_x \sigma_x \sigma_x \rangle = -1 \ \frac{1}{2} \left[ |000 \rangle + |011 \rangle + |101 \rangle + |110 \rangle \right] \ \langle \sigma_x \sigma_y \sigma_y \rangle = 1 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.18%3A_Another_Simulation_of_a_GHZ_Gedanken_Experiment.txt
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Three photons are created in a single event (Nature, February 3, 2000 pp. 515-519) and move apart in the horizontal y-z plane. The goal of this exercise is to demonstrate that an analysis of measurements in the diagonal and circular polarization basis reveals the impossibility of assigning definite values to the polarization states of the photons prior to and independent of measurement.
First some definitions:
Realism - experiments yield values for properties that exist independent of experimental observation
Locality - the experimental results obtained at location A at time t, do not depend on the results at some other remote location B at time t. The diagonal and circular polarization matrix operators:
$\begin{matrix} D = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} C = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
The eigenvalues of the matrices are +/- 1:
$\begin{matrix} \text{eigenvals(D)} = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvals(C)} = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The eigenvectors of the operators:
$\begin{matrix} \text{eigenvecs(D)} = \begin{pmatrix} 0.707 & -0.707 \ 0.707 & 0.707 \end{pmatrix} & \text{eigenvecs(C)} = \begin{pmatrix} -0.707i & 0.707 \ 0.707 & -0.707i \end{pmatrix} \end{matrix} \nonumber$
The following operators represent the measurement protocols for spins 1, 2 and 3. For example, the first operator designates that diagonal polarization is measured on the first photon and circular polarization on the second and third photons.
$\begin{matrix} D_1 C_2 C_3 & C_1 D_2 C_3 & C_1 C_2 D_3 & D_1 D_2 D_3 \end{matrix} \nonumber$
The four operators are constructed in matrix format using tensor multiplication.
$\begin{matrix} \text{DCC} = \text{kronecker(D, kronecker(C, C))} & \text{CDC} = \text{kronecker(C, kronecker(D, C))} \ \text{CCD} = \text{kronecker(C, kronecker(C, D))} & \text{DDD} = \text{kronecker(D, kronecker(D, D))} \end{matrix} \nonumber$
1. The operators are unitary. Demonstrate this for DDC and DDD.
$\begin{matrix} \text{DCC}^2 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} & \text{DDD}^2 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
2. The operators are also Hermitian. Demonstrate this for CDC. What is the significance of being Hermitian?
$\begin{matrix} \text{CDC} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \text{CDC}^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
Hermitian operators have real eigenvalues.
3. The operators have eigenvalues +/- 1 and the same set of eigenvectors. Demonstrate this for DCC and CCD.
$\begin{matrix} \text{eigenvals(CDC)} = \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} & \text{eigenvecs(DCC)} = \begin{pmatrix} 0.707 & 0.707 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0.707 & 0.707 \ 0 & 0 & 0.707 & 0.707 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & -0.707 & -0.707 & 0 & 0 \ 0 & 0 & 0 & 0 & 0.707 & -0.707 & 0 & 0 \ 0 & 0 & 0.707 & -0.707 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0.707 & -0.707 \ -0.707 & 0.707 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{eigenvals(CDC)} = \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} & \text{eigenvecs(DCC)} = \begin{pmatrix} 0.707 & 0.707 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & -0.707 & 0.707 \ 0 & 0 & 0.707 & 0.707 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0.707 & -0.707 & 0 & 0 \ 0 & 0 & 0 & 0 & 0.707 & 0.707 & 0 & 0 \ 0 & 0 & 0.707 & -0.707 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0.707 & 0.707 \ -0.707 & 0.707 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
4. One of the eigenvectors is Ψ. Calculate its expectation values for the four operators.
$\Psi = \begin{pmatrix} 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 \end{pmatrix} \nonumber$
$\begin{matrix} \Psi^T = \text{DCC} \Psi = 1 & \Psi^T \text{CDC} \Psi = -1 & \Psi^T \text{CCD} \Psi = -1 & \Psi^T \text{DDD} \Psi = -1 \ \Psi^T \text{DCC CDC CCD} \Psi = 1 \end{matrix} \nonumber$
5. Demonstrate that Ψ is a maximally entangle three-photon state. These are called GHZ states.
In decimal notation:
$\Psi = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} (3 - 4) \nonumber$
6. Demonstrate that these operators mutually commute (use the symbolic processor -->). What is the physical significance of this fact?
$\begin{matrix} \text{DCC CDC} - \text{CDC DCC} \rightarrow 0 & \text{DCC CCD} - \text{CCD DCC} \rightarrow 0 & \text{DCC DDD} - \text{DDD DCC} \rightarrow 0 \ \text{CDC CCD} - \text{CCD CDC} \rightarrow 0 & \text{CDC DDD} - \text{DDD CDC} \rightarrow 0 & \text{CCD DDD} - \text{DDD CCD} \rightarrow 0 \end{matrix} \nonumber$
That these operators mutually commute means that they can have simultaneous eigenstates with simultaneous eigenvalues. In other words, they can be in well-defined states at the same time.
7. Display $\text{DCC CDC CCD$ and $\text{DDD}$ in matrix format (use the traditional =).
$\begin{matrix} \text{DCC CDC CCD} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \text{DDD} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
8. Write this result in simple algebraic format.
$\text{DCC CDC CCD} = - \text{DDD} \nonumber$
9. Using arguments based on local realism demonstrate that this result leads to the following contradiction.
$\text{DDD} = - \text{DDD} \nonumber$
Local realism assumes that objects have definite properties independent of measurement. In this example it assumes the diagonal and circular polarization states have definite values prior to measurement. This position leads to a contradiction with the above result. There is no way to assign eigenvalues (+/-1) to the operators that is consistent with the above result.
$\text{DCC CDC CCD} = - \text{DDD} \nonumber$
Concentrating on the composite operator on the left side, we notice that there is a C measurement on the first photon in the second and third operators (green). If the photon state is well-defined before measurement those results have to be the same, either both +1 or both -1, so that the product of the two measurements is +1. There is a C measurement on the second photon in operators one and three (blue). By similar arguments those results will lead to a product of +1 also. Finally there is a C measurement on the third photon in operators one and two (red). By similar arguments those results will lead to a product of +1 also. Incorporating these observations into the expression above leads to the following contradiction.
$\text{DDD} = - \text{DDD} \nonumber$
This result should cause all mathematically literate local realists to renounce and recant their heresy.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.19%3A_A_Surgical_Refutation_of_the_Local_Realism_Heresy.txt
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This tutorial analyzes the results reported in "Experimental Violation of Local Realism by Four-Photon GHZ Entanglement" by Zhao, et al. and published in Physical Review Letters on October 31, 2003.
The null vector and required photon polarization states:
$\begin{matrix} N = \begin{pmatrix} 0 \ 0 \end{pmatrix} & H = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V = \begin{pmatrix} 0 \ 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} V = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} & L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \end{matrix} \nonumber$
To facilitate tensor vector multiplication the polarization states are stored in the left column of a 2x2 matrix using the null vector
$\begin{matrix} H = \text{augment(H, N)} & V = \text{augment(V, N)} & H' = \text{augment(H', N)} & V'N = \text{augment(V', N)} & R = \text{augment(R, N)} & L = \text{augment(L, N)} \ H = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} & V = \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix} & H' = \begin{pmatrix} 0.707 & 0 \ 0.707 & 0 \end{pmatrix} & V' = \begin{pmatrix} 0.707 & 0 \ -0.707 & 0 \end{pmatrix} & R = \begin{pmatrix} 0.707 & 0 \ 0.707i & 0 \end{pmatrix} & L = \begin{pmatrix} 0.707 & 0 \ -0.707i & 0 \end{pmatrix} \end{matrix} \nonumber$
Operators:
$\begin{matrix} H'V' = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & RL = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
Eigenvaues of various polarization states:
$\begin{matrix} \text{H'V' H'} = \begin{pmatrix} 0.707 & 0 \ 0.707 & 0 \end{pmatrix} & \text{H'V' V'} = \begin{pmatrix} -0.707 & 0 \ 0.707 & 0 \end{pmatrix} & \text{RL R} = \begin{pmatrix} 0.707 & 0 \ 0.707i & 0 \end{pmatrix} & \text{RL L} = \begin{pmatrix} -0.707 & 0 \ 0.707i & 0 \end{pmatrix} \ \end{matrix} \nonumber$
The initial GHZ four-photon entangled state:
$\Psi = \frac{1}{ \sqrt{2}} (H_1 V_2 V_3 H_4 + V_1 H_2 H_3 V_4) \nonumber$
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \nonumber$
Initial state set up in tensor format.
$\Psi_i = \frac{1}{ \sqrt{2}} \text{(submatrix(kronecker(H, kronecker(V, kronecker(V, H)))} + \text{kronecker(V, kronecker(H, kronecker(H, V)))},~1,~16,~1,~1)) \nonumber$
$\Psi^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0.707 & 0 & 0 & 0.707 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
The authors initially consider three measurements which are summarized below. It is shown that the inital photon state is an eigenstate of each of the operators with eigenvalue +1. In the operators x refers to a linear polarization measurement (H'V') and y refers to a circular polarization measurement (RL). The experimental results are reported in Figure 3 of the paper. Below quantum mechanical (QM) calculations (predictions) are compared with experimental outcomes. QM agrees with experiment.
The following calculations are facilitated by the following general expression for the measurement eigenstates.
$\Psi \text{(a, b, c, d)} = \text{submatrix(kronecker(a, kronecker(b, kronecker(c, d)))},~1,~16,~1,~1) \nonumber$
σxxxx experiment:
Operator:
$\sigma_{xxxx} = \text{kronecker(H'V', kronecker(H'V', kronecker(H'V', H'V')))} \nonumber$
Eigenvalue/Expectation Value:
$\Psi_i^T \sigma_{xxxx} \Psi_i = 1 \nonumber$
Observed: H'H'H'H', H'H'V'V', H'V'H'V', H'V'V'H', V'H'H'V', V'H'V'H', V'V'H'H' and V'V'V'V'.
$\begin{matrix} \left( \left| \Psi ( \text{H', H', H', H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', H', V', V'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', V', V', H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', V', V', H'})^T \Psi_i \right| \right)^2 = 0.125 \ \left( \left| \Psi ( \text{V', H', H', V'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', H', V', H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', V', H', H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', V', V', V'})^T \Psi_i \right| \right)^2 = 0.125 \end{matrix} \nonumber$
σxyxy experiment:
Operator:
$\sigma_{xyxy} = \text{kronecker(H'V', kronecker(RL, kronecker(H'V', RL)))} \nonumber$
Eigenvalue/Expectation Value:
$\Psi_i^T \sigma_{xyxy} \Psi_i = 1 \nonumber$
Observed: H'RH'R, H'RV'L, H'LH'L, H'LV'R, V'RH'L, V'R'V'R, V'LH'R and V'LV'L.
$\begin{matrix} \left( \left| \Psi ( \text{H', R, H', R})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', R, V', L})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', L, H', L})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', L, V', R}) ^T \Psi_i \right| \right)^2 = 0.125 \ \left( \left| \Psi ( \text{V', R, H', L})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', R, V', R})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', L, H', R})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', L, V', L})^T \Psi_i \right| \right)^2 = 0.125 \end{matrix} \nonumber$
σxxyy experiment:
Operator:
$\sigma_{xxyy} = \text{kronecker(H'V', kronecker(H'V', kronecker(RL, RL)))} \nonumber$
Eigenvalue/Expectation Value:
$\Psi_i^T \sigma_{xxyy} \Psi_i = 1 \nonumber$
Observed: H'H'RR, H'H'LL, H'V'RL, H'V'LR, H'V'RL, V'H'LR, V'V'RR and V'V'LL.
$\begin{matrix} \left( \left| \Psi ( \text{H', H', R, R})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', H', L, L})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', V', R, L})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', V', L, R}) ^T \Psi_i \right| \right)^2 = 0.125 \ \left( \left| \Psi ( \text{V', H', R, L})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', H', L, R})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', V', R, R})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', V', L, L})^T \Psi_i \right| \right)^2 = 0.125 \end{matrix} \nonumber$
This analysis shows that quantum mechanics (QM) is in agreement with experimental results. The next step is to perform an experiment that shows that local realism (LR) is not in agreement with experimental results.
The fact that the eigenvalues of the individual operators examined above is +1, guarantees that the same is true for their product.
$\left( x_1x_2x_3x_4 \right) \left( x_1y_2x_3y_4 \right) \left( x_1x_2y_3y_4 \right) = 1 \nonumber$
Local realism assumes that physical properties exist independent of measurement. Because commuting operators have simultaneous eigenvalues x1x1 = x2x2 = x3x3 = y4y4. It follows that,
$\left( x_1y_2y_3x_4 \right) = 1 \nonumber$
The following results are consistent with this local realism analysis: H'RRH', H'RLV', H'LRV', H'LLH', V'RRV', V'RLH', V'LRH' and V'LLV'. As shown below, this is in complete disagreement with quantum mechanics and the experimental data. QM shows that the eigenvalue of the operator is actually -1, and, furthermore none of LR predicted results are observed. QM, however, is in agreement with the experimental results.
σxyyx experiment:
Operator:
$\sigma_{xyyx} = \text{kronecker(H'V', kronecker(RL, kronecker(RL, H'V')))} \nonumber$
Eigenvalue/Expectation Value:
$\Psi_i^T \sigma_{xyyx} \Psi_i = 1 \nonumber$
Observed: H'H'RR, H'H'LL, H'V'RL, H'V'LR, H'V'RL, V'H'LR, V'V'RR and V'V'LL.
$\begin{matrix} \left( \left| \Psi ( \text{H', R, R, V'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', R, L, H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', L, R, H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{H', L, L, V'})^T \Psi_i \right| \right)^2 = 0.125 \ \left( \left| \Psi ( \text{V', R, R, H'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', R, L, V'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', L, R, V'})^T \Psi_i \right| \right)^2 = 0.125 & \left( \left| \Psi ( \text{V', L, L, H'})^T \Psi_i \right| \right)^2 = 0.125 \end{matrix} \nonumber$
Appendix
All four operators commute with each other allowing them to have simultaneous eigenvalues.
$\begin{matrix} \sigma_{xxxx} \sigma_{xyxy} - \sigma_{xyxy} \sigma_{xxxx} \rightarrow 0 & \sigma_{xxxx} \sigma_{xyyx} - \sigma_{xyyx} \sigma_{xxxx} \rightarrow 0 & \sigma_{xxxx} \sigma_{xxyy} - \sigma_{xxyy} \sigma_{xxxx} \rightarrow 0 \ \sigma_{xxyy} \sigma_{xyxy} - \sigma_{xyxy} \sigma_{xxyy} \rightarrow 0 & \sigma_{xxyy} \sigma_{xyyx} - \sigma_{xyyx} \sigma_{xxyy} \rightarrow 0 & \sigma_{xyxy} \sigma_{xyyx} - \sigma_{xyyx} \sigma_{xyxy} \rightarrow 0 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.20%3A_GHZ_Four-Photon_Entanglement_Analyzed_Using_Tensor_Algebra.txt
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This tutorial demonstrates the conflict between quantum theory and realism in an experiment described in Physical Review Letters on October 31, 2003 by Zhao, et al. titled ʺExperimental Violation of Local Realism by Four‐Photon GHZ Entanglement.ʺ It draws on the methodology outlined by N. David Mermin in two articles in the general physics literature: Physics Today, June 1990; American Journal of Physics, August 1990.
The experiment involves the measurement of the diagonal and circular polarization states of a four‐photon entangled state using the following measurement protocols.
$\begin{matrix} \sigma_d^1 \otimes \sigma_d^2 \otimes \sigma_d^3 \otimes \sigma_d^4 & \sigma_d^1 \otimes \sigma_c^2 \otimes \sigma_d^3 \otimes \sigma_c^4 & \sigma_d^1 \otimes \sigma_d^2 \otimes \sigma_c^3 \otimes \sigma_c^4 & \sigma_d^1 \otimes \sigma_c^2 \otimes \sigma_c^3 \otimes \sigma_d^4 \end{matrix} \nonumber$
The individual polarization operators and their eigenvalues are:
$\begin{matrix} D = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{eigenvals}(D) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & C = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & \text{eigenvals}(C) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The composite operators are formed by tensor matrix multiplication, where kronecker is Mathcadʹs command for tensor multiplication.
$\begin{matrix} \sigma_{dddd} = \text{kronecker(D, kronecker(D, kronecker(D, D)))} & \sigma_{dcdc} = \text{kronecker(D, kronecker(C, kronecker(D, C)))} \ \sigma_{ddcc} = \text{kronecker(D, kronecker(D, kronecker(C, C)))} & \sigma_{dccd} = \text{kronecker(D, kronecker(C, kronecker(C, D)))} \end{matrix} \nonumber$
These operators commute with each other allowing them to have simultaneous eigenvalues.
$\begin{matrix} \sigma_{dddd} \sigma_{dcdc} - \sigma_{dcdc} \sigma_{dddd} \rightarrow 0 & \sigma_{dddd} \sigma_{dccd} - \sigma_{dccd} \sigma_{dddd} \rightarrow 0 & \sigma_{dddd} \sigma_{ddcc} - \sigma_{ddcc} \sigma_{dddd} \rightarrow 0 \ \sigma_{ddcc} \sigma_{dcdc} - \sigma_{dcdc} \sigma_{ddcc} \rightarrow 0 & \sigma_{ddcc} \sigma_{dccd} - \sigma_{dccd} \sigma_{ddcc} \rightarrow 0 & \sigma_{dcdc} \sigma_{dccd} - \sigma_{dccd} \sigma_{dcdc} \rightarrow 0 \end{matrix} \nonumber$
Next we show that the null matrix results when the fourth operator is added to the product of the first three.
$\sigma_{dddd} \sigma_{dcdc} \sigma_{ddcc} + \sigma_{dccd} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
To facilitate further analysis, the null result is written as follows.
$\sigma_{dddd} \sigma_{dcdc} \sigma_{ddcc} = - \sigma_{dccd} \nonumber$
Realism maintains that objects have values for observable properties that exist prior to measurement and independent of the choice of measurement (noncontextual). If this assumption is valid, then the operators highlighted with the same color must have the same eigenvalues (+1 or ‐1) and therefore the product of their eigenvalues must be unity.
$\left( \textcolor{red}{ \sigma_d^1} \otimes \textcolor{blue}{ \sigma_d^2} \otimes \textcolor{green}{ \sigma_d^3} \otimes \sigma_d^4 \right) \left( \textcolor{red}{ \sigma_d^1} \otimes \sigma_d^2 \otimes \textcolor{green}{ \sigma_d^3} \otimes \textcolor{yellow}{ \sigma_d^4} \right) \left( \sigma_d^1 \otimes \textcolor{blue}{ \sigma_d^2} \otimes \sigma_d^3 \otimes \textcolor{yellow}{ \sigma_d^4} \right) = - \left( \sigma_d^1 \sigma_c^2 \sigma_c^3 \sigma_d^4 \right) \nonumber$
Thus, applying a classical concept (noncontextual realism) to the above quantum mechanical equation leads to the following contradiction.
$\sigma_d^1 \otimes \sigma_c^2 \otimes \sigma_c^3 \otimes \sigma_d^4 = - \sigma_d^1 \otimes \sigma_c^2 \otimes \sigma_c^3 \otimes \sigma_d^4 \nonumber$
The experimental results reported by Zhao, et al. validate the quantum mechanical analysis, and contradict the realistic interpretation. See the preceding tutorial for a summary of their experimental results.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.21%3A_Quantum_v._Realism.txt
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Twenty years ago N. David Mermin published two articles (Physics Today, June 1990; American Journal of Physics, August 1990) in the general physics literature on a Greenberger-Horne-Zeilinger (American Journal of Physics, December 1990; Nature, 3 February 2000) gedanken experiment involving spins that sharply revealed the clash between local realism and the quantum view of reality. In what follows I present Mermin's gedanken experiment using tensor algebra.
Three spin-1/2 particles are created in a single event and move apart in the horizontal y-z plane. Subsequent spin measurements will be carried out in units of h/4Ψ with spin operators in the x- and y-directions.
The z-basis eigenfunctions are:
$\begin{matrix} Sz_{up} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & Sz_{down} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The x- and y-direction spin operators and eigenvalues:
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{eigenvals}( \sigma_x ) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & \text{eigenvals}( \sigma_y ) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The initial spin state for the three spin-1/2 particles in tensor notation is:
$\begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -1 \end{pmatrix} & \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The Appendix shows how to carry out vector tensor products in Mathcad.
The following operators represent the actual measurements to be carried out on spins 1, 2 and 3, in that order.
$\begin{matrix} \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3\end{matrix} \nonumber$
The matrix tensor product is also known as the Kronecker product, which is available in Mathcad. The three operators in tensor format are formed as follows.
$\begin{matrix} \sigma_{xyy} = \text{kronecker}( \sigma_x \text{, kronecker}( \sigma_y, ~ \sigma_y)) & \sigma_{xyy} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \sigma_{yxy} = \text{kronecker}( \sigma_y \text{, kronecker}( \sigma_x, ~ \sigma_y)) & \sigma_{yxy} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \sigma_{yyx} = \text{kronecker}( \sigma_y \text{, kronecker}( \sigma_y, ~ \sigma_x)) & \sigma_{yyx} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
That the initial state is an eigenfunction of these operators with eigenvalue +1 is now demonstrated.
$\begin{matrix} \Psi^T \sigma_{xyy} \Psi = 1 & \Psi^T \sigma_{yxy} \Psi = 1 & \Psi^T \sigma_{yyx} \Psi = 1 \end{matrix} \nonumber$
The fact that the operators commute means that they can have simultaneous eigenvalues.
$\begin{matrix} \sigma_{xyy} \sigma_{yxy} - \sigma_{yxy} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{yyx} - \sigma_{yyx} \sigma_{xyy} \rightarrow 0 & \sigma_{yxy} \sigma_{yyx} - \sigma_{yyx} \sigma_{yxy} \rightarrow 0 \end{matrix} \nonumber$
The significance of this is evident when we consider the eigenvalue of the product of the three operators which obviously must be +1.
$\begin{matrix} \left( \sigma_x^1 \sigma_y^2 \sigma_y^3 \right) \left( \sigma_y^1 \sigma_x^2 \sigma_y^3 \right) \left( \sigma_y^1 \sigma_y^2 \sigma_x^3 \right) = 1 & \Psi^T \sigma_{xyy} \sigma_{yxy} \sigma_{yyx} \Psi = 1 \end{matrix} \nonumber$
If it is assumed that this result occurs because the particles are in well-defined spin states (sx, sy) prior to measurement the following must be accepted,
$\left( s_x^1 \textcolor{blue}{ s_y^2} \textcolor{green}{ s_y^3} \right) \left( \textcolor{red}{s_y^1} \textcolor{black}{ s_x^2} \textcolor{green}{ s_y^3} \right) \left( \textcolor{red}{s_y^1} \textcolor{blue}{ s_y^2} \textcolor{black}{ s_x^3} \right) = 1 \nonumber$
Given that the spin measurement values can be +/- 1 and that the individual operators can have simultaneous eigenvalues, the following must be true,
$s_y^1 s_y^1 = s_y^2 s_y^2 = s_y^3 s_y^3 = 1 \nonumber$
This reduces the previous equation involving the three operators to the following local realistic prediction.
$\left( s_x^1 s_x^2 s_x^3 \right) = \left( \sigma_x^1 \sigma_x^2 \sigma_x^3 \right) = 1 \nonumber$
The disagreement between quantum mechanics and the local realistic view becomes starkly apparent when the σx1σx2σx3 measurement outcome is calculated.
$\begin{matrix} \sigma_{xxx} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_x,~ \sigma_x)) & \Psi^T \sigma_{xxx} \Psi = -1 \end{matrix} \nonumber$
Thus we see absolute disagreement between local realism and quantum mechanics. Local realism predicts an eigenvalue of +1 and quantum mechanics an eigenvalue of -1.
The validity of the reasoning above requires that σx1σx2σx3 commutes with the other operators, which we now demonstrate.
$\begin{matrix} \sigma_{xyy} \sigma_{xxx} - \sigma_{xxx} \sigma_{xyy} \rightarrow 0 & \sigma_{yxy} \sigma_{xxx} - \sigma_{xxx} \sigma_{yxy} \rightarrow 0 & \sigma_{yyx} \sigma_{xxx} - \sigma_{xxx} \sigma_{yyx} \rightarrow 0 \end{matrix} \nonumber$
Appendix
The tensor product of three vectors is shown below.
$\begin{pmatrix} a \ b \end{pmatrix} \otimes \begin{pmatrix} c \ d \end{pmatrix} \otimes \begin{pmatrix} e \ f \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix} \otimes \begin{pmatrix} ce \ cf \ de \ df \end{pmatrix} = \begin{pmatrix} ace \ acf \ ade \ adf \ bce \ bcf \ bde \ bdf \end{pmatrix} \nonumber$
Mathcad does not have a command for the vector tensor product, so it is necessary to develop a way of implementing it using kronecker, which requires square matrices. For this reason the spin vector is stored in the left column of a 2x2 matrix by augmenting the spin vector with the null vector. After all the matrix tensor products have been carried out using kronecker the final spin vector resides in the left column of the final square matrix. Next the submatrix cammand is used to save this column, discarding the rest of the matrix.
The Mathcad syntax for the tensor multiplication of three vectors is as follows.
$\Psi (a,~ b,~ c) = \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ a,~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right],~ \text{kronecker} \left[ \text{augment} \left[ b,~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right],~ \text{augment} \left[ c,~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] \right] \right], 1,~ 8,~ 1,~ 1 \right] \nonumber$
The initial spin state:
$\frac{1}{ \sqrt{2}} \left( \Psi \left( Sz_{up},~ Sz_{up},~Sz_{up} \right) - \Psi \left( Sz_{down},~Sz_{down},~Sz_{down} \right) \right) = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.707 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.22%3A_Elements_of_Reality-_Another_GHZ_Gedanken_Experiment_Analyzed.txt
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In the '90s N. David Mermin published two articles in the general physics literature (Physics Today, June 1990; American Journal of Physics, August 1990) on the Greenberger-Horne-Zeilinger (GHZ) gedanken experiment (American Journal of Physics, December 1990; Nature, 3 February 2000) involving three spin-1/2 particles that illustrated the clash between local realism and the quantum view of reality for the quantum nonspecialist.
The three spin-1/2 particles are created in a single event and move apart in the horizontal y-z plane. It will be shown that a consideration of spin measurements (in units of h/4π in the x- and y-directions reveals the impossibility of assigning values to the spin observables independent of measurement.
The x- and y-direction spin operators are the Pauli matrices:
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
The eigenvalues of the Pauli matrices are +/- 1:
$\begin{matrix} \text{eigenvals}( \sigma_x ) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvals}( \sigma_y ) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The following operators represent the measurement protocols for spins 1, 2 and 3.
$\begin{matrix} \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 & \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \end{matrix} \nonumber$
The tensor matrix product, also known as the Kronecker product, is available in Mathcad. The four operators in tensor format are formed as follows.
$\begin{matrix} \sigma_{xyy} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_y,~ \sigma_y)) & \sigma_{yxy} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_x,~ \sigma_y)) \ \sigma_{yyx} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_y,~ \sigma_x)) & \sigma_{xxx} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_x,~ \sigma_x)) \end{matrix} \nonumber$
These operators mutually commute, meaning that they can be assigned simultaneous eigenstates with simultaneous eigenvalues.
$\begin{matrix} \sigma_{xyy} \sigma_{yxy} - \sigma_{yxy} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{yyx} - \sigma_{yyx} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{xxx} - \sigma_{xxx} \sigma_{xyy} \rightarrow 0 \ \sigma_{yxy} \sigma_{yyx} - \sigma_{yyx} \sigma_{yxy} \rightarrow 0 & \sigma_{yxy} \sigma_{xxx} - \sigma_{xxx} \sigma_{yxy} \rightarrow 0 & \sigma_{yyx} \sigma_{xxx} - \sigma_{xxx} \sigma_{yyx} \rightarrow 0 \end{matrix} \nonumber$
The next step is to compare the matrix for the product of the first three operators (σxyy σyxyσyyx) with that of the fourth (σxxx).
$\begin{matrix} \sigma_{xyy} \sigma_{yxy} \sigma_{yyx} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \sigma_{xxx} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
This indicates the following relationship between the four operators and leads quickly to a refutation of the concept of noncontextual, hidden values for quantum mechanical observables.
$\left( \textcolor{black}{ \sigma_x^1} \otimes \textcolor{red}{ \sigma_y^2} \otimes \textcolor{green}{ \sigma_y^3} \right) \left( \textcolor{blue}{ \sigma_y^1} \otimes \textcolor{black}{ \sigma_x^2} \otimes \textcolor{green}{ \sigma_y^3} \right) \left( \textcolor{blue}{ \sigma_y^1} \otimes \textcolor{red}{ \sigma_y^2} \otimes \textcolor{black}{ \sigma_x^3} \right) = - \left( \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \right) \nonumber$
Local realism assumes that objects have definite properties independent of measurement. In this example it assumes that the x- and y-components of the spin have definite values prior to measurement. This position leads to a contradiction with the above result. There is no way to assign eigenvalues (+/-1) to the operators that is consistent with the above result.
Concentrating on the operator on the left side, we notice that there is a σy measurement on the first spin in the second and third term (blue). If the spin state is well-defined before measurement those results have to be the same, either both +1 or both -1, so that the product of the two measurements is +1. There is a σy measurement on the second spin in terms one and three (red). By similar arguments those results will lead to a product of +1 also. Finally there is a σy measurement on the third spin in terms one and two (green). By similar arguments those results will lead to a product of +1 also. Incorporating these observations into the expression above leads to the following contradiction.
$\sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 = - \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \nonumber$
This result should cause all mathematically literate local realists to renounce and recant their heresy.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.23%3A_Brief_Elements_of_Reality.txt
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Twenty years ago N. David Mermin published two articles (Physics Today, June 1990; American Journal of Physics, August 1990) in the general physics literature on a Greenberger-Horne-Zeilinger (American Journal of Physics, December 1990; Nature, 3 February 2000) gedanken experiment involving spins that sharply revealed the clash between local realism and the quantum view of reality.
Three spin-1/2 particles are created in a single event and move apart in the horizontal y-z plane. Subsequent spin measurements will be carried out in units of h/4π in the z-basis with spin operators in the x- and y-directions.
The z-basis eigenfunctions are:
$\begin{matrix} Sz_{up} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & Sz_{down} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The x- and y-direction spin operators in the z-basis are the Pauli matrices:
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
The initial entangled spin state for the three spin-1/2 particles in tensor notation is:
$\begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -1 \end{pmatrix} & \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The following operators represent the measurements to be carried out on spins 1, 2 and 3, in that order.
$\begin{matrix} \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 & \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \end{matrix} \nonumber$
The matrix tensor product is also known as the Kronecker product, which is available in Mathcad. The four operators in tensor format are formed as follows.
$\begin{matrix} \sigma_{xyy} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_y,~ \sigma_y)) & \sigma_{yxy} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_x,~ \sigma_y)) \ \sigma_{yyx} = \text{kronecker}( \sigma_y,~ \text{kronecker}( \sigma_y,~ \sigma_x)) & \sigma_{xxx} = \text{kronecker}( \sigma_x,~ \text{kronecker}( \sigma_x,~ \sigma_x)) \end{matrix} \nonumber$
The expectation values of the operators are now calculated.
$\begin{matrix} \Psi^T \sigma_{xyy} \Psi = 1 & \Psi^T \sigma_{yxy} \Psi = 1 & \Psi^T \sigma_{yyx} \Psi = 1 & \Psi^T \sigma_{xxx} \Psi = -1 \end{matrix} \nonumber$
Consequently the product of the four operators has the expectation value of -1.
$\Psi^T \sigma_{xyy} \sigma_{yxy} \sigma_{yyx} \sigma_{xxx} \Psi = -1 \nonumber$
Local realism assumes that objects have definite properties independent of measurement. In this example it assumes that the x- and y-components of the spin have definite values prior to measurement. This position leads to a contradiction with the above result. The following analysis is taken from "Quantum Information Science" by Seth Lloyd.
Looking again at the measurement operators, notice that there is a σx measurement on the first spin in the first and fourth experiment. If the spin state is well-defined before measurement those results have to be the same, either both +1 or both -1, so that the product of the two measurements is +1.
$\begin{matrix} \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 & \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3\end{matrix} \nonumber$
Likewise there is a σy measurement on the second spin in experiments one and three. By similar arguments those results will lead to a product of +1 also. Continuing with all the pairs in the total operator using local realistic reasoning unambiguously shows that its expectation value should be +1, in sharp disagreement with the quantum mechanical result of -1.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.24%3A_A_Brief_Analysis_of_Mermin%27s_GHZ_Thought_Experiment.txt
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Hardy created a two-photon thought experiment for which a local hidden-variable (EPR) model for the photon states is not consistent with all the predictions of quantum theory.
A source emits two photons in the following entangled state, $\Psi = \frac{2}{ \sqrt{3}} \left( H_A H_B - \frac{1}{2} H'_A H'_B \right)$, with the first photon traveling to Alice at a detector to the left of the source and the second to Bob at a detector on the right.
The following diagram shows the directions that linear polarization measurements will be made on the entangled two-photon system. The detectors can be set to measure in either the H-V or H'-V' basis.
Definition of polarization eigenstates:
$\begin{matrix} H = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V = \begin{pmatrix} 0 \ 1 \end{pmatrix} & H' = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & V' = \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Ψ is expressed in tensor format:
$\begin{matrix} \Psi = \frac{2}{ \sqrt{3}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} - \frac{1}{4} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} \right] & \Psi = \begin{pmatrix} 0.866 \ -0.289 \ -0.289 \ -0.289 \end{pmatrix} & \Psi^T \Psi = 1 \end{matrix} \nonumber$
The first photon goes to Alice, the second to Bob. They record the results of a large number of independent, random polarization measurements on their photon pairs. They could for example enter their results in the following table which identifies all possible measurement outcomes.
$\begin{pmatrix} \text{V'V'} & \text{V'V} & \text{V'H'} & \text{V'H} \ \text{VV'} & \text{VV} & \text{VH'} & \text{VH} \ \text{H'V'} & \text{H'V} & \text{H'H'} & \text{H'H} \ \text{HV'} & \text{HV} & \text{HH'} & \text{HH} \end{pmatrix} \nonumber$
The next step is to calculate the probability that these observations will be made given Ψ as the initial state. To this end we form the product state vectors in tensor format using the definitions of H, V, H' and V' provided earlier.
$\begin{matrix} \text{V'V'} = \frac{1}{2} \begin{pmatrix} 1 \ -1 \ -1 \ 1 \end{pmatrix} & \text{V'V} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ -1 \ 0 \ 1 \end{pmatrix} & \text{V'H'} = \frac{1}{2} \begin{pmatrix} -1 \ -1 \ 1 \ 1 \end{pmatrix} & \text{V'H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 0 \ 1 \ 0 \end{pmatrix} \ \text{VV'} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 0 \ -1 \ 1 \end{pmatrix} & \text{VV} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} & \text{VH'} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 0 \ 1 \ 1 \end{pmatrix} & \text{VH} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \ \text{H'V'} = \frac{1}{2} \begin{pmatrix} -1 \ 1 \ -1 \ 1 \end{pmatrix} & \text{H'V} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 0 \ 1 \end{pmatrix} & \text{H'H'} = \frac{1}{2} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} & \text{H'H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} \ \text{HV'} = \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \ 0 \ 0 \end{pmatrix} & \text{HV} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \text{HH'} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \ 0 \ 0 \end{pmatrix} & \text{HH} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
Using these two-photon state functions we now calculate the probability of occurrence for each possible measurement outcome, as displayed in the following matrix.
$\begin{bmatrix} \left( \left| V'V'^T \Psi \right| \right)^2 & \left( \left| V'V^T \Psi \right| \right)^2 & \left( \left| V'H'^T \Psi \right| \right)^2 & \left( \left| V'H^T \Psi \right| \right)^2 \ \left( \left| VV'^T \Psi \right| \right)^2 & \left( \left| VV^T \Psi \right| \right)^2 & \left( \left| VH'^T \Psi \right| \right)^2 & \left( \left| VH^T \Psi \right| \right)^2 \ \left( \left| H'V'^T \Psi \right| \right)^2 & \left( \left| H'V^T \Psi \right| \right)^2 & \left( \left| H'H'^T \Psi \right| \right)^2 & \left( \left| H'H^T \Psi \right| \right)^2 \ \left( \left| HV'^T \Psi \right| \right)^2 & \left( \left| HV^T \Psi \right| \right)^2 & \left( \left| HH'^T \Psi \right| \right)^2 & \left( \left| HH^T \Psi \right| \right)^2 \ \end{bmatrix} = \begin{pmatrix} 0.333 & 0 & 0.333 & 0.667 \ 0 & 0.083 & 0.167 & 0.083 \ 0.333 & 0.167 & 0 & 0.167 \ 0.667 & 0.083 & 0.167 & 0.75 \end{pmatrix} \nonumber$
So, where's the paradox, where's the problem? The paradox/problem is revealed by concentrating on four entries in the matrix above.
$\begin{pmatrix} \text{Alice} & \text{Bob} & \text{Result} \ V & V' & \text{Never} \ V' & V & \text{Never} \ V & V & \text{Sometimes} \ H' & H' & \text{Never} \end{pmatrix} \nonumber$
In any run the detectors might be set to measure [H'-V']/[H-V] or [H-V]/[H'-V']. So if one photon triggers a [H-V] detector to register V, its partner must require a [H'-V'] detector to register H' according to the first two rows of the matrix above. According to the local hidden-variable model (EPR) this means H' is an element of reality.
It follows that any [H-V]/[H-V] run in which both detectors register V (probability 0.083) each photon must require a [H'-V'] detector to register H'. Therefore, if a [H'-V']/[H'-V'] run had been selected both detectors would have registered H'.
However, given the initial photon state function the result H'H' should never be observed. In other words it is impossible to assign specific polarization states (instruction sets) to the photons prior to measurement that are in agreement with all quantum mechanical predictions for this thought experiment.
An algebraic analysis shows that destructive interference eliminates the H'H' term. It also gives the correct probabilities for V'V', V'H' and H'V' measurements. Algebraic analysis could be used to produce the entire measurement outcome matrix calculated above.
$\begin{matrix} \frac{2}{ \sqrt{3}} \left[ | H \rangle_A |H \rangle_B - \frac{1}{2} H' \rangle_A |H' \rangle_B \right] \ \text{Substitute} \ |H \rangle = \frac{1}{ \sqrt{2}} \left( |H' \rangle - |V' \rangle \right) \ \downarrow \ \frac{2}{ \sqrt{3}} \left[ \frac{1}{ \sqrt{2}} \left( |H' \rangle_A - |V' \rangle_B \right) \frac{1}{ \sqrt{2}} \left( |H' \rangle_A - |V' \rangle_B \right) - \frac{1}{2} |H' \rangle_A |H' \rangle_B \right] \ \downarrow \ \frac{1}{ \sqrt{3}} \left[ |V' \rangle_A |V' \rangle_B - |V' \rangle_A |H' \rangle_B - |H' \rangle_A |V' \rangle_B \right] \end{matrix} \nonumber$
Sources:
Lucien Hardy, "Spooky Action at a Distance in Quantum Mechanics," Contemporary Physics 39, 419-429 (1998).
N. David Mermin, "Quantum Mysteries Refined," American Journal of Physics 62, 880-887 (1994). C. C. Gerry and K. M. Bruno, The Quantum Divide, Oxford U. P., 2013, pp. 124-29.
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Hardy created a two-photon thought experiment for which a local hidden-variable (EPR) model for the photon states is not consistent with all the predictions of quantum theory.
A source emits two photons in the following entangled state, $\Psi = \frac{2}{ \sqrt{3}} \left( H_A H_B - \frac{1}{2} H'_A H'_B \right)$, with the first photon traveling to Alice at a detector to the left of the source and the second to Bob at a detector on the right.
The following diagram shows the directions that linear polarization measurements will be made on the entangled two-photon system. The detectors can be set to measure in either the H-V or H'-V' basis.
Alice and Bob record the results of a large number of independent, random polarization measurements on their photon pairs. The following table identifies all possible measurement outcomes. The probabilites of these outcomes are determined algebraically below.
$\begin{pmatrix} \text{V'V'} & \text{V'V} & \text{V'H'} & \text{V'H} \ \text{VV'} & \text{VV} & \text{VH'} & \text{VH} \ \text{H'V'} & \text{H'V} & \text{H'H'} & \text{H'H} \ \text{HV'} & \text{HV} & \text{HH'} & \text{HH} \end{pmatrix} \nonumber$
Alice and Bob both make H-V measurements:
$\Psi = \frac{2}{ \sqrt{3}} \left( H_A H_B - \frac{1}{2} H'_A H'_B \right) \begin{array}{|l} \text{substitute, } H'_A = \frac{1}{ \sqrt{2}} (H'_A - V'_A) \ \text{substitute, } H'_B = \frac{1}{ \sqrt{2}} (H'_B - V'_B) \ \text{simplify} \end{array} \rightarrow \Psi = \frac{\sqrt{3} V'_A V'_B}{3} - \frac{\sqrt{3} H'_B V'_A}{3} - \frac{\sqrt{3} H'_A V'_B}{3} \nonumber$
Measurement probabilities: V'V' = V'H' = H'V' = 0.333
Alice and Bob both make H'-V' measurements:
$\Psi = \frac{2}{ \sqrt{3}} \left( H_A H_B - \frac{1}{2} H'_A H'_B \right) \begin{array}{|l} \text{substitute, } H'_A = \frac{1}{ \sqrt{2}} (H_A + V_A) \ \text{substitute, } H_B = \frac{1}{ \sqrt{2}} (H_B - V_B) \ \text{simplify} \end{array} \rightarrow \Psi = \frac{\sqrt{3} \left( H_A V_B - 3 H_A H_B + H_B V_A + V_A V_B \right)}{6} \nonumber$
Measurement probabilities: HV = VH = VV = 0.083 HH = 0.75
Alice makes H-V measurements and Bob makes H'-V' measurements:
$\Psi = \frac{2}{ \sqrt{3}} \left( H_A H_B - \frac{1}{2} H'_A H'_B \right) \begin{array}{|l} \text{substitute, } H'_A = \frac{1}{ \sqrt{2}} (H_A + V_A) \ \text{substitute, } H_B = \frac{1}{ \sqrt{2}} (H'_B - V'_B) \ \text{simplify} \end{array} \rightarrow \Psi = \frac{\sqrt{6} H_A H'_B}{6} - \frac{\sqrt{6} H'_B V_A}{6} - \frac{\sqrt{6} H_A V'_B}{6} \nonumber$
Measurement probabilities: HH' = VH' = 0.167 HV' = 0.667
Alice makes H'-V' measurements and Bob makes H-V measurements:
$\Psi = \frac{2}{ \sqrt{3}} \left( H_A H_B - \frac{1}{2} H'_A H'_B \right) \begin{array}{|l} \text{substitute, } H_A = \frac{1}{ \sqrt{2}} (H'_A + V'_A) \ \text{substitute, } H'_B = \frac{1}{ \sqrt{2}} (H_B + V_B) \ \text{simplify} \end{array} \rightarrow \Psi = \frac{\sqrt{6} H_B H'_A}{6} - \frac{\sqrt{6} H'_A V_B}{6} - \frac{\sqrt{6} H_B V'_A}{3} \nonumber$
Measurement probabilities: H'H = H'V = 0.167 V'H = 0.667
The following measurement outcomes never occur: VV' = V'V = H'H' = 0
The results are collected in the following table.
$\begin{bmatrix} \left( \left| V'V'^T \Psi \right| \right)^2 & \left( \left| V'V^T \Psi \right| \right)^2 & \left( \left| V'H'^T \Psi \right| \right)^2 & \left( \left| V'H^T \Psi \right| \right)^2 \ \left( \left| VV'^T \Psi \right| \right)^2 & \left( \left| VV^T \Psi \right| \right)^2 & \left( \left| VH'^T \Psi \right| \right)^2 & \left( \left| VH^T \Psi \right| \right)^2 \ \left( \left| H'V'^T \Psi \right| \right)^2 & \left( \left| H'V^T \Psi \right| \right)^2 & \left( \left| H'H'^T \Psi \right| \right)^2 & \left( \left| H'H^T \Psi \right| \right)^2 \ \left( \left| HV'^T \Psi \right| \right)^2 & \left( \left| HV^T \Psi \right| \right)^2 & \left( \left| HH'^T \Psi \right| \right)^2 & \left( \left| HH^T \Psi \right| \right)^2 \ \end{bmatrix} = \begin{pmatrix} 0.333 & 0 & 0.333 & 0.667 \ 0 & 0.083 & 0.167 & 0.083 \ 0.333 & 0.167 & 0 & 0.167 \ 0.667 & 0.083 & 0.167 & 0.75 \end{pmatrix} \nonumber$
Hardy's paradox is revealed by concentrating on four entries in the table.
$\begin{pmatrix} \text{Alice} & \text{Bob} & \text{Result} \ V & V' & \text{Never} \ V' & V & \text{Never} \ V & V & \text{Sometimes} \ H' & H' & \text{Never} \end{pmatrix} \nonumber$
In any run the detectors might be set to measure [H'-V']/[H-V] or [H-V]/[H'-V']. So if one photon triggers a [H-V] detector to register V, its partner must require a [H'-V'] detector to register H' according to the first two rows of the table above. According to the local hidden-variable model (EPR) this means H' is an element of reality.
It follows that any [H-V]/[H-V] run in which both detectors register V (probability 0.083) each photon must require a [H'-V'] detector to register H'. Therefore, if a [H'-V']/[H'-V'] run had been selected both detectors would have registered H'. However, given the initial photon state function the result H'H' should never be observed. In other words it is impossible to assign specific polarization states (instruction sets) to the photons prior to measurement that are in agreement with all quantum mechanical predictions for this thought experiment.
C. C. Gerry and K. M. Bruno, The Quantum Divide, Oxford U. P., 2013, pp. 124-29.
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This tutorial employs a tensor algebra approach to a gedanken experiment published by P. K. Aravind in the 2004 October issue of the American Journal of Physics. Aravind's thought experiment demonstrates how quantum entanglement leads directly to bizarre nonclassical correlations.
A source emits the following four-particle entangled state, with particles 1 and 3 going to Alice and particles 2 and 4 going to Bob. α and β are the eigenstates of the Pauli σz operator.
$\begin{matrix} \alpha = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \beta = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left( | \alpha \rangle_1 | \alpha \rangle_2 + | \beta \rangle_1 | \beta \rangle_2 \right) \otimes \frac{1}{ \sqrt{2}} \left( | \alpha \rangle_3 | \alpha \rangle_4 + | \beta \rangle_3 | \beta \rangle_4 \right) \ = \frac{1}{2} \left( | \alpha \rangle_1 | \alpha \rangle_2 | \alpha \rangle_3 | \alpha_4 + | \alpha \rangle_1 | \alpha \rangle_2 | \beta \rangle_3 | \beta \rangle_4 + | \beta \rangle_1 | \beta \rangle_2 | \alpha \rangle_3 | \alpha \rangle_4 + | \beta \rangle_1 | \beta \rangle_2 | \beta \rangle_3 | \beta \rangle_4 \right) \ = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}^T \ \Psi = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}^T \end{matrix} \nonumber$
Alice and Bob each have six measurement choices: R1, R2, R3, C1, C2, and C3. Each choice consists of a sequence of three measurements on the entangled spin pair they receive. These are shown in the table below. A sequence of measurements is possible because the operators in each row and each column mutually commute, as will be shown later.
$\begin{array}{|c| |c| |c| |c|} \hline \ ~ & C1 & C2 & C3 \ \hline \ R1 & I \otimes \sigma_z & \sigma_z \otimes I & \sigma_z \otimes \sigma_z \ \hline \ R2 & \sigma_x \otimes I & I \otimes \sigma_x & \sigma_x \otimes \sigma_x \ \hline \ R3 & \sigma_x \otimes \sigma_z & \sigma_z \otimes \sigma_x & \sigma_y \otimes \sigma_y \ \hline \end{array} \nonumber$
Alice and Bob independently and randomly set their detectors to one of the six possible settings each time the source emits the entangled particles, and record the result (+1 or -1) for each panel. After a statistically meaningful number of events they compare their results.
The operators required for this exercise are as follows:
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
Alice and Bob's measurement operators are constructed in tensor format:
$\begin{matrix} A(a,~b,~c,~d) = \text{kronecker(a, kronecker(b, kronecker(c, d)))} & B(a,~b,~c,~d) = \text{kronecker(a, kronecker(b, kronecker(c, d)))} \end{matrix} \nonumber$
Because Alice gets particles 1 and 3, and Bob particles 2 and 4, their measurement operators in tensor notation are as shown below.
Alice
$\begin{pmatrix} A(I,~I,~ \sigma_z,~I) & A( \sigma_z,~I,~I,~I) & A( \sigma_z,~I,~ \sigma_z,~ I) \ A( \sigma_x,~I,~I,~I) & A(I,~I,~ \sigma_x,~I) & A( \sigma_x,~I,~ \sigma_x,~I) \ A( \sigma_x,~I,~ \sigma_z,~I) & A( \sigma_z,~I,~\sigma_x,~I) & A( \sigma_y,~I,~ \sigma_y,~I) \end{pmatrix} \nonumber$
Bob
$\begin{pmatrix} B(I,~I,~I,~ \sigma_z) & B(I,~ \sigma_z,~ I, I) & B(I,~ \sigma_z,~ I,~ \sigma_z ) \ B(I,~ \sigma_x,~ I,~ I) & B(I,~ I,~ I,~ \sigma_x) & B(I,~ \sigma_x,~ I,~ \sigma_x) \ B(I,~ \sigma_x,~ I,~ \sigma_z) & B(I,~ \sigma_z,~I,~\sigma_z) & B(I, \sigma_y,~ I,~ \sigma_y) \end{pmatrix} \nonumber$
Where $A(I,~I,~ \sigma_z,~I)$ stands for $I \otimes I \otimes \sigma_z \otimes I$ and means Alice's operator for her particle is $I \otimes \sigma_z$. Using a representative row and column, we show that the measurement operators in the rows and columns of the measurement grid commute.
First row:
$\begin{pmatrix} A(I,~ I,~ \sigma_z,~I) A( \sigma_z,~I,~I~,I) - A( \sigma_z, ~I,~I~,I) A(I,~ I,~ \sigma_z,~I) \ A(I,~ I,~ \sigma_z,~I) A( \sigma_z,~I,~ \sigma_z~,I) - A( \sigma_z, ~I,~ \sigma_z~,I) A(I,~ I,~ \sigma_z,~I) \ A( \sigma_z,~ I,~ I,~I) A( \sigma_z,~I,~ \sigma_z~,I) - A( \sigma_z, ~I,~ \sigma_z~,I) A( \sigma_z,~ I,~ I,~I) \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \nonumber$
Third column:
$\begin{pmatrix} A( \sigma_z,~ I,~ \sigma_z,~I) A( \sigma_x,~I,~ \sigma_x~,I) - A( \sigma_x, ~I,~ \sigma_x~,I) A( \sigma_z,~ I,~ \sigma_z,~I) \ A( \sigma_z,~ I,~ \sigma_z,~I) A( \sigma_y,~I,~ \sigma_y~,I) - A( \sigma_y, ~I,~ \sigma_y~,I) A( \sigma_z,~ I,~ \sigma_z,~I) \ A( \sigma_x,~ I,~ \sigma_x,~I) A( \sigma_y,~I,~ \sigma_y~,I) - A( \sigma_y, ~I,~ \sigma_y~,I) A( \sigma_x,~ I,~ \sigma_x,~I) \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \nonumber$
These results establish the validity of doing sequential measurements in any row or column.
The eigenvalues of the individual operators in each panel are +/-1, and for Ψ the expectation values for the operators of the individual panels making up the rows and columns are zero. This is demonstrated for both Alice and Bob.
Alice
$\begin{pmatrix} \Psi^T A( I,~ I,~ \sigma_z,~I) \Psi & \Psi^T A( \sigma_z,~ I,~ I,~I) \Psi & \Psi^T A( \sigma_z,~ I,~ \sigma_z,~I) \Psi \ \Psi^T A( \sigma_x,~ I,~ I,~I) \Psi & \Psi^T A( I,~ I,~ \sigma_x,~I) \Psi & \Psi^T A( \sigma_x,~ I,~ \sigma_x,~I) \Psi \ \Psi^T A( \sigma_x,~ I,~ \sigma_z,~I) \Psi & \Psi^T A( \sigma_z,~ I,~ \sigma_x,~I) \Psi & \Psi^T A( \sigma_y,~ I,~ \sigma_y,~I) \Psi \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} \nonumber$
Bob
$\begin{pmatrix} \Psi^T B( I,~ I,~ I,~ \sigma_z) \Psi & \Psi^T B( I,~ \sigma_z,~ I,~I) \Psi & \Psi^T B( I,~ \sigma_z,~ I,~ \sigma_z) \Psi \ \Psi^T B( I,~ \sigma_x,~ I,~I) \Psi & \Psi^T B( I,~ I,~ I,~ \sigma_x) \Psi & \Psi^T B( I,~ \sigma_x,~ I,~ \sigma_x) \Psi \ \Psi^T B( I,~ \sigma_x,~ I,~ \sigma_z) \Psi & \Psi^T B( I,~ \sigma_z,~ I,~ \sigma_x) \Psi & \Psi^T B( I,~ \sigma_y,~ I,~ \sigma_y) \Psi \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} \nonumber$
These calculations indicate that the individual panels on the measurement grids will flash +1 and -1 with equal frequency. However, if Alice and Bob measure the same observable pair, they always obtain the same eigenvalue, suggesting at this point a classical correlation between their individual results.
$\begin{pmatrix} \Psi^T A( I,~ I,~ \sigma_z,~I) B( I,~ I,~ I,~ \sigma_z) \Psi & \Psi^T A( \sigma_z,~ I,~ I,~I) B( I,~ \sigma_z,~ I,~I) \Psi & \Psi^T A( \sigma_z,~ I,~ \sigma_z,~I) B( I,~ \sigma_z,~ I,~ \sigma_z) \Psi \ \Psi^T A( \sigma_x,~ I,~ I,~I) B( I,~ \sigma_x,~ I,~I) \Psi & \Psi^T A( I,~ I,~ \sigma_x,~I) B( I,~ I,~ I,~ \sigma_x) \Psi & \Psi^T A( \sigma_x,~ I,~ \sigma_x,~I) B( I,~ \sigma_x,~ I,~ \sigma_x) \Psi \ \Psi^T A( \sigma_x,~ I,~ \sigma_z,~I) B( I,~ \sigma_x,~ I,~ \sigma_z) \Psi & \Psi^T A( \sigma_z,~ I,~ \sigma_x,~I) B( I,~ \sigma_z,~ I,~ \sigma_x) \Psi & \Psi^T A( \sigma_y,~ I,~ \sigma_y,~I) B( I,~ \sigma_y,~ I,~ \sigma_y) \Psi \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} \nonumber$
This striking result appears to require that Alice and Bob's observables are "elements of reality" and therefore represent preexisting properties of their spin-1/2 particles. In other words, the particles carry instruction sets to their detectors which determine how the nine measurement panels respond. Not only that, it requires that the instruction sets for both detectors be identical.
However, the calculations below show that the expectation values for the sequence of measurements for rows 1, 2, 3, and columns 1 and 2 are +1. For column 3 the expectation value is -1.
Alice
$\begin{pmatrix} \Psi^T A( I,~ I,~ \sigma_z,~I) A( \sigma_z,~ I,~ I,~I) A( \sigma_z,~ I,~ \sigma_z,~I) \Psi & \Psi^T A( I,~ I,~ \sigma_z,~I) A( \sigma_x,~ I,~ I,~I) A( \sigma_x,~ I,~ \sigma_z,~I) \Psi \ \Psi^T A( \sigma_x,~ I,~ I,~I) A( I,~ I,~ \sigma_x,~I) A( \sigma_x,~ I,~ \sigma_x,~I) \Psi & \Psi^T A( \sigma_z,~ I,~ I,~I) A( I,~ I,~ \sigma_x,~I) A( \sigma_z,~ I,~ \sigma_x,~I) \Psi \ \Psi^T A( \sigma_x,~ I,~ \sigma_z,~I) A( \sigma_z,~ I,~ \sigma_x,~I) A( \sigma_y,~ I,~ \sigma_y,~I) \Psi & \Psi^T A( \sigma_z,~ I,~ \sigma_z,~I) A( \sigma_x,~ I,~ \sigma_x,~I) A( \sigma_y,~ I,~ \sigma_y,~I) \Psi\end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
Bob
$\begin{pmatrix} \Psi^T B( I,~ I,~ I,~ \sigma_z) B( I,~ \sigma_z,~ I,~I) B( I,~ \sigma_z,~ I,~ \sigma_z) \Psi & \Psi^T B( I,~ I,~ I,~ \sigma_z) B( I,~ \sigma_x,~ I,~I) B( I,~ \sigma_x,~ I,~ \sigma_z) \Psi \ \Psi^T B( I,~ \sigma_x,~ I,~I) B( I,~ I,~ I,~ \sigma_x) B( I,~ \sigma_x,~ I,~ \sigma_x) \Psi & \Psi^T B( I,~ \sigma_z,~ I,~I) B( I,~ I,~ I,~ \sigma_x) B( I,~ \sigma_z,~ I,~ \sigma_x) \Psi \ \Psi^T B( I,~ \sigma_x,~ I,~ \sigma_z) B( I,~ \sigma_z,~ I,~ \sigma_x) B( I,~ \sigma_y,~ I,~ \sigma_y) \Psi & \Psi^T B( I,~ \sigma_z,~ I,~ \sigma_z) B( I,~ \sigma_x,~ I,~ \sigma_x) B( I,~ \sigma_y,~ I,~ \sigma_y) \Psi \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & 1 \ 1 & -1 \end{pmatrix} \nonumber$
In order to validate a classical correlation between Alice and Bob's results based on the concept of "elements of reality", it is necessary to determine a set of instructions for the detectors that is in agreement with the highlighted results shown above. Unfortunately this is impossible; there is no way to assign definite eigenvalues (+1 or -1) to each of the nine panels of the measurement grid that satisfy these results.
This contradiction shows that there is no solution to our puzzle based on instruction sets. A willingness to accept the notion of instruction sets (or "elements of reality") to begin with, followed by the recognition that they cannot provide a solution to our puzzle, amounts to an informal appreciation of the central point of Bell's theorem. P. K. Aravind, AJP 72, 1305 (2004).
On the basis of this gedanken experiment we must reject the position that quantum entities, quons (thank you Nick Herbert), have well-defined properties independent of measurement; in other words, that measurement simply reveals a preexisting state. Quantum entanglement is, as Schrödinger noted in 1935, "...the essential trait of the new theory, the one which forces a complete departure from all classical concepts."
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A source emits the following six cubit state (|0> and |1> are orthogonal states), with cubits 1,3 and 5 going to Alice, and cubits 2, 4 and 6 going to Bob. This example is due to P. K. Aravind and available at arXiv:quant-ph/0701031v1.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} (| 0 \rangle_1 |0 \rangle_2 + |1 \rangle_1 |1 \rangle_2 ) \otimes \frac{1}{ \sqrt{2}} ( |0 \rangle_3 |0 \rangle_4 + |1 \rangle_3 |1 \rangle_4 ) \otimes \frac{1}{ \sqrt{2}} (|0 \rangle_5 |0 \rangle_6 + |1 \rangle_5 |1 \rangle_6) = \frac{1}{ 2 \sqrt{2}} \begin{bmatrix} |0 \rangle_1 |0 \rangle_2 |0 \rangle_3 |0 \rangle_4 |0 \rangle_5 |0 \rangle_6 + |0 \rangle_1 |0 \rangle_2 |1 \rangle_3 |1 \rangle_4 |0 \rangle_5 |0 \rangle_6 + |1 \rangle_1 |1 \rangle_2 |0 \rangle_3 |0 \rangle_4 |0 \rangle_5 |0 \rangle_6 + |1 \rangle_1 |1 \rangle_2 |1 \rangle_3 |1 \rangle_4 |0 \rangle_5 |0 \rangle_6 \ |0 \rangle_1 |0 \rangle_2 |0 \rangle_3 |0 \rangle_4 |1 \rangle_5 |1 \rangle_6 + |0 \rangle_1 |0 \rangle_2 |1 \rangle_3 |1 \rangle_4 |1 \rangle_5 |1 \rangle_6 + |1 \rangle_1 |1 \rangle_2 |0 \rangle_3 |0 \rangle_4 |1 \rangle_5 |1 \rangle_6 + |1 \rangle_1 |1 \rangle_2 |1 \rangle_3 |1 \rangle_4 |1 \rangle_5 |1 \rangle_6 \end{bmatrix} \nonumber$
Ψ can be written in condensed binary format as,
$| \Psi \rangle = \frac{1}{ 2 \sqrt{2}} \left[ |000000 \rangle + |001100 \rangle + |110000 \rangle + |111100 \rangle + |000011 \rangle + |001111 \rangle + |110011 \rangle +|111111 \rangle \right] \nonumber$
In decimal notation the kets contain 0, 12, 48, 60, 3, 15, 51, and 63. All other vector elements are zero.
$\begin{matrix} i = 0 ... 63 & \Psi_i = 0 \end{matrix} \nonumber$
$\begin{matrix} \Psi_0 = \frac{1}{ 2 \sqrt{2}} & \Psi_3 = \frac{1}{2 \sqrt{2}} & \Psi_{12} = \frac{1}{2 \sqrt{2}} & \Psi_{15} = \frac{1}{2 \sqrt{2}} \ \Psi_{48} = \frac{1}{2 \sqrt{2}} & \Psi_{51} = \frac{1}{2 \sqrt{2}} & \Psi_{60} = \frac{1}{2 \sqrt{2}} & \Psi_{63} = \frac{1}{2 \sqrt{2}} \end{matrix} \nonumber$
The following pentagram describes the measurement protocols followed by Alice and Bob.
Alice and Bob independently and randomly select one of five measurement protocols (E1, E2, E3, E4 and E5) shown on the edges of the pentagram above each time the source emits the entangled particles, and record the result (+1 or -1) for each vertex. After a statistically meaningful number of events they compare their results.
Each vertex represents a three qubit measurement sequence shown above and given in the following table. Note that Alice's sequence uses the identity operator for photons 2, 4 and 6 because she receives photons 1, 3 and 5, while Bob uses the identity operator for photons 1, 3 and 5 because he receives photons 2, 4 and 6.
$\begin{pmatrix} \text{Operator} & \text{Alice} & \text{Bob} \ 1 & A( - \sigma_z,~ I,~ \sigma_z,~I,~ \sigma_z,~I) & B(I,~ - \sigma_z,~I,~ \sigma_z,~I,~ \sigma_z) \ 2 & A( \sigma_z,~ I,~ I,~I,~ I,~I) & B(I,~ \sigma_z,~I,~ I,~I,~ I) \ 3 & A( I,~ I,~ \sigma_x,~I,~ I,~I) & B(I,~ I,~I,~ \sigma_x,~I,~ I) \ 4 & A(I,~ I,~ I,~I,~ \sigma_z,~I) & B(I,~ I,~I,~ I,~I,~ \sigma_z) \ 5 & A(I,~ I,~ \sigma_z,~I,~ I,~I) & B(I,~ I,~I,~ \sigma_z,~I,~ I) \ 6 & A( \sigma_x,~ I,~ I,~I,~ I,~I) & B(I,~ \sigma_x,~I,~ I,~I,~ I) \ 7 & A( - \sigma_x,~ I,~ \sigma_x,~I,~ \sigma_x,~I) & B(I,~ - \sigma_x,~I,~ \sigma_x,~I,~ \sigma_z) \ 8 & A( - \sigma_x,~ I,~ \sigma_z,~I,~ \sigma_x,~I) & B(I,~ - \sigma_x,~I,~ \sigma_z,~I,~ \sigma_x) \ 9 & A(I,~ I,~ I,~I,~ \sigma_x,~I) & B(I,~ I,~I,~ I,~I,~ \sigma_x) \ 10 & A( - \sigma_z,~ I,~ \sigma_x,~I,~ \sigma_x,~I) & B(I,~ - \sigma_z,~I,~ \sigma_x,~I,~ \sigma_x) \ \end{pmatrix} \nonumber$
The measurement operators required are:
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \text{eigenvals}( \sigma_x) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvals} ( \sigma_z) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{A(a, b, c, d, e, f)} = \text{kronecker(a, kronecker(b, kronecker(c, kronecker(d, kronecker(e, f)))))} \ \text{B(a, b, c, d, e, f)} = \text{kronecker(a, kronecker(b, kronecker(c, kronecker(d, kronecker(e, f)))))} \end{matrix} \nonumber$
The eigenvalues of the Pauli operators are +/- 1. The first thing to note is that the individual vertices (measurement sites) flash +1 and -1 one randomly giving expectation values of zero at each vertex.
$\begin{pmatrix} \Psi^T A( - \sigma_z,~ I,~ \sigma_z,~I,~ \sigma_z,~I) \Psi & \Psi^T B(I,~ - \sigma_z,~I,~ \sigma_z,~I,~ \sigma_z) \Psi \ \Psi^T A( \sigma_z,~ I,~ I,~I,~ I,~I) \Psi & \Psi^T B(I,~ \sigma_z,~I,~ I,~I,~ I) \Psi \ \Psi^T A( I,~ I,~ \sigma_x,~I,~ I,~I) \Psi & \Psi^T B(I,~ I,~I,~ \sigma_x,~I,~ I) \Psi \ \Psi^T A(I,~ I,~ I,~I,~ \sigma_z,~I) \Psi & \Psi^T B(I,~ I,~I,~ I,~I,~ \sigma_z) \Psi \ \Psi^T A(I,~ I,~ \sigma_z,~I,~ I,~I) \Psi & \Psi^T B(I,~ I,~I,~ \sigma_z,~I,~ I) \Psi \ \Psi^T A( \sigma_x,~ I,~ I,~I,~ I,~I) \Psi & \Psi^T B(I,~ \sigma_x,~I,~ I,~I,~ I) \Psi \ \Psi^T A( - \sigma_x,~ I,~ \sigma_x,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ - \sigma_x,~I,~ \sigma_x,~I,~ \sigma_z) \Psi \ \Psi^T A( - \sigma_x,~ I,~ \sigma_z,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ - \sigma_x,~I,~ \sigma_z,~I,~ \sigma_x) \Psi \ \Psi^T A(I,~ I,~ I,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ I,~I,~ I,~I,~ \sigma_x) \Psi \ \Psi^T A( - \sigma_z,~ I,~ \sigma_x,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ - \sigma_z,~I,~ \sigma_x,~I,~ \sigma_x) \Psi \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
However, if Alice and Bob chose the same measurement protocol they always get the same eigenvalue, as is shown by the calculations below.
$\begin{pmatrix} \Psi^T A( - \sigma_z,~ I,~ \sigma_z,~I,~ \sigma_z,~I) \Psi & \Psi^T B(I,~ - \sigma_z,~I,~ \sigma_z,~I,~ \sigma_z) \Psi \ \Psi^T A( \sigma_z,~ I,~ I,~I,~ I,~I) \Psi & \Psi^T B(I,~ \sigma_z,~I,~ I,~I,~ I) \Psi \ \Psi^T A( I,~ I,~ \sigma_x,~I,~ I,~I) \Psi & \Psi^T B(I,~ I,~I,~ \sigma_x,~I,~ I) \Psi \ \Psi^T A(I,~ I,~ I,~I,~ \sigma_z,~I) \Psi & \Psi^T B(I,~ I,~I,~ I,~I,~ \sigma_z) \Psi \ \Psi^T A(I,~ I,~ \sigma_z,~I,~ I,~I) \Psi & \Psi^T B(I,~ I,~I,~ \sigma_z,~I,~ I) \Psi \ \Psi^T A( \sigma_x,~ I,~ I,~I,~ I,~I) \Psi & \Psi^T B(I,~ \sigma_x,~I,~ I,~I,~ I) \Psi \ \Psi^T A( - \sigma_x,~ I,~ \sigma_x,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ - \sigma_x,~I,~ \sigma_x,~I,~ \sigma_z) \Psi \ \Psi^T A( - \sigma_x,~ I,~ \sigma_z,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ - \sigma_x,~I,~ \sigma_z,~I,~ \sigma_x) \Psi \ \Psi^T A(I,~ I,~ I,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ I,~I,~ I,~I,~ \sigma_x) \Psi \ \Psi^T A( - \sigma_z,~ I,~ \sigma_x,~I,~ \sigma_x,~I) \Psi & \Psi^T B(I,~ - \sigma_z,~I,~ \sigma_x,~I,~ \sigma_x) \Psi \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \end{pmatrix} \nonumber$
This result is somewhat surprising given the result immediately above. It suggests (at this point) that the particles involved in this experiment carry instruction sets telling the detectors how to operate, and that Alice and Bob's detectors receive the same sets of instructions.
The edges (E1 through E5) shown on the pentagram identify a sequence of four mutually commuting operators (see Appendix). The net eigenvalues for these sequences of measurements for Alice and Bob are now calculated.
$\begin{pmatrix} \Psi^T A( - \sigma_z,~ I,~ \sigma_z,~I,~ \sigma_z,~I) A( \sigma_z,~ I,~ I,~I,~ I,~I) A( I,~ I,~ I,~I,~ \sigma_z,~I) A(I,~ I,~ \sigma_z,~I,~ I,~I) \Psi \ \Psi^T A(I,~ I,~ \sigma_x,~I,~ I,~I) A(I,~ I,~ I,~I,~ \sigma_z,~I) A( \sigma_x,~ I,~ I,~I,~ I,~I) A( -\sigma_x,~ I,~ \sigma_x,~I,~ \sigma_z,~I) \Psi \ \Psi^T A(I,~ I,~ \sigma_x,~I,~ I,~I) A( \sigma_x,~ I,~ I,~I,~ I,~I) A( - \sigma_x,~ I,~ \sigma_x,~I,~ \sigma_x,~I) A(I,~ I,~ I,~I,~ \sigma_x,~I) \Psi \ \Psi^T A(I,~ I,~ \sigma_z,~I,~ I,~I) A( \sigma_x,~ I,~ I,~I,~ I,~I) A( - \sigma_x,~ I,~ \sigma_z,~I,~ \sigma_x,~I) A(I,~ I,~ I,~I,~ \sigma_x,~I) \Psi \ \Psi^T A(- \sigma_z,~ I,~ \sigma_z,~I,~ \sigma_z,~I) A( \sigma_z,~ I,~ \sigma_x,~I,~ \sigma_x,~I) A( - \sigma_x,~ I,~ \sigma_z,~I,~ \sigma_x,~I) A(- \sigma_x,~ I,~ \sigma_x,~I,~ \sigma_z,~I) \Psi \ \end{pmatrix} = \begin{pmatrix} -1 \ -1 \ -1 \ -1 \ -1 \end{pmatrix} \nonumber$
$\begin{pmatrix} \Psi^T B(I,~ - \sigma_z,~I,~ \sigma_z,~I,~ \sigma_z) & B(I,~ \sigma_z,~I,~I,~I,~I) & B(I,~I,~I,~I,~I,~ \sigma_z) & B(I,~I,~I,~ \sigma_z,~I,~I) \ \Psi^T B(I,~I,~I,~ \sigma_x,~I,~I) & B(I,~I,~I,~I,~I,~ \sigma_z) & B(I,~ \sigma_x,~I,~I,~I,~I) & B(I,~ - \sigma_x,~I,~ \sigma_x,~I,~ \sigma_z) \ \Psi^T B(I,~I,~I,~ \sigma_x,~I,~I) & B(I,~ \sigma_z,~I,~I,~I,~I) & B(I,~ - \sigma_z,~I,~ \sigma_z,~I,~ \sigma_x) & B(I,~ I,~I,~ I,~I,~ \sigma_x) \ \Psi^T B(I,~I,~I,~ \sigma_z,~I,~I) & B(I,~ \sigma_x,~I,~I,~I,~I) & B(I,~ - \sigma_x,~I,~ \sigma_z,~I,~ \sigma_x) & B(I,~ I,~I,~ I,~I,~ \sigma_x) \ \Psi^T B(I,~ - \sigma_z,~I,~ \sigma_z,~I,~ \sigma_z) & B(I,~ - \sigma_z,~I,~ \sigma_x,~I,~ \sigma_x) & B(I,~ - \sigma_x,~I,~ \sigma_z,~I,~ \sigma_x) & B(I,~ - \sigma_x,~I,~ \sigma_x,~I,~ \sigma_z) \ \end{pmatrix} \begin{pmatrix} -1 \ -1 \ -1 \ -1 \ -1 \end{pmatrix} \nonumber$
We now have the problem of reconciling these results with those immediately prior. How is it possible to assign +1s and -1s to the measurement vertices such that they satisfy the results immediately above. The answer is that it is not possible. The ideas of instruction sets and elements of reality are not capable of explaining these results.
For this assignment of vertex eigenvalues we see that the composite eigenvalue (-1) is satisfied for E1, E2, E3 and E4, but is violated for E5. All attempts to assign +/-1 values to the vertices fail to satisfy the composite eigenvalue for one of the measurement protocols.
Appendix
The four operators comprising the five measurement protocols mutually commute. This is demonstrated below for the E1 protocol.
$\begin{matrix} ( - \sigma_z \sigma_z \sigma_z ) ( \sigma_z \text{I I} ) - (\sigma_z \text{I I})( -\sigma_z \sigma_z \sigma_z) = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} & ( - \sigma_z \sigma_z \sigma_z ) (\text{I I} \sigma_z) - (\text{I I} \sigma_z)( -\sigma_z \sigma_z \sigma_z) = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \ ( - \sigma_z \sigma_z \sigma_z ) (\text{I} \sigma_z \text{I} ) - (\text{I} \sigma_z \text{I})( -\sigma_z \sigma_z \sigma_z) = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} & ( \sigma_z \text{I I}) (\text{I I} \sigma_z) - (\text{I I} \sigma_z)( \sigma_z \text{I I}) = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \ ( \sigma_z \text{I I}) (\text{I} \sigma_z \text{I} ) - (\text{I} \sigma_z \text{I})( \sigma_z \text{I I}) = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} & ( \text{I I} \sigma_z) (\text{I} \sigma_z \text{I}) - (\text{I} \sigma_z \text{I})( \text{I I} \sigma_z) = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
It is also the case that the measurement protocols commute. This is demonstrated for E1 and E5.
$\left[ ( - \sigma_z \sigma_z \sigma_z) (\sigma_z \text{I I} ( \text{I I} \sigma_z) ( \text{I} \sigma_z \text{I}) \right] \left[ ( - \sigma_z \sigma_z \sigma_z ) (- \sigma_z \sigma_x \sigma_x ) ( - \sigma_x \sigma_z \sigma_x )( - \sigma_x \sigma_x \sigma_z) \right] ... + - \left[ ( -\sigma_z \sigma_z \sigma_z) ( - \sigma_z \sigma_x \sigma_x) ( -\sigma_x \sigma_z \sigma_x) ( - \sigma_x \sigma_x \sigma_z ) \right] \left[ ( - \sigma_z \sigma_z \sigma_z) ( \sigma_z \text{I I}) ( \text{I I} \sigma_z) ( \text{I} \sigma_z \text{I}) \right] = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.28%3A_Nonclassial_Correlations_Revealed_with_Mermin%27s_Pentagram.txt
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Quantum theory is a very successful mathematical description of the structure of the atomic and molecular world. But unlike other scientific theories of wide applicability, its interpretation and meaning have been vigorously debated since its inception over seventy years ago. To clarify the unusual stature of quantum theory it might be helpful first to make a list of some of the characteristics we associate with a successful scientific theory.
• Experimental confirmation; predictive success
• Explanatory power; intelligibility
• Internal coherence; aesthetic appeal
• Breadth of applicability; philosophic implications
Scientists agree that quantum theory has received ample experimental confirmation, and that it is a coherent and mathematically appealing theory. In fact quantum mechanics has provided the computational methods and conceptual infra-structure that have catalyzed the impressive progress the physical sciences have experienced in the 20th century. However, in spite of its experimental validation, many would argue that quantum mechanics does not explain in the traditional sense, that it is not really intelligible, and that it merely offers a mathematical algorithm for making experimental predictions. And, if it has broad philosophic implications they are strange and unsettling to most people.
Basic Postulates of Quantum Mechanics
Before proceeding to the main issue, which is the analysis of an important experiment, it might be helpful to review some of the basic postulates of quantum theory.
1. The wave function, Y, provides a complete description of the physical properties of the system it represents.
2. Systems with the same wave function are identical and indistinguishable.
3. For every observable quantity there is an associated operator.
4. If the wave function is an eigenfunction of an operator the observable quantity represented by the operator has a definite value. Every time a measurement for the observable quantity is made the same value is obtained. $\hat{O} | \Psi \rangle = o | \Psi \rangle \nonumber$
5. If the wave function is not an eigenfunction of a particular operator the system is in a state which does not possess a definite value for the observable quantity associated with that operator. In this case quantum mechanics does not predict with certainty the results of a measurement, but only allows one to calculate (a) the "average value" when many determinations are made, or (b) predict the probability that a particular result will occur. For example, the two equations that follow show how to calculate the average position for a particle in state |Y> and the probability that the particle will be found in the spatial interval from x to x + dx. $\begin{matrix} a) & \langle x \rangle = \langle \Psi | \hat{X} | \Psi \rangle \ b) & \Psi (x)^2 dx \end{matrix} \nonumber$ Richard Feynman captured the significance of this postulate in the following quotation. "A philosopher once said, 'It is necessary for the very existence of science that the same conditions always produce the same results.' Well, they don't!"
6. After a measurement the wave function of the system is an eigenfunction of the measurement operator.
• As Pascal Jordan has remarked, "Observations not only disturb what has to be measured, they produce it ... We compel the electron to assume a definite position ... We ourselves produce the result of the experiment."
7. Quantum mechanics does not provide a description of what actually happens between two consecutive observations.
• As Heisenberg has succinctly put it, "If we want to describe what happens in an atomic event, we have to realize that the word 'happens' can apply only to the observation, not to the state of affairs between two observations."
The indeterministic nature of quantum mechanics is inherent in the basic principles listed above. Quantum theory says that a system can be in a well-defined state described by its wave function and yet the results of subsequent experiments are not necessarily uniquely determined. For example, if the wave function of a system prior to a measurement is not an eigenfunction of the measurement operator, then a certain prediction of the measurement outcome cannot be made.
This is fundamentally different from the classical view in which an experimental result is always uniquely determined by the state of the system and the forces acting on it. According to classical physics, prior to measurement, every property of a system possesses a definite but unknown value, and experiment or observation merely reveals the value of that previously unknown property. By comparison, quantum mechanics denies the existence of certain physical properties independent of measurement. However, it does not deny, as is frequently erroneously claimed, the existence of the physical object itself in the absence of observation.
What is the reaction to such quantum weirdness? Here are some quotations from those who contributed to the foundation of quantum theory.
• Any one who is not shocked by quantum mechanics has not fully understood it. - Bohr
• I think it is safe to say that no one understands quantum mechanics. Do not keep saying to yourself, if you can possibly avoid it, 'But how can it possibly be like that?' because you will go down the drain into a blind alley from which nobody has yet escaped. Nobody know show it can be like that. - Feynman
• We have always had a great deal of difficulty understanding the world view that quantum mechanics represents. At least I do, because I'm an old enough man that I haven't got to the point that this stuff is obvious to me. Okay, I still get nervous with it... You know how it always is, every new idea, it takes a generation or two until it becomes obvious that there's no real problem. I cannot define the real problem, therefore I suspect that there is no real problem, but I'm not sure there's no real problem. - Feynman
• Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the real thing. The theory says a lot, but does not really bring us any closer to the secret of the "old one." I, at any rate, am convinced that He is not playing at dice. - Einstein
• It seems hard to look at God's cards. But I cannot for a moment believe that he plays dice and makes use of 'telepathic' means as the current quantum theory alleges He does. - Einstein
• The mathematical predictions of quantum mechanics yield results that are in agreement with experimental findings. That is the reason we use quantum theory. That quantum theory fits experiment is what validates the theory, but why experiment should give such peculiar results is a mystery. This is the shock to which Bohr referred. - Marvin Chester
However strange it may seem, quantum theory has survived all the experimental tests it has been subjected to. Therefore, if quantum theory is strange, it is because the nano-world of electrons, atoms, and molecules is strange. Quantum mechanics is an accurate mathematical account of the behavior of the physical world at the nano scale.
In what follows an important experiment, stimulated by a thought experiment designed originally by Einstein, Podolsky, and Rosen in 1935, is described. Classical and quantum mechanical interpretations of the results and are offered. The classical interpretation fails to account fully for the experimental results, while the quantum mechanical analysis is in complete agreement with the experiment.
Apparatus
A schematic representation of an apparatus is shown in the figure below in which laser radiation is used to raise calcium atoms to an excited state. On returning to the ground state the calcium atoms emit two photons in opposite directions. The polarization of the photons is determined with a combination of polarizing film and a detector as illustrated in the figure. The polarizing films are oriented randomly in three directions as shown. The relative orientation of the polarizers is 60o.
The polarizers 1, 2, and 3 are oriented at angles of -60o, 0o, and 60o, and selected randomly. Thus the angle between any two polarizers is 60o.
Experiment
As the calcium atoms are excited by the laser radiation the polarizing films are oriented in a random way, and the detectors record whether or not a photon has arrived. The coincidence counter shown in the figure records whether or not both photons passed through the polarizing films. Note that there are nine possible orientations of the polarizing films: 11, 22, 33, 12, 21, 23, 32, 13, 31.
Results
Case 1. In runs for which the polarizing films are oriented at the same angle (11, 22, 33) the photons behave the same: they both either pass through the polarizers or they are both stopped by the polarizers. In other words, if the left detector registers a photon, so does the right detector, and vice versa.
Case 2. In runs for which the polarizing films are oriented at different angles (12, 21, 23, 32, 13, 31) the photons behave in the same way only 25% of the time. That is they both pass the films 25% of the time, while the other 75% of the time one passes and the other doesn't.
Overall. Because in 1/3 of the runs the photons behave the same and in 2/3 of the runs they behave the same way only 25% of the time, in a large number of runs it is found that the photons behave the same way 50% of the time.
Classical Interpretation
Belief in the doctrine of realism is considered by most scientists to be a necessary pre-condition for scientific inquiry. Realism, for the present purposes, might be described briefly as the philosophical belief that an experimental result is determined by a physical property of the system under study whose existence is independent of the experiment or the experimenter.
Thus, the results reported in Case 1 indicate that the photons have the same polarization. In all cases in which the film orientations are the same both photons either pass or do not pass through the polarizing films.
This can lead one to assume that an intrinsic property of a photon is its ability to pass the polarizing film at the three angles chosen. If this assumption is correct, and it convincingly and simply accounts for the Case 1 data, then eight states are required to account for the behavior of the photons. Each photon can come in eight varieties: YYY, YYN, YNY, NYY, YNN, NYN, NNY, NNN. Here Y (yes) and N (no) specify whether or not the photon will pass through the polarizing film when it is oriented at angles 1, 2, or 3. This view that photons or other systems under study have intrinsic properties which are independent of their measurement, but determine the result of a measurement, is one of the cornerstones of science.
The belief that photons carry information on their polarization along the three orientations of the films would yield the results shown in the table below. There are eight photon states and nine film orientations. Thus there are 72 types of encounters between the photons and the polarizing films as recorded by the detectors and the coincidence counter. The realist position (the one adopted by Einstein, for example) demands that the photons behave the same way 67% of the time (48/72), as is shown in the table. This is clearly not in agreement with the experimental result that overall the photons behave the same way only 50% of the time.
$\begin{array}{|c | c | c | c | c | c | c | c | c | c|} \hline ~ & 11 & 22 & 33 & 12 & 21 & 13 & 31 & 23 & 32 \ \hline \text{YYY} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} \ \hline \text{YYN} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Different} & \text{Different} & \text{Different} & \text{Different} \ \hline \text{YNY} & \text{Same} & \text{Same} & \text{Same} & \text{Different} & \text{Different} & \text{Same} & \text{Same} & \text{Different} & \text{Different} \ \hline \text{NNY} & \text{Same} & \text{Same} & \text{Same} & \text{Different} & \text{Different} & \text{Different} & \text{Different} & \text{Same} & \text{Same} \ \hline \text{YNN} & \text{Same} & \text{Same} & \text{Same} & \text{Different} & \text{Different} & \text{Different} & \text{Different} & \text{Same} & \text{Same} \ \hline \text{NYN} & \text{Same} & \text{Same} & \text{Same} & \text{Different} & \text{Different} & \text{Same} & \text{Same} & \text{Different} & \text{Different} \ \hline \text{NNY} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Different} & \text{Different} & \text{Different} & \text{Different} \ \hline \text{NNN} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} & \text{Same} \ \hline \end{array} \nonumber$
Note that the realist position is in agreement with Case 1 results, but in serious disagreement with Case 2 results. For Case 2 it predicts that 50% of the time the behavior of the photons is the same, when the experimental result is that the photons behave the same only 25% of the time.
Quantum Interpretation
According to quantum theory a photon polarized at some angle q to the vertical direction can be written as a linear combination of a vertically and horizontally polarized photon.
$| \Theta \rangle = | v \rangle \langle v | \Theta \rangle + | h \rangle \langle h | \Theta \rangle = | v \rangle \cos \theta + | h \rangle \sin \Theta \nonumber$
Figure 2. $| \langle v | \theta \rangle |^2 = \cos^2 \theta \nonumber$
Thus the probability that a photon polarized at an angle q will pass a film with vertical polarization is cos2q. In general we can state that the probability that a photon will pass a polarizer is given by the square of the cosine of the angle its plane of polarization makes with the direction of the polarization of the film.
Now to a quantum mechanical explanation for the results of the experiment. For Case 1results, the fact that the photons behave in the same way when the detector settings are the same (11, 22, 33) indicates that the photons have the same polarization and, therefore, the same wave function. They are in the same state. This is basically the same as the classical interpretation.
However, for the Case 2 results where the detector settings are different (12, 21, 23, 32, 13, 31) quantum mechanics gives a different explanation. Let's take a specific example. First of all, as Figure 1 shows, the three polarizing films have relative angles of 60o. Now suppose for the 12 detector orientation that photon A moving to the left passes through its film indicating that it is polarized in that direction. Photon moving to the right has the same polarization and, therefore, the same wave function. It is approaching a film that is polarized at an angle of 60oto the one that A just passed. Thus, the probability that B will pass its film is cos260 or 25% in agreement with the Case 2 experimental results described earlier.
Let's take a look at another example. Photon A is absorbed by a vertical polarizer indicating that it is horizontally polarized. Since the photons are in the same polarization state, B is also horizontally polarized. The probability it will pass a polarizer oriented 60o relative to vertical is,
$| \langle \theta = 60^o | h \rangle |^2 = \sin^2 60^o = 0.75 \nonumber$
Thus, the probability that photons A and B behave differently (one is absorbed by its polarizer, the other passes its polarizer) is 0.75, again in agreement with Case 2 results.
Of course this also means that the quantum interpretation is in agreement with the overall result that the photons behave the same way 50% of the time. The films have the same setting 1/3 of time and different settings 2/3 of the time. When the films are set the same the photons behave the same way 100% of the time (either they both pass or they both don't pass). When the film settings are different the photons behave the same way only 25% of the time [(1/3) 100% + (2/3) 25% = 50%].
References:
• N. D. Mermin, "Spooky Actions at a Distance: Mysteries of the Quantum Theory," The Great Ideas of Today, Encyclopedia Britannica, Chicago, 1988, pp 2-53.
• N. D. Mermin, "Bringing home the atomic world: Quantum mysteries for anybody," American Journal of Physics, 49, 940-943, (1981).
• N. D. Mermin, "Is the moon there when nobody looks," Physics Today , 38(4), 38-47, (1985).
• N. D. Mermin, "Quantum mysteries revisited," American Journal of Physics, 58, 731-734, (1990).
• J. C. Polkinghorne, The Quantum World, Princeton Science Library, Princeton, N.J.,1984.
• A. I. M Rae, Quantum Mechanics, 3rd Ed., Institute of Physics Publishing, Bristol, UK, 1992.
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Quantum theory is both stupendously successful as an account of the small‐scale structure of the world and it is also the subject of an unresolved debate and dispute about its interpretation. J. C. Polkinghorne, The Quantum World, p. 1.
In 1951 David Bohm proposed a gedanken experiment that further illuminated the conflict between local realism and quantum mechanics first articulated by Einstein, Podolsky and Rosen (EPR) in 1935. In this thought experiment a spin‐1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions on the y‐axis to a pair of observers set up to measure spin in either the x‐ or z‐direction.
In this summary tensor algebra will be used to analyze Bohmʹs thought experiment. The vector states and matrix operators required are provided below.
Spin eigenvectors:
$\begin{matrix} S_{zu} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & S_{zd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & S_{xu} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & S_{xd} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
Spin operators in units of h/4π:
$\begin{matrix} S_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & S_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
According to established quantum principles the singlet state for fermions is an entangled superposition written as follows in tensor format in the z-direction spin axis.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \left| \uparrow \rangle_1 \right| \downarrow \rangle_2 - \left| \downarrow \rangle_1 \right| \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
First we calculate the expectation values for spin measurements in the z‐ and x‐directions using the eigenvectors of the respective spin operators. We see that spin‐up yields +1 and spin‐down ‐1.
$\begin{matrix} S_{zu}^T S_z S_{zu} = 1 & S_{zd}^T S_z S_{zd} = -1 & S_{xu}^T S_x S_{xu} = 1 & S_{xd}^T S_x S_{xd} = -1 \end{matrix} \nonumber$
Consequently the expectation value observed when both observers jointly measure the same spin direction is ‐1, because in the singlet state the spins have opposite orientations (kronecker performs the tensor multiplication of two matrices, as illustrated in the Appendix).
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~S_z) \Psi = -1 & \Psi^T \text{kronecker}(S_x,~ S_x) \Psi = -1 \end{matrix} \nonumber$
Note that while the entangled singlet spin state shown above is written in the z‐basis, it is the same to an over‐all phase using the x‐direction eigenvectors as is demonstrated in the Appendix. Next we show that in spite of the strong correlation shown above, individual spin measurements are totally random yielding expectation values of zero in both the z‐ and x‐directions.
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~ S_z) \Psi = 0 \ \Psi^T \text{kronecker}(S_x,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~ S_x) \Psi = 0 \end{matrix} \nonumber$
Quantum mechanics predicts the following results when the observers make different spin measurements on the particles.
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~S_x) \Psi = 0 & \Psi^T \text{kronecker}(S_x,~ S_z) \Psi = 0 \end{matrix} \nonumber$
From a classical realist position the results for all the previous quantum calculations can be explained by assigning specific x‐ and z‐spin states to the particles, as shown in the table below on the left (see Townsendʹs A Modern Approach to Quantum Mechanics, page 135). Each particle can be in any one of four equally probable spin states, and taken together the particles form four equally probable joint spin states. In other words, the particles are in well‐defined, although unknown, spin states prior to measurement. Measurement, according to a realist, simply reveals a pre‐existing state. The right‐hand part of the table gives the joint measurement results expected given the spin states specified on the left. The bottom row gives the expectation (average) values under the assumptions stated above.
$\begin{pmatrix} \text{Particle 1} & \text{Particle 2} & ' & S_z(1) S_z(2) & S_x(1) S_x(2) & S_z(1) S_x(2) & S_x(1) S_z(2) \ S_{zu} S_{xu} & S_{zd} S_{xd} & ' & -1 & -1 & -1 & -1 \ S_{zu} S_{xd} & S_{zd} S_{xu} & ' & -1 & -1 & 1 & 1 \ S_{zd} S_{xu} & S_{zu} S_{xd} & ' & -1 & -1 & 1 & 1 \ S_{zd} S_{xd} & S_{zu} S_{xu} & ' & -1 & -1 & -1 & -1 \ \text{Expectation} & \text{Value} & ' & -1 & -1 & 0 & 0 \end{pmatrix} \nonumber$
At this point one may ask, ʺWhereʹs the problem? The quantum and classical pictures are in agreement on the prediction of experimental results.ʺ The difficulty is that quantum mechanics does not accept the legitimacy of the states shown in the table on the left. One way to state the problem is to note that Sx and Sz are noncommuting operators. This means that according to quantum mechanics spin in the x‐ and z‐directions cannot simultaneously have well‐defined values (like position and momentum, they are conjugate observables).
The Sx-Sz commutator:
$S_z S_x - S_x S_z = \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix} \nonumber$
For example, if measurement reveals that particle 1 has Szu then particle 2 is definitely Szd. But that means it cannot have a well‐defined value for the x‐direction spin. In fact Szd is a superposition of Sxu and Sxd, as shown below, with <Sx> = 0.
$\begin{matrix} S_{zd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \frac{1}{ \sqrt{2}} (S_{xu} - S_{xd}) = \begin{pmatrix} 0 \ 1 \end{pmatrix} & S_{sd}^T S_x S_{zd} = 0 \end{matrix} \nonumber$
Or, suppose particle 1 is found to have Sxd, then particle 2 is Sxu which is a superposition of spin up and down in the z‐direction and therefore <Sx> = 0.
$\begin{matrix} S_{xu} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & \frac{1}{ \sqrt{2}} (S_{zu} + S_{zd}) = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & S_{xu}^T S_z S_{xu} = 0 \end{matrix} \nonumber$
To summarize, the states in the table are not valid, in spite of their agreement with the predictions of quantum mechanics, because they give well‐defined values to incompatible (according to quantum theory) observables. The quantum objection to the classical spin states can also be expressed using the Uncertainty Principle. For a particle known to have spin‐up in the z‐direction, quantum theory requires that its x‐direction spin be uncertain, as shown by the following calculation.
$\begin{matrix} \Delta S_x = \sqrt{S_{zu}^T S_x^2 S_{zu} - (S_{zu}^T S_x S_{zu})^2} & \Delta S_x = 1 ~ \text{in units of}~ \frac{h}{4 \pi} \end{matrix} \nonumber$
Of course, the realist says that these arguments simply indicate that quantum mechanics is not a complete theory because it cannot assign definite values to all elements of reality independent of measurement.
While Bohmʹs 1951 gedanken experiment clarified the conflict between quantum theory and classical realism, it did not provide for a direct experimental adjudication of the disagreement. That changed in 1964 with the theoretical analysis of John Bell who recognized the potential of Bohmʹs thought experiment to decide the issue one way or the other empirically. As is well known, the subsequent experimental work based on Bellʹs theorem decided the conflict between the two views in favor of quantum theory. See the other entries in this section for examples of the impact of Bellʹs work and the experimentalist who verified the implications of his theorem.
Appendix
Tensor multiplication of several of the spin operators using Mathcadʹs kronecker command:
$\begin{matrix} \text{kronecker}(S_z,~ S_z) = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} & \text{kronecker}(S_x,~ S_x) = \begin{pmatrix} 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \end{pmatrix} \ \text{kronecker}(S_z,~ S_x) = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 \ 0 & 0 & -1 & 0 \end{pmatrix} & \text{kronecker}(S_z,~I) = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
The following shows that the singlet state used in this summary can be expressed in terms of either the z‐ or x‐direction eigenvectors.
Mathcadʹs kronecker command is only useful for matrix tensor multiplication, but can be adapted to carry out vector tensor multiplication in the manner shown below. Two matrices are created using the null vector, tensor multiplied and everything but the first column of the product matrix is discarded.
Null vector:
$\begin{matrix} N = \begin{pmatrix} 0 \ 0 \end{pmatrix} & \Psi (a,~ b) = \text{submatrix(kronecker(augment(a, N), augment(b, N)), 1, 4, 1, 1)} \end{matrix} \nonumber$
$\begin{matrix} \frac{\Psi (S_{zu},~ S_{zd}) - \Psi(S_{zd},~ S_{zu})}{ \sqrt{2}} = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} & \frac{\Psi (S_{xd},~ S_{xu}) - \Psi(S_{xu},~ S_{xd})}{ \sqrt{2}} = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.30%3A_David_Bohm%27s_EPR_Gedanken_Experiment.txt
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Quantum theory is both stupendously successful as an account of the small‐scale structure of the world and it is also the subject of an unresolved debate and dispute about its interpretation. J. C. Polkinghorne, The Quantum World, p. 1.
In 1951 David Bohm proposed a gedanken experiment that further illuminated the conflict between local realism and quantum mechanics first articulated by Einstein, Podolsky and Rosen (EPR) in 1935. In this thought experiment a spin‐1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions on the y‐axis to a pair of observers set up to measure spin in either the x‐ or z‐direction.
In this summary tensor algebra will be used to analyze Bohmʹs thought experiment. The vector states and matrix operators required are provided below.
Spin eigenvectors:
$\begin{matrix} S_{zu} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & S_{zd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & S_{xu} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & S_{xd} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
Spin operators in units of h/4π:
$\begin{matrix} S_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & S_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Identity:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
According to established quantum principles the singlet state for fermions is an entangled superposition written as follows in tensor format in the z-direction spin axis.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \left| S_{zu} \rangle_1 \right| S_{zd} \rangle_2 - \left| S_{zd} \rangle_1 \right| S_{zu} \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
First we calculate the expectation values for spin measurements in the z‐ and x‐directions using the eigenvectors of the respective spin operators. We see that spin‐up yields +1 and spin‐down ‐1.
$\begin{matrix} S_{zu}^T S_z S_{zu} = 1 & S_{zd}^T S_z S_{zd} = -1 & S_{xu}^T S_x S_{xu} = 1 & S_{xd}^T S_x S_{xd} = -1 \end{matrix} \nonumber$
Consequently the expectation value observed when both observers jointly measure the same spin direction is ‐1, because in the singlet state the spins have opposite orientations (kronecker performs the tensor multiplication of two matrices, as illustrated in the Appendix).
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~S_z) \Psi = -1 & \Psi^T \text{kronecker}(S_x,~ S_x) \Psi = -1 \end{matrix} \nonumber$
Note that while the entangled singlet spin state shown above is written in the z‐basis, it is the same to an over‐all phase using the x‐direction eigenvectors.
$\Psi = \frac{1}{ \sqrt{2}} \left[ \left| S_{xd} \rangle_1 \right| S_{xu} \rangle_2 - \left| S_{xu} \rangle_1 \right| S_{xd} \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} - \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
Next we show that in spite of the strong correlation shown above, individual spin measurements are totally random yielding expectation values of zero in both the z‐ and x‐directions.
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~ S_z) \Psi = 0 \ \Psi^T \text{kronecker}(S_x,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~ S_x) \Psi = 0 \end{matrix} \nonumber$
Quantum mechanics predicts the following results when the observers make different spin measurements on the particles.
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~S_x) \Psi = 0 & \Psi^T \text{kronecker}(S_x,~ S_z) \Psi = 0 \end{matrix} \nonumber$
From a classical realist position the results for all the previous quantum calculations can be explained by assigning specific x‐ and z‐spin states to the particles, as shown in the table below on the left (see Townsendʹs A Modern Approach to Quantum Mechanics, page 135). Each particle can be in any one of four equally probable spin states, and taken together the particles form four equally probable joint spin states. In other words, the particles are in well‐defined, although unknown, spin states prior to measurement. Measurement, according to a realist, simply reveals a pre‐existing state. The right‐hand part of the table gives the joint measurement results expected given the spin states specified on the left. The bottom row gives the expectation (average) values under the assumptions stated above.
$\begin{pmatrix} \text{Particle 1} & \text{Particle 2} & ' & S_z(1) S_z(2) & S_x(1) S_x(2) & S_z(1) S_x(2) & S_x(1) S_z(2) \ S_{zu} S_{xu} & S_{zd} S_{xd} & ' & -1 & -1 & -1 & -1 \ S_{zu} S_{xd} & S_{zd} S_{xu} & ' & -1 & -1 & 1 & 1 \ S_{zd} S_{xu} & S_{zu} S_{xd} & ' & -1 & -1 & 1 & 1 \ S_{zd} S_{xd} & S_{zu} S_{xu} & ' & -1 & -1 & -1 & -1 \ \text{Expectation} & \text{Value} & ' & -1 & -1 & 0 & 0 \end{pmatrix} \nonumber$
At this point one may ask, ʺWhereʹs the problem? The quantum and classical pictures are in agreement on the prediction of experimental results.ʺ The difficulty is that quantum mechanics does not accept the legitimacy of the states shown in the table on the left. One way to state the problem is to note that Sx and Sz are noncommuting operators. This means that according to quantum mechanics spin in the x‐ and z‐directions cannot simultaneously have well‐defined values (like position and momentum, they are conjugate observables).
The Sx-Sz commutator:
$S_z S_x - S_x S_z = \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix} \nonumber$
For example, if measurement reveals that particle 1 has Szu then particle 2 is definitely Szd. But that means it cannot have a well‐defined value for the x‐direction spin. In fact Szd is a superposition of Sxu and Sxd, as shown below, with <Sx> = 0.
$\begin{matrix} S_{zd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \frac{1}{ \sqrt{2}} (S_{xu} - S_{xd}) = \begin{pmatrix} 0 \ 1 \end{pmatrix} & S_{sd}^T S_x S_{zd} = 0 \end{matrix} \nonumber$
Or, suppose particle 1 is found to have Sxd, then particle 2 is Sxu which is a superposition of spin up and down in the z‐direction and therefore <Sx> = 0.
$\begin{matrix} S_{xu} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & \frac{1}{ \sqrt{2}} (S_{zu} + S_{zd}) = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & S_{xu}^T S_z S_{xu} = 0 \end{matrix} \nonumber$
To summarize, the states in the table are not valid, in spite of their agreement with the predictions of quantum mechanics, because they give well‐defined values to incompatible (according to quantum theory) observables. The quantum objection to the classical spin states can also be expressed using the Uncertainty Principle. For a particle known to have spin‐up in the z‐direction, quantum theory requires that its x‐direction spin be uncertain, as shown by the following calculation.
While Bohmʹs 1951 gedanken experiment clarified the conflict between quantum theory and classical realism, it did not provide for a direct experimental adjudication of the disagreement. That changed in 1964 with the theoretical analysis of John Bell who recognized the potential of Bohmʹs thought experiment to decide the issue one way or the other empirically. As is well known, the subsequent experimental work based on Bellʹs theorem decided the conflict between the two views in favor of quantum theory. See the other entries in this section for examples of the impact of Bellʹs work and the experimentalist who verified the implications of his theorem.
A modified version of the thought experiment shows that there are experiments involving entangled spin systems for which a local hidden‐variable theory gives predictions which are incompatible with those of quantum theory. Instead of measuring the spins in the z‐ and x‐directions, measure one in the z‐direction and the other at some non‐orthogonal angle to the z‐axis, say 45 degrees (π/4). The appropriate spin operator for this diagonal direction (an even superposition of Sx and Sz) and its eigenvalues and eigenvectors are given below, and the singlet spin state is written in the diagonal spin basis.
$\begin{matrix} S_d = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{eigenvals}(S_d) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvecs}(S_d) = \begin{pmatrix} 0.924 & -0.383 \ 0.383 & 0.924 \end{pmatrix} \end{matrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \left[ \left| S_{du} \rangle_A \right| S_{dd} \rangle_B - \left| S_{dd} \rangle_A \right| S_{du} \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} \otimes \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} - \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} \otimes \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
As expected, perfect anti‐correlation is observed if both spins are measured in the diagonal direction while the individual spin measurements are totally random with an expectation value of zero.
$\begin{matrix} \Psi^T \text{kronecker}(S_d,~S_d) \Psi = -1 & \Psi^T \text{kronecker}(S_d,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~S_d) \Psi = 0 \end{matrix} \nonumber$
Up to this point the calculations are consistent with those for the x‐ and z‐directions, and would appear to justify the local realist in providing the following hidden‐variable interpretation of the results.
$\begin{pmatrix} \text{Particle 1} & \text{Particle 2} & ' & S_z(1) S_z(2) & S_d(1) S_d(2) & S_z(1) S_d(2) & S_d(1) S_z(2) \ S_{zu} S_{du} & S_{zd} S_{dd} & ' & -1 & -1 & -1 & -1 \ S_{zu} S_{dd} & S_{zd} S_{du} & ' & -1 & -1 & 1 & 1 \ S_{zd} S_{du} & S_{zu} S_{dd} & ' & -1 & -1 & 1 & 1 \ S_{zd} S_{dd} & S_{zu} S_{du} & ' & -1 & -1 & -1 & -1 \ \text{Expectation} & \text{Value} & ' & -1 & -1 & 0 & 0 \end{pmatrix} \nonumber$
However when the spins are measured in different directions (the z‐ and d‐directions) quantum mechanics predicts that the expectation values are not zero as predicted by the hidden‐variable model. According to quantum theory there is significant anti‐correlation in the joint spin measurements.
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~S_d) \Psi = -0.707 & \Psi^T \text{kronecker}(S_d,~S_z) \Psi = -0.707 \end{matrix} \nonumber$
This disagreement has been examined experimentally and the experimental evidence supports quantum theory.
Appendix: Vector and Matrix Math
This addendum reviews basic vector and matrix operations. At the end it illustrates how vector and matrix tensor multiplication are implemented in Mathcad.
Tensor multiplication of several of the spin operators using Mathcadʹs kronecker command:
$\begin{matrix} \text{kronecker}(S_z,~ S_z) = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} & \text{kronecker}(S_x,~ S_x) = \begin{pmatrix} 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \end{pmatrix} \ \text{kronecker}(S_d,~ S_d) = \begin{pmatrix} 0.5 & 0.5 & 0.5 & 0.5 \ 0.5 & -0.5 & 0.5 & -0.5 \ 0.5 & 0.5 & -0.5 & -0.5 \ 0.5 & -0.5 & -0.5 & 0.5 \end{pmatrix} & \text{kronecker}(S_z,~S_x) = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 \ 0 & 0 & -1 & 0 \end{pmatrix} \ \text{kronecker}(S_z,~ S_d) = \begin{pmatrix} 0.707 & 0.707 & 0 & 0 \ 0.707 & -0.707 & 0 & 0 \ 0 & 0 & -0.707 & -0.707 \ 0 & 0 & -0.707 & 0.707 \end{pmatrix} & \text{kronecker}(S_z,~I) = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix} \ \text{kronecker}(S_x,~I) = \begin{pmatrix} 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} & \text{kronecker}(S_d,~ I) = \begin{pmatrix} 0.707 & 0 & 0.707 & 0 \ 0 & 0.707 & 0 & 0.707 \ 0.707 & 0 & -0.707 & 0 \ 0 & 0.707 & 0 & -0.707 \end{pmatrix} \end{matrix} \nonumber$
Mathcadʹs kronecker command is only useful for matrix tensor multiplication, but can be adapted to carry out vector tensor multiplication in the manner shown below. Two matrices are created using the null vector, tensor multiplied and everything but the first column of the product matrix is discarded.
Null vector:
$\begin{matrix} N = \begin{pmatrix} 0 \ 0 \end{pmatrix} & \Psi (a,~ b) = \text{submatrix(kronecker(augment(a, N), augment(b, N)), 1, 4, 1, 1)} \end{matrix} \nonumber$
$\begin{matrix} \frac{\Psi (S_{zu},~ S_{zd}) - \Psi(S_{zd},~ S_{zu})}{ \sqrt{2}} = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} & \frac{\Psi (S_{xd},~ S_{xu}) - \Psi(S_{xu},~ S_{xd})}{ \sqrt{2}} = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} \ S_{du} = \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} ~~~ S_{dd} = \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} & \frac{\Psi (S_{du},~ S_{dd}) - \Psi(S_{dd},~ S_{du})}{ \sqrt{2}} = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.31%3A_An_Extension_of_Bohm%27s_EPR_Experiment.txt
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In 1951 David Bohm (Quantum Theory, pp. 614-623) proposed a gedanken experiment that further illuminated the conflict between local realism and quantum mechanics first articulated by Einstein, Podolsky and Rosen (EPR) in 1935. In his thought experiment a spin-1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions on the y-axis to a pair of observers set up to measure spin in either the x- or z-direction. While Bohm's thought experiment clarified the conflict between quantum theory and classical realism, it did not provide for a direct experimental adjudication of the disagreement.
However, a slightly modified version of Bohm's thought experiment shows that there are experiments involving entangled spin systems for which a local hidden-variable theory makes predictions which are incompatible with those of quantum mechanics. For example, instead of measuring the spins in the xand z-directions, use the z-direction and another direction at some non-orthogonal angle to the z-axis, say 45 degrees, the diagonal or d-direction. My goal in the following analysis is to bring the conflict between quantum theory and local realism into sharp focus as quickly as possible.
The z-direction spin operator and its eigenvalues and eigenvectors:
$\begin{matrix} S_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \text{eigenvals}(S_z) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvecs}(S_z) = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The singlet spin state written using the z-direction eigenstates:
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \left| \uparrow \rangle_1 \right| \downarrow \rangle_2 - \left| \downarrow \rangle_1 \right| \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
$\Psi = \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The d-direction spin operator and its eigenvalues and eigenvectors:
$\begin{matrix} S_d = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{eigenvals}(S_d) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvecs}(S_d) = \begin{pmatrix} 0.924 & -0.383 \ 0.383 & 0.924 \end{pmatrix} \end{matrix} \nonumber$
The singlet spin state written using the d-direction eigenstates:
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \left| \nearrow \rangle_1 \right| \swarrow \rangle_2 - \left| \swarrow \rangle_1 \right| \nearrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} \otimes \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} - \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} \otimes \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The calculations below show that the individual spin measurements are totally random yielding expectation values of zero in both the z- and d-directions. The identity operator, do nothing, is required for these calculations. However, the expectation value observed when both observers jointly measure the same spin direction is -1, because in the singlet state the spins have opposite orientations in both spin bases.
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~S_z) \Psi = 0 & \Psi^T \text{kronecker}(S_z,~S_z) \Psi = -1 \ \Psi^T \text{kronecker}(S_d,~I) \Psi = 0 & \Psi^T \text{kronecker}(I,~S_d) \Psi = 0 & \Psi^T \text{kronecker}(S_d,~S_d) \Psi = -1 \end{matrix} \nonumber$
The first four columns of the following table provide a local realist's explanation of these calculations. Specific z- and d-spin states are assigned to the particles in the first two columns, with each particle in one of four equally probable spin orientations consistent with the composite singlet state. The next two columns show that these assignments agree with the quantum calculations.
$\begin{pmatrix} \text{Particle 1} & \text{Particle 2} & \hat{S}_z(1) \hat{S}_z(2) & \hat{S}_d(1) \hat{S}_d(2) & \hat{S}_z(1) \hat{S}_d(2) & \hat{S}_d(1) \hat{S}_z(2) \ | \uparrow \rangle | \nearrow \rangle & | \downarrow \rangle | \swarrow \rangle & -1 & -1 & -1 & -1 \ | \uparrow \rangle | \swarrow \rangle & | \downarrow \rangle | \nearrow \rangle & -1 & -1 & 1 & 1 \ | \downarrow \rangle | \nearrow \rangle & | \uparrow \rangle | \swarrow \rangle & -1 & -1 & 1 & 1 \ | \downarrow \rangle | \swarrow \rangle & | \uparrow \rangle | \nearrow \rangle & -1 & -1 & -1 & -1 \ \text{Expectation} & \text{Value} & -1 & -1 & 0 & 0 \end{pmatrix} \nonumber$
At this point one may ask, "Where's the problem? The quantum and classical pictures are in agreement on the prediction of experimental results." The difficulty is that quantum mechanics does not accept the legitimacy of the states shown in the table on the left. One way to state the problem is to note that Sd and Sz are non-commuting operators. This means that spin in the d- and z-directions cannot simultaneously have well-defined values because like position and momentum they are conjugate observables.
The Sz-Sd commutator:
$S_z S_d - S_d S_z = \begin{pmatrix} 0 & 1.414 \ -1.414 & 0 \end{pmatrix} \nonumber$
In other words, the z-eigenstates are superpositions of d-eigenstates and vice versa.
$\begin{matrix} .924 \begin{pmatrix} .924 \ .383 \end{pmatrix} - .383 \begin{pmatrix} -.383 \ .924 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & .383 \begin{pmatrix} .924 \ .383 \end{pmatrix} + .924 \begin{pmatrix} -.383 \ .924 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \ .924 \begin{pmatrix} 1 \ 0 \end{pmatrix} + .383 \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} & .924 \begin{pmatrix} 0 \ 1 \end{pmatrix} - .383 \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} \end{matrix} \nonumber$
Of course, the realist says this argument simply reveals that quantum mechanics does not provide a complete representation of reality because it is unable to assign definite values to all observables prior to and independent of measurement.
Fortunately another set of measurements can settle the dispute. If one spin is measured in the z-direction and the other in the d-direction, local realism predicts an expectation value of zero as shown in the last two columns of the table above. Quantum theory, however, predicts -0.707.
$\begin{matrix} \Psi^T \text{kronecker}(S_z,~S_d) \Psi = -0.707 & \Psi^T \text{kronecker}(S_d,~S_z) \Psi = -0.707 \end{matrix} \nonumber$
This brief analysis demonstrates that there are conceptually simple, Stern-Gerlach like, experiments on spin-1/2 systems for which the predictions of quantum mechanics and local realism are in disagreement.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.32%3A_A_Surgical_Adjudication_of_the_Conflict_Between_Quantum_Theory_and_Local_Realism.txt
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In order to explore the conflict between quantum mechanics and local realism a spin-1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions on the y-axis to a pair of observers set up to measure spin with Stern-Gerlach magnets oriented in either the d- or z-direction, where d (diagonal) refers to a 45 degree rotation clockwise from the vertical z-axis. The entangled singlet spin state can be written using either the z- or d-direction spin eigenstates.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \left| \uparrow \rangle_1 \right| \downarrow \rangle_2 - \left| \downarrow \rangle_1 \right| \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \left| \nearrow \rangle_1 \right| \swarrow \rangle_2 - \left| \swarrow \rangle_1 \right| \nearrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix}^T \nonumber$
The spin-up and spin-down eigenstates in the x-z plane are shown below in general and explicity for the z- and d-directions.
Spin-up Eigenvalue +1
$\varphi_u ( \varphi) = \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \nonumber$
Spin-down Eigenvalue -1
$\varphi_d ( \varphi) = \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \nonumber$
$\begin{matrix} z = 0 \text{deg} & d = 45 \text{deg} & \varphi_u(z) = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \varphi_d(z) = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \varphi_u(d) = \begin{pmatrix} 0.924 \ 0.383 \end{pmatrix} & \varphi_d(d) = \begin{pmatrix} -0.383 \ 0.924 \end{pmatrix} \end{matrix} \nonumber$
If the observers measure their spins in the same direction quantum mechanics predicts they will get opposite values due to the singlet nature of the spin state. In other words, the combined expectation value is -1 for these measurements.
A realist believes that objects have well-defined properties prior to and independent of observation. The first four columns of the following table provide a local realist's explanation of this result. Specific z- and d-spin states are assigned to the particles in the first two columns, with each particle in one of four equally probable spin orientations consistent with the composite singlet state. The next two columns show that these assignments agree with the quantum prediction. Unfortunately quantum mechanics does not accept the legitimacy of the local realist's spin states because Sz and Sd are noncommuting spin operators and therefore cannot have simultaneous eigenvalues.
$\begin{pmatrix} \text{Particle 1} & \text{Particle 2} & \hat{S}_z(1) \hat{S}_z(2) & \hat{S}_d(1) \hat{S}_d(2) & \hat{S}_z(1) \hat{S}_d(2) & \hat{S}_d(1) \hat{S}_z(2) \ | \uparrow \rangle | \nearrow \rangle & | \downarrow \rangle | \swarrow \rangle & -1 & -1 & -1 & -1 \ | \uparrow \rangle | \swarrow \rangle & | \downarrow \rangle | \nearrow \rangle & -1 & -1 & 1 & 1 \ | \downarrow \rangle | \nearrow \rangle & | \uparrow \rangle | \swarrow \rangle & -1 & -1 & 1 & 1 \ | \downarrow \rangle | \swarrow \rangle & | \uparrow \rangle | \nearrow \rangle & -1 & -1 & -1 & -1 \ \text{Realist} & \text{Value} & -1 & -1 & 0 & 0 \ \text{Quantum} & \text{Value} & -1 & -1 & -0.707 & -0.707 \end{pmatrix} \nonumber$
Putting this issue asside momentarily, if one spin is measured in the z-direction and the other in the d-direction, a realist predicts an expectation value of zero as shown in the last two columns of the table. However, quantum theory predicts an expectation value of -0.707. If one of the particles is observed to be spin-up in the z-direction (eigenvalue +1), the other is spin-down in the z-direction. The probability it will be found on measurement to be spin-up in the d-direction yielding a composite eigenvalue of +1 is 0.146. The probability it will be found to be spin-down in the d-direction yielding a composite eigenvalue of -1 is 0.854.
$\begin{pmatrix} \left( \left| \varphi_u(d)^T \varphi_d(z) \right| \right)^2 = 0.146 & \left( \left| \varphi_d(d)^T \varphi_d(z) \right| \right)^2 = 0.854 & \left( \left| \varphi_u(d)^T \varphi_d(z) \right| \right)^2 - \left( \left| \varphi_d(d)^T \varphi_d(z) \right| \right)^2 = -0.707 \end{pmatrix} \nonumber$
This brief analysis demonstrates that there are conceptually simple, Stern-Gerlach like, experiments on spin-1/2 systems which can adjudicate the conflict between local realism and quantum mechanics.
A Quantum Simulation
This thought experiment is simulated using the following quantum circuit. As shown below the results are in agreement with the previous theoretical quantum calculations. The initial Hadamard and CNOT gates create the singlet state from the |11> input. Rz(θ) rotates spin B. The final Hadamard gates prepare the system for measurement. See arXiv:1712.05642v2 for further detail.
$\begin{matrix} \text{Spin A} & |1 \rangle & \triangleright & \text{H} & \cdot & \cdots & \text{H} & \triangleright & \text{Measure 0 or 1: Eigenvalue +1 or -1} \ ~ & ~ & ~ & ~ | \ \text{Spin B} & |1 \rangle & \triangleright & \cdots & \oplus & \text{R})_z (\theta) & \text{H} & \triangleright & \text{Measure 0 or 1: Eigenvalue +1 or -1} \end{matrix} \nonumber$
The quantum gates required to execute this circuit:
$\begin{matrix} \text{Identity} & \text{Hadamard gate} & \text{R}_z~ \text{Rotation} & \text{Controlled NOT} \ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & R_z ( \theta) = \begin{pmatrix} 1 & 0 \ 0 & e^{i \theta} \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
The operator representing the circuit is constructed from the matrix operators provided above.
$\text{Op}( \theta) = \text{kronecker(H, H) kronecker(I, R)}_z ( \theta)) \text{CNOT kronecker(H, I)} \nonumber$
There are four equally likely measurement outcomes with the eigenvalues and overal expectation values shown below.
$\begin{array}{r & l} |00 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0 & |01 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.5 \ |10 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.5 & |11 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0 \ \text{Expectation value}: & 0 - 0.5 - 0.5 + 0 = -1 \ |00 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{ Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.073 & |01 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{ Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.427 \ |10 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{ Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.427 & |11 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{ Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.073 \ \text{Expectation value}: & 0.073 - 0.427 - 0.427 + 0.073 = -0.708 \end{array} \nonumber$
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The following provides an alternative mathematical analysis of the annihilation of positronium as presented in section 18-3 in Volume III of The Feynman Lectures on Physics.
Positronium is an analog of the hydrogen atom in which the proton is replaced by a positron, the electron's anti-particle. The electron-positron pair undergoes annihilation in $10^{-10}$ seconds producing two γ-ray photons. Positronium's effective mass is 1/2, yielding a ground state energy (excluding the magnetic interactions between the spin 1/2 anti-particles)
$E = -0.5μEh = -0.25 Eh. \nonumber$
Considering spin the ground state is four-fold degenerate, but this degeneracy is split by the magnetic spin-spin hyperfine interaction shown below. See "The Hyperfine Splitting in the Hydrogen Atom" for further detail.
\begin{align} \hat{H}_{SpinSpin} &= A \sigma^e \sigma^p \[4pt] &= A \left( \sigma_x^e \sigma_x^p + \sigma_y^e \sigma_y^p + \sigma_z^e \sigma_z^p \right) \end{align} \nonumber
The spin-spin Hamiltonian has the following eigenvalues (top row) and eigenvectors (columns beneath the eigenvalues), showing a singlet ground state and triplet excited state. The electron-positron spin states are to the right of the table with their m quantum numbers, showing that the singlet (j = 0, m = 0) is a superposition state as is one of the triplet states (j = 1, m = 0). The parameter A is much larger for positronium than for the hydrogen atom because the positron has a much larger magnetic moment than the proton.
$\begin{pmatrix} -3A & A & A & A \ 0 & 0 & 0 & 1 \ \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} & 0 & 0 \ - \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} & 0 & 0 \ 0 & 0 & 1 & 0 \end{pmatrix} \begin{matrix} ~ \ |++ \rangle & m = 1 \ | +- \rangle & m = 0 \ |-+ \rangle & m = 0 \ |-- \rangle & m = -1 \end{matrix} \nonumber$
Feynman shows that when the singlet ground state (J = 0, m = 0) annihilates, conservation of momentum requires that the photons emitted in opposite directions (A and B) must have the same circular polarization state, either both in the right or both in left circular state in their direction of motion. This leads to the following entangled superposition. The negative sign is required by parity conservation. The positronium ground state has negative parity (see above), therefore the final photon state must have negative parity.
\begin{align} | \Psi \rangle &= \frac{1}{ \sqrt{2}} \left[ |R \rangle_A | R \rangle_B - |L \rangle_A |L \rangle_B \right] \[4pt] &= \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B - \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] \[4pt] &= \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} - \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{i}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \end{align} \nonumber
The circular polarization states are:
$\begin{matrix} R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} & L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \end{matrix} \nonumber$
The appropriate operators are formed below:
$\begin{matrix} RC = R( \overline{R})^T \rightarrow \begin{pmatrix} \frac{1}{2} & - \frac{1}{2} i \ \frac{1}{2}i & \frac{1}{2} \end{pmatrix} & LC = L( \overline{L})^T \rightarrow \begin{pmatrix} \frac{1}{2} & - \frac{1}{2} i \ \frac{1}{2}i & \frac{1}{2} \end{pmatrix} & RLC = R( \overline{R})^T - L( \overline{L})^T \rightarrow \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
RLC is the angular momentum operator for photons. Below it is shown that $| R \rangle$ and $| L \rangle$ are eigenstates with eigenvalues of +1 and -1, respectively.
$\begin{matrix} \text{eigenvals(RLC)} = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{RLC R} \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} i}{2} \end{pmatrix} & \text{RLC L} \rightarrow \begin{pmatrix} - \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} i}{2} \end{pmatrix} \end{matrix} \nonumber$
Now we can consider some of the measurements that Feynman discusses in his analysis of positronium annihilation. Because the photons in state Ψ are entangled the measurements of observers A and B are correlated. For example, if observers A and B both measure the circular polarization of their photons and compare their results they always agree that they have measured the same polarization state. Their composite expectation value is 1.
$\begin{matrix} \Psi = \frac{i}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} & ( \overline{ \Psi})^T \text{kronecker(RLC, RLC)} \Psi = 1\end{matrix} \nonumber$
The identity operation, do nothing, is now needed:
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
But their individual results are a random sequence of +1 and -1 outcomes averaging to an expectation value of zero.
$\begin{matrix} ( \overline{ \Psi})^T \text{kronecker(I, RLC)} \Psi = 0 & ( \overline{ \Psi})^T \text{kronecker(RLC, I)} \Psi = 0 \end{matrix} \nonumber$
The probability that both observers will measure $| R \rangle$ or both measure $| L \rangle$ is 0.5. The probability that one will measure $| R \rangle$ and the other $| L \rangle$, or vice versa is zero.
$\begin{matrix} ( \overline{ \Psi})^T \text{kronecker(RC, RC)} \Psi = 0.5 & ( \overline{ \Psi})^T \text{kronecker(LC, LC)} \Psi = 0.5 & ( \overline{ \Psi})^T \text{kronecker(LC, RC)} \Psi = 0 \end{matrix} \nonumber$
Because $| R \rangle$ and $| L \rangle$ are superpositions of |V> and |H>, the photon wave function can also be written in the V-H plane polarization basis as is shown below. See the Appendix for an alternative justification.
\begin{align} | \Psi \rangle &= \frac{i}{ \sqrt{2}} \left[ |V \rangle_A | H \rangle_B - |H \rangle_A |V \rangle_B \right] \[4pt] &= \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B - \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B \right] \[4pt] &= \frac{i}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{i}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \end{align} \nonumber
The eigenstates for plane polarization are:
$\begin{matrix} V = \begin{pmatrix} 1 \ 0 \end{pmatrix} & H = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The appropriate measurement operators are:
$\begin{matrix} V_{op} = \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} & H_{op} = \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} & VH = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
As VH is diagonal it is obvious that its eigenvalues are +1 and -1, and that V is the eigenstate with eigenvalue +1 and H is the eigenstate with eigenvalue -1.
$\begin{matrix} \text{VH V} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{VH H} = \begin{pmatrix} 0 \ -1 \end{pmatrix} \end{matrix} \nonumber$
Just as for the circular polarization measurements, the observers individual plane polarization measurements are totally random, but when they compare their results they find perfect anti-correlation, always observing the opposite polarization state.
$\begin{matrix} ( \overline{ \Psi} )^T \text{kronecker(VH, I)} \Psi = 0 & ( \overline{ \Psi} )^T \text{kronecker(I, VH)} \Psi = 0 & ( \overline{ \Psi} )^T \text{kronecker(VH, VH)} \Psi = -1 \ ( \overline{ \Psi} )^T \text{kronecker(Vop, Hop)} \Psi = 0.5 & ( \overline{ \Psi} )^T \text{kronecker(Hop, Vop)} \Psi = 0.5 \ ( \overline{ \Psi} )^T \text{kronecker(Vop, Vop)} \Psi = 0 & ( \overline{ \Psi} )^T \text{kronecker(Hop, Hop)} \Psi = 0 \end{matrix} \nonumber$
If one observer measures circular polarization and the other measures plane polarization the expectation value is 0. In other words there is no correlation between the measurements.
$\begin{matrix} ( \overline{ \Psi} )^T \text{kronecker(RLC, VH)} \Psi = 0 & ( \overline{ \Psi} )^T \text{kronecker(VH, RLC)} \Psi = 0 \end{matrix} \nonumber$
Classical reasoning (according to Feynman) is in disagreement with the highlighted result. Earlier it was demonstrated that the photons are either $| L \rangle$ or $| R \rangle$ polarized. However, suppose photon A is measured in the V-H basis and found to be |V>, and given that B is either $| L \rangle$ or $| R \rangle$, which are superpositions of |V> and |H> (see Appendix), measurement of B in the V-H basis should yield |V> 50% of the time and |H> 50% of the time. There should be no correlation between the A and B measurements. The expectation value should be zero.
Feynman put it this way (parenthetical material added):
Surely you (A) cannot alter the physical state of his (B) photons by changing the kind of observation you make on your photons. No matter what measurements you make on yours, his must still be either RHC ($| R \rangle$) or LHC ($| L \rangle$).
But according to quantum mechanics the photons are entangled in the R-L and V-H bases as shown above, and therefore measurement of |V> at A collapses the wave function to |H> at B.
The highlighted prediction is confirmed experimentally leading to the conclusion that reasoning classically in this manner about the photons created in positronium annihilation is not valid.
While this analysis of positronium annihilation clarifies the conflict between quantum theory and classical realism, it does not lead to an experimental adjudication of the disagreement. In 1964 John Bell demonstrated that entangled systems, like the positronium decay products, could be used to decide the conflict one way or the other empirically. As is well known the subsequent experimental work based on Bell's theorem decided the conflict between the two views in favor of quantum theory.
Appendix
The relationships between plane and circularly polarized light.
$\begin{matrix} | R \rangle & |L \rangle & |V \rangle & |H \rangle \ \frac{1}{ \sqrt{2}} (V + iH) \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ \frac{ \sqrt{2} i}{2} \end{pmatrix} & \frac{1}{ \sqrt{2}} (V - iH) \rightarrow \begin{pmatrix} \frac{ \sqrt{2}}{2} \ - \frac{ \sqrt{2} i}{2} \end{pmatrix} & \frac{1}{ \sqrt{2}} (L + R) \rightarrow \begin{pmatrix} 1 \ 0 \end{pmatrix} & \frac{i}{ \sqrt{2}} (L - R) \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Transforming Ψ from the R-L basis to the V-H basis using the superpositions above.
$\psi = \frac{1}{ \sqrt{2}} (R_A R_B - L_A L_B)~ \begin{array}{|l} \text{substitute,}~ R_A = \frac{1}{ \sqrt{2}} (V_A + iH_A) \ \text{substitute,}~ R_B = \frac{1}{ \sqrt{2}} (V_B + iH_B) \ \text{substitute,}~ L_A = \frac{1}{ \sqrt{2}} (V_A + iH_A) \ \text{substitute,}~ R_A = \frac{1}{ \sqrt{2}} (V_B - iH_B) \ \text{simplify} \end{array} \rightarrow \psi = \sqrt{2} \left( \frac{H_A V_B}{2} + \frac{H_B V_A}{2} \right)i \nonumber$
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The purpose of this tutorial is to summarize a gedanken experiment that reveals a conflict between the predictions of local realism and quantum mechanics. The thought experiment was presented by N. David Mermin in the American Journal of Physics (October 1981, pp 941-943) and Physics Today (April 1985, pp 38-47).
A spin-1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions to a pair of detectors which are set up to measure spin in three directions in x-z plane: along the z-axis, and angles of 120 and 240 degrees with respect to the z-axis. The detector settings are labeled 1, 2 and 3, respectively.
The switches on the detectors are set randomly so that all nine possible settings of the two detectors occur with equal frequency.
Local realism holds that objects have properties independent of measurement and that measurements at one location on a particle cannot influence measurements of another particle at another distant location even if the particles were created in the same event. Local realism maintains that the spin-1/2 particles carry instruction sets which dictate the results of subsequent measurements. That is, prior to measurement the particles are in an unknown but well-defined state.
The following table presents the experimental results expected on the basis of local realism. Singlet spin states have opposite spin values for each of the three measurement directions. For example, if A's spin state is (+-+), then B's spin state is (-+-). If A's detector is set to spin direction "1" and B's detector is set to spin direction "3" the measured result will be recorded as +-.
Note that there are eight spin states and nine possible detector settings, giving 72 possible measurement outcomes all of which are equally probable.
The table shows that the assumption that the singlet-state particles have well-defined spin states prior to measurement requires that the probability the detectors will register opposite spin values is 0.67 (48/72). If the detectors are set to the same direction, they always register different spin values (24/24), and if they are set to different directions the probability they will register different spin values is 0.50 (24/48).
We now show that quantum mechanics disagrees with the local realistic view just presented. The analysis that follows is based on material in chapters 6 and 11 of A. I. M. Rae's Quantum Mechanics.
The singlet state produced by the source is the following entangled superposition, where the arrows indicate the spin orientation for any direction in the x-z plane. As noted above the directions used here are 0, 120 and 240 degrees, relative to the z-axis.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 \downarrow \rangle_2 - | \downarrow \rangle_1 | \uparrow \rangle_2 \right] \nonumber$
The spin operator in the x-z plane is constructed from the Pauli spin operators in the x- and z-directions. φ is the angle of orientation of the measurement magnet with the z-axis. Note that the Pauli operators measure spin in units of h/4π. This provides for some mathematical clarity in the forthcoming analysis.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & S( \varphi) = \cos (\varphi) \sigma_z + \sin ( \varphi) \sigma_x \rightarrow \begin{pmatrix} \cos \varphi & \sin \varphi \ \sin \varphi & - \cos \varphi \end{pmatrix} \end{matrix} \nonumber$
The spin-up and spin-down eigenvectors for this operator are as follow:
$\begin{matrix} \text{Eigenvalue +1} & \varphi_u (\varphi) = \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} & S ( \varphi) \varphi_u ( \varphi) \text{simplify} \rightarrow \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \ \text{Eigenvalue -1} & \varphi_d (\varphi) = \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} & S ( \varphi) \varphi_d ( \varphi) \text{simplify} \rightarrow \begin{pmatrix} \sin \left( \frac{ \varphi}{2} \right) \ - \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \end{matrix} \nonumber$
Note that we get the expected eigenvectors for the z- and x-directions for angles of 0 and 90 degrees, respectively.
$\begin{matrix} \varphi_u (0) = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \varphi_d (0) = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \varphi_u \left( \frac{ \pi}{2} \right) = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & \varphi_d \left( \frac{ \pi}{2} \right) = \begin{pmatrix} -0.707 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
For spin measurements in which the detectors are set at the same angle quantum mechanics agrees with local realism that the detectors will register opposite spins. To illustrate what quantum mechanics predicts when the detectors are set at different angles we will consider a specific example. Suppose the particle at detector A is found to be spin-up in the z-direction, $\begin{pmatrix} 1 \ 0 \end{pmatrix}$. It follows from the singlet spin state that the particle at B is spin-down in the z-direction, $\begin{pmatrix} 0 \ 1 \end{pmatrix}$. But as shown below this state can be written as a superposition of φu and φd.
$\sin \left( \frac{ \varphi}{2} \right) \varphi_u ( \varphi) + \cos \left( \frac{ \varphi}{2} \right) \varphi_d (\varphi) \text{simplify} \rightarrow \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
The quantum mechanical prediction of the spin measurement at an angle φ on particle B is given by the squares of the coefficients of φu and φd. Therefore the probability that the measurement on particle B will be opposite of that on particle A is cos2(φ/2), which is 0.25 for angles of 120 and 240 degrees.
Thus over-all quantum mechanics predicts that opposite spins will be recorded [(1/3)100% + (2/3)25% ] 50% of the time, in sharp disagreement with the 67% calculated on the basis on local realism.
Next it is shown that the result of 25% is general for detector angle differences of 120 degrees (2π/3). First the 0-120 degree result is calculated using a different but equivalent method. The particle at detector A was found to be spin-up in the z-direction (φ = 0 deg) requiring that particle B is spin-down in the z-direction. The probability that particle B will be found spin-down when its detector is set to 120 deg is the square of the projection of φd(0) onto φd(2π/3). The state vectors are real so it is not necessary to square the absolute magnitude.
$\left( \varphi_d (120~ \text{deg})^T \varphi_d(0) \right)^2 = 0.25 \nonumber$
This calculation is now repeated for the 30-150, 60-180, 90-210 and 120-240 detector orientations.
$\begin{matrix} \left( \varphi_d (150~ \text{deg})^T \varphi_d(30~ \text{deg}) \right)^2 = 0.25 & \left( \varphi_d (180~ \text{deg})^T \varphi_d(60~ \text{deg}) \right)^2 = 0.25 \ \left( \varphi_d (210 ~\text{deg})^T \varphi_d(90~ \text{deg}) \right)^2 = 0.25 & \left( \varphi_d (120~ \text{deg})^T \varphi_d(120~ \text{deg}) \right)^2 = 0.25 \end{matrix} \nonumber$
This graphic shows the probability calculation for all detector orientations.
Summary
Local realism maintains that objects have well-defined properties independent of observation, and that the acquisition of a definite value for a property by an object at B due to a measurement carried out on a distant object at A is "spooky action at a distance" and physically unintelligible and therefore impossible.
By contrast quantum theory teaches that quantum particles do not in general have well-defined properties independent of measurement, and that particles with a common origin are in an entangled state and therefore are not independent, no matter how far apart they may be. Together they are in a well-defined correlated state, but their individual properties are uncertain. When measurement determines the state of the particle at A, the correlated property of its distant partner at B becomes known instantaneously.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.35%3A_Mermin%27s_Version_of_Bohm%27s_EPR_Gedanken_Experiment.txt
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In Bohm's EPR thought experiment (Quantum Theory, 1951, pp. 614-619), both local realism and quantum mechanics were shown to be consistent with the experimental data. However, the local realistic explanation used composite spin states that were invalid according to quantum theory. The local realist countered that this was an indication that quantum mechanics was incomplete, because it couldn't assign well-defined values to all observable properties. In this example a thought experiment is presented in which the predictions of local realism and quantum mechanics differ. The thought experiment was presented by N. David Mermin in the American Journal of Physics (October 1981, pp 941-943) and Physics Today (April 1985, pp 38-47).
A spin-1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions to a pair of detectors which are set up to measure spin in three directions in x-z plane: along the z-axis, and angles of 120 and 240 degrees with respect to the z-axis. The detector settings are labeled 1, 2 and 3, respectively.
The switches on the detectors are set randomly so that all nine possible settings of the two detectors occur with equal frequency.
Local realism holds that objects have properties independent of measurement and that measurements at one location on a particle cannot influence measurements of another particle at another distant location even if the particles were created in the same event. Local realism maintains that the spin-1/2 particles carry instruction sets (hidden variables) which dictate the results of subsequent measurements. Prior to measurement the particles are in an unknown but well-defined state.
The following table presents the experimental results expected on the basis of local realism. Singlet spin states have opposite spin values for each of the three measurement directions. For example, if A's spin state is (+-+), then B's spin state is (-+-). If A's detector is set to spin direction "1" and B's detector is set to spin direction "3" the measured result will be recorded as +-.
Note that there are eight spin states and nine possible detector settings, giving 72 possible measurement outcomes all of which are equally probable.
The table shows that the assumption that the singlet-state particles have well-defined spin states prior to measurement requires that the probability the detectors will register opposite spin values is 0.67 (48/72). If the detectors are set to the same direction, they always register different spin values (24/24), and if they are set to different directions the probability they will register different spin values is 0.50 (24/48). Quantum mechanics disagrees with this local realistic analysis.
The singlet state produced by the source is the following entangled superposition, where the arrows indicate the spin orientation for any direction in the x-z plane. See the Appendix for a proof of this assertion. As noted above the directions used here are 0, 120 and 240 degrees, relative to the z-axis.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 | \downarrow \rangle_2 - | \downarrow \rangle_1 | \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The orthonormal spin-up and spin-down vectors in the x-z plane are:
$\begin{matrix} \varphi_u ( \varphi) = \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} & \varphi_d ( \varphi) = \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \varphi_u ( \varphi)^T \varphi_u ( \varphi) \text{simplify} \rightarrow 1 & \varphi_d ( \varphi)^T \varphi_d ( \varphi) \text{simplify} \rightarrow 1 & \varphi_d ( \varphi)^T \varphi_u ( \varphi) \text{simplify} \rightarrow 0 \end{matrix} \nonumber$
For spin measurements in which the detectors are set at the same angle, quantum mechanics and local realism agree that the detectors will register opposite spins. Therefore we will concentrate on what quantum mechanics predicts when the detectors are set at different angles. Recall that local realism predicts that the detectors will behave differently 50% of the time.
If the particle at detector A is found to be spin-up in the z-direction, it follows from the singlet spin state that the particle at B is spin-down in the z-direction, φd(0). The probability that it will be detected spin down if detector B is rotated clockwise or counter-clockwise by 120 degrees is,
$\begin{matrix} \left( \varphi_d (120~ \text{deg})^T \varphi_d (0~ \text{deg}) \right)^2 = 0.25 & \left( \varphi_d (240~ \text{deg})^T \varphi_d (0~ \text{deg}) \right)^2 = 0.25 \end{matrix} \nonumber$
Thus for those cases for which the detectors are set at different angles there is a sharp disagreement between local realism (50%) and quantum mechanics (25%) as to the percentage of the time the detectors behave differently. For all detector settings quantum mechanics predicts that opposite spins will be recorded [(1/3)100% + (2/3)25% ] 50% of the time, compared with the 67% calculated on the basis of local realism.
Quantum theory maintains that the discrepancy in the predictions is due to the fact that the local realistic spin states in the left column of the table above are invalid because they assign definite values to incompatible observables. For example, if the z-axis spin value of the first particle is known to be +, then the state of the composite system is not +++/--- or ++-/--+ or +-+/-+- or +--/-++, but +??/-??. The Appendix demonstrates that the spin operators for 0, 120 and 240 degrees do not commute, which means that the observables associated with these operators cannot simultaneously be well defined.
Summary
Local realism maintains that objects have well-defined properties independent of observation, and that the acquisition of a definite value for a property by an object at B due to a measurement carried out on a distant object at A is "spooky action at a distance" and physically unintelligible and therefore impossible.
By contrast quantum theory teaches that quantum particles do not in general have well-defined properties independent of measurement, and that particles with a common origin are in an entangled state and therefore are not independent, no matter how far apart they may be. Together they are in a well-defined correlated state, but their individual properties are uncertain. When measurement determines the state of the particle at A, the correlated property of its distant partner at B becomes known instantaneously.
Appendix
The spin operator in the x-z plane is constructed from the Pauli spin operators in the x- and z-directions. ϕ is the angle of orientation of the measurement magnet with the z-axis. Note that the Pauli operators measure spin in units of h/4π.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & S( \varphi) = \cos (\varphi) \sigma_z + \sin (\varphi) \sigma_x \rightarrow \begin{pmatrix} \cos (\varphi) & \sin (\varphi) \ \sin (\varphi) & - \cos (\varphi) \end{pmatrix} \end{matrix} \nonumber$
The following calculations demonstrate that the spin operators do not commute and therefore represent incompatible observables. In other words, they are observables that cannot simultaneously be in well-defined states.
$\begin{matrix} S( 0 \text{ deg}) ~S(120 \text{ deg}) - S(120 \text{ deg}) ~S(0 \text{ deg}) = \begin{pmatrix} 0 & 1.732 \ -1.732 & 0 \end{pmatrix} \ S( 0 \text{ deg}) ~S(240 \text{ deg}) - S(240 \text{ deg}) ~S(0 \text{ deg}) = \begin{pmatrix} 0 & -1.732 \ 1.732 & 0 \end{pmatrix} \ S( 1200 \text{ deg}) ~S(240 \text{ deg}) - S(240 \text{ deg}) ~S(120 \text{ deg}) = \begin{pmatrix} 0 & 1.732 \ -1.732 & 0 \end{pmatrix} \end{matrix} \nonumber$
For the singlet state produced by the source the arrows indicate the spin orientation for any direction in the x-z plane.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 | \downarrow \rangle_2 - | \uparrow \rangle_1 | \downarrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \otimes \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} - \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \otimes \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.36%3A_A_Concise_Version_of_Mermin%27s_EPR_Gedanken_Experiment.txt
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Quantum theory is both stupendously successful as an account of the small-scale structure of the world and it is also the subject of an unresolved debate and dispute about its interpretation. J. C. Polkinghorne, The Quantum World, p. 1.
In Bohm's EPR thought experiment (Quantum Theory, 1951, pp. 611-623), both local realism and quantum mechanics were shown to be consistent with the experimental data. However, the local realistic explanation used composite spin states that were invalid according to quantum theory. The local realists countered that this was an indication that quantum mechanics was incomplete because it couldn't assign well-defined values to all observable properties prior to or independent of observation. In the 1980s N. David Mermin presented a related thought experiment [American Journal of Physics (October 1981, pp 941-943) and Physics Today (April 1985, pp 38-47)] in which the predictions of local realism and quantum mechanics disagree. As such Mermin's thought experiment represents a specific illustration of Bell's theorem.
A spin-1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions to detectors which are set up to measure spin in three directions in x-z plane: along the z-axis, and angles of 120 and 240 degrees with respect to the z-axis. The detector settings are labeled 1, 2 and 3, respectively.
The switches on the detectors are set randomly so that all nine possible settings of the two detectors occur with equal frequency.
Local realism holds that objects have properties independent of measurement and that measurements at one location on a particle cannot influence measurements of another particle at a distant location even if the particles were created in the same event. Local realism maintains that the spin-1/2 particles carry instruction sets (hidden variables) which dictate the results of subsequent measurements. Prior to measurement the particles are in an unknown but well-defined state.
The following table presents the experimental results expected on the basis of local realism. Singlet spin states have opposite spin values for each of the three measurement directions. If A's spin state is (+-+), then B's spin state is (-+-). A '+' indicates spin-up and a measurement eigenvalue of +1. A '-' indicates spin-down and a measurement eigenvalue of -1. If A's detector is set to spin direction "1" and B's detector is set to spin direction "3" the measured result will be recorded as +-,with an eigenvalue of -1.
There are eight spin states and nine possible detector settings, giving 72 possible measurement outcomes all of which are equally probable. The next to bottom line of the table shows the average (expectation) value for the nine possible detector settings given the local realist spin states. When the detector settings are the same there is perfect anti-correlation between the detectors at A and B. When the detectors are set at different spin directions there is no correlation.
As will now be shown quantum mechanics (bottom line of the table) disagrees with this local realistic analysis. The singlet state produced by the source is the following entangled superposition, where the arrows indicate the spin orientation for any direction in the x-z plane. As noted above the directions used are 0, 120 and 240 degrees, relative to the z-axis.
$\begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 | \downarrow \rangle_2 - | \uparrow \rangle_1 | \downarrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \otimes \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} - \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \ \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \otimes \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \ \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The single particle spin operator in the x-z plane is constructed from the Pauli spin operators in the x and z-directions. ϕ is the angle of orientation of the measurement magnet with the z-axis. Note that the Pauli operators measure spin in units of h/4π. This provides for some mathematical clarity in the forthcoming analysis.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & S( \varphi) = \cos (\varphi) \sigma_z + \sin (\varphi) \sigma_x \rightarrow \begin{pmatrix} \cos (\varphi) & \sin (\varphi) \ \sin (\varphi) & - \cos (\varphi) \end{pmatrix} \end{matrix} \nonumber$
The joint spin operator for the two-spin system in tensor format is,
$\begin{pmatrix} \cos \varphi_A & \sin \varphi_A \ \sin \varphi_A & - \cos \varphi_A \end{pmatrix} \otimes \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} = \begin{pmatrix} \cos \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} & \sin \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} \ \sin \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} & - \cos \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} \end{pmatrix} \nonumber$
In Mathcad syntax this operator is:
$\text{kronecker}(S( \varphi_A),~S( \varphi_B)) \nonumber$
When the detector settings are the same quantum theory predicts an expectation value of -1, in agreement with the analysis based on local realism.
$\begin{matrix} \Psi^T \text{kronecker(S, (0 deg), S(0 deg)} \Psi = -1 & \Psi^T \text{kronecker(S, (120 deg), S(120 deg)} \Psi = -1 & \Psi^T \text{kronecker(S, (240 deg), S(240 deg)} \Psi = -1 \end{matrix} \nonumber$
However, when the detector settings are different quantum theory predicts an expectation value of 0.5, in disagreement with the local realistic value of 0.
$\begin{matrix} \Psi^T \text{kronecker(S, (0 deg), S(120 deg)} \Psi = 0.5 & \Psi^T \text{kronecker(S, (0 deg), S(240 deg)} \Psi = 0.5 & \Psi^T \text{kronecker(S, (120 deg), S(240 deg)} \Psi = 0.5 \end{matrix} \nonumber$
Considering all detector settings local realism predicts an expectation value of -1/3 [2/3(0) + 1/3(-1)], while quantum theory predicts an expectation value of 0 [2/3(1/2) + 1/3(-1)]. (See the two bottom rows in the table above.)
Furthermore, the following calculations demonstrate that the various spin operators do not commute and therefore represent incompatible observables. In other words, they are observables that cannot simultaneously be in well-defined states. Thus, quantum theory also rejects the realist's spin states used in the table.
$\begin{matrix} S( 0 \text{ deg}) ~S(120 \text{ deg}) - S(120 \text{ deg}) ~S(0 \text{ deg}) = \begin{pmatrix} 0 & 1.732 \ -1.732 & 0 \end{pmatrix} \ S( 0 \text{ deg}) ~S(240 \text{ deg}) - S(240 \text{ deg}) ~S(0 \text{ deg}) = \begin{pmatrix} 0 & -1.732 \ 1.732 & 0 \end{pmatrix} \ S( 1200 \text{ deg}) ~S(240 \text{ deg}) - S(240 \text{ deg}) ~S(120 \text{ deg}) = \begin{pmatrix} 0 & 1.732 \ -1.732 & 0 \end{pmatrix} \end{matrix} \nonumber$
The local realist is undeterred by this argument and the disagreement with the quantum mechanical predictions, asserting that the fact that quantum theory cannot assign well-defined states to all elements of reality independent of observation is an indication that it provides an incomplete description of reality.
However, results available for experiments of this type with photons support the quantum mechanical predictions and contradict the local realists analysis shown in the table above. Thus, there appears to be a non-local interaction between the two spins at their measurement sites. Nick Herbert provides a memorable and succinct description of such non-local influences on page 214 of Quantum Reality.
A non-local interaction links up one location with another without crossing space, without decay, and without delay. A non-local interaction is, in short, unmediated, unmitigated, and immediate.
Jim Baggott puts it this way (The Meaning of Quantum Theory, page 135):
The predictions of quantum theory (in this experiment) are based on the properties of a two-particle state vector which ... is 'delocalized' over the whole experimental arrangement. The two particles are, in effect, always in 'contact' prior to measurement and can therefore exhibit a degree of correlation that is impossible for two Einstein separable particles.
"...if [a hidden-variable theory] is local it will not agree with quantum mechanics, and if it agrees with quantum mechanics it will not be local. This is what the theorem says." John S. Bell
Further Information
The eigenvectors of the single particle spin operator, S(φ), in the x-z plane are given below along with their eigenvalues.
$\begin{matrix} \varphi_u ( \varphi) = \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \ \begin{pmatrix} \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} & \varphi_d (\varphi) = \begin{pmatrix} - \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \ \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \varphi_u ( \varphi)^T \varphi_u (\varphi) \text{simplify} \rightarrow 1 & \varphi_d ( \varphi)^T \varphi_d (\varphi) \text{simplify} \rightarrow 1 & \varphi_d ( \varphi)^T \varphi_u (\varphi) \text{simplify} \rightarrow 0 \end{matrix} \nonumber$
$\begin{matrix} \text{Eigenvalue +1} & \text{Eigenvalue -1} \ S( \varphi) \varphi_u ( \varphi) \text{simplify} \rightarrow \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \ S( \varphi) \varphi_d ( \varphi) \text{simplify} \rightarrow \begin{pmatrix} \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} & \begin{pmatrix} \sin \left( \frac{ \varphi}{2} \right) \end{pmatrix} \ \begin{pmatrix} \cos \left( \frac{ \varphi}{2} \right) \end{pmatrix} \end{matrix} \nonumber$
A summary of the quantum mechanical calculations:
$\begin{pmatrix} \Psi^T \text{kronecker(S, (0 deg), S(120 deg)} \Psi \ \Psi^T \text{kronecker(S, (0 deg), S(240 deg)} \Psi \ \Psi^T \text{kronecker(S, (120 deg), S(0 deg)} \Psi \ \Psi^T \text{kronecker(S, (120 deg), S(240 deg)} \Psi \ \Psi^T \text{kronecker(S, (240 deg), S(0 deg)} \Psi \ \Psi^T \text{kronecker(S, (240 deg), S(120 deg)} \Psi \ \Psi^T \text{kronecker(S, (0 deg), S(0 deg)} \Psi \ \Psi^T \text{kronecker(S, (120 deg), S(120 deg)} \Psi \ \Psi^T \text{kronecker(S, (240 deg), S(240 deg)} \Psi \end{pmatrix}^T = \begin{pmatrix} 0.5 & 0.5 & 0.5 & 0.5 & 0.5 & 0.5 & -1 & -1 & -1 \end{pmatrix} \nonumber$
Calculation of the overall spin expectation value:
$\sum_{i = 0}^2 \sum_{j = 0}^2 \left[ \Psi^T \text{kronecker[S [i (120 deg)], S[j (120 deg)]]} \Psi \right] = 0 \nonumber$
The expectation value as a function of the relative orientation of the detectors reveals the level of correlation between the two spin measurements. For θ = 00 there is perfect anti-correlation; for θ = 1800 perfect correlation; for θ = 900 no correlation; for θ = 600 intermediate anti-correlation (-0.5) and for θ = 1200 intermediate correlation (0.5).
A Quantum Simulation: This thought experiment is simulated using the following quantum circuit. As shown below the results are in agreement with the previous theoretical quantum calculations. The initial Hadamard and CNOT gates create the singlet state from the |11> input. Rz(θ) rotates spin B. The final Hadamard gates prepare the system for measurement. See arXiv:1712.05642v2 for further detail.
$\begin{matrix} \text{Spin A} & |1 \rangle & \triangleright & H & \cdot & \cdots & H & \triangleright & \text{Measure 0 or 1: Eigenvalue +1 or -1} \ ~ & ~ & ~ & ~ & | \ \text{Spin B} & |1 \rangle & \triangleright & \cdots & \oplus & R_z ( \theta) & H & \triangleright & \text{Measure 0 or 1: Eigenvalue +1 or -1} \end{matrix} \nonumber$
The quantum gates required to execute this circuit:
$\begin{matrix} \text{Identity} & \text{Hadamard gate} & \text{R}_z ~ \text{Rotation} & \text{Controlled NOT} \ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & R_z ( \theta) = \begin{pmatrix} 1 & 0 \ 0 & e^{i \theta} \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
The operator representing the circuit is constructed from the matrix operators provided above.
$\text{Op} (\theta) = \text{kronecker(H, H) kronecker(I, R}_z (\theta)) \text{CNOT kronecker(H, I)} \nonumber$
There are four equally likely measurement outcomes with the eigenvalues and overall expectation values shown below for relative measurement angles 0 and 120 deg (2π/3).
$\begin{array}{r & l} |00 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0 & |01 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.5 \ |10 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.5 & |11 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{ Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0 \ \text{Expectation value}: & 0 - 0.5 - 0.5 + 0 = -1 \ |00 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{ Op \left( \frac{2 \pi}{3} \right)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.375 & |01 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{ \left( \frac{2 \pi}{3} \right)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.125 \ |10 \rangle \text{eigenvalue -1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{ \left( \frac{2 \pi}{3} \right)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.125 & |11 \rangle \text{eigenvalue +1 } \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{ \left( \frac{2 \pi}{3} \right)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right] ^2 = 0.375 \ \text{Expectation value}: & 0.375 - 0.125 + 0.375 - 0.125 = 0.5 \end{array} \nonumber$
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The purpose of this tutorial is to summarize a gedanken experiment that reveals a conflict between the predictions of local realism and quantum mechanics. The thought experiment was presented by N. David Mermin in the American Journal of Physics (October 1981, pp 941-943) and Physics Today (April 1985, pp 38-47). In this summary the quantum mechanical analysis employs tensor algebra.
A spin-1/2 pair is prepared in a singlet state and the individual particles travel in opposite directions to a pair of detectors which are set up to measure spin in three directions in x-z plane: along the z-axis, and angles of 120 and 240 degrees with respect to the z-axis. The detector settings are labeled 1, 2 and 3, respectively.
The switches on the detectors are set randomly so that all nine possible settings of the two detectors occur with equal frequency.
Local realism holds that objects have properties independent of measurement and that measurements at one location on a particle cannot influence measurements of another particle at another distant location even if the particles were created in the same event. Local realism maintains that the spin-1/2 particles carry instruction sets which dictate the results of subsequent measurements. That is, prior to measurement the particles are in a well-defined state.
The following table presents the experimental results expected on the basis of local realism. Singlet spin states have opposite spin values for each of the three measurement directions. For example, if A's spin state is (+-+), then B's spin state is (-+-). If A's detector is set to spin direction "1" and B's detector is set to spin direction "3" the measured result will be recorded as +-.
Note that there are eight spin states and nine possible detector settings, giving 72 possible measurement outcomes all of which are equally probable.
The table shows that the assumption that the singlet-state particles have well-defined spin states prior to measurement requires that the probability the detectors will register opposite spin values is 0.67 (48/72). If the detectors are set to the same direction, they always register different spin values (24/24), and if they are set to different directions the probability they will register different spin values is 0.50 (24/48).
Now we show that a quantum mechanical analysis is in sharp disagreement with the local realistic view just presented.
The natural base states for this quantum analysis are spin-up and spin-down in the z-direction.
$\begin{matrix} S_{zu} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & S_{zd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The singlet state produced by the source (fermions have anti-symmetric wave functions and are indistinguishable) in tensor format is,
$\begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 | \downarrow \rangle_2 - | \downarrow \rangle_1 | \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The single particle spin operator in the x-z plane is constructed from the Pauli spin operators in the xand z-directions. ϕ is the angle of orientation of the measurement magnet with the z-axis. Note that the Pauli operators measure spin in units of h/4π. This provides for some mathematical clarity in the forthcoming analysis.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & S( \varphi) = \cos (\varphi) \sigma_z + \sin (\varphi) \sigma_x \rightarrow \begin{pmatrix} \cos (\varphi) & \sin (\varphi) \ \sin (\varphi) & - \cos (\varphi) \end{pmatrix} \end{matrix} \nonumber$
Obviously if ϕ or π/2 we recover σz and σx.
$\begin{matrix} S(0) = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & S \left( \frac{ \pi}{2} \right) = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
The spin operator for the two-particle system in tensor format is,
$\begin{pmatrix} \cos \varphi_A & \sin \varphi_A \ \sin \varphi_A & - \cos \varphi_A \end{pmatrix} \otimes \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} = \begin{pmatrix} \cos \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} & \sin \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} \ \sin \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} & - \cos \varphi_A \begin{pmatrix} \cos \varphi_B & \sin \varphi_B \ \sin \varphi_B & - \cos \varphi_B \end{pmatrix} \end{pmatrix} \nonumber$
In Mathcad the two-spin operator is written as,
$\text{kronecker}(S( \varphi_A),~S( \varphi_B)) \nonumber$
In the z-basis the eigenvalues for spin-up and spin-down are +1 and -1 respectively.
$\begin{matrix} S_{zu}^T \sigma_z S_{zu} \rightarrow 1 & S_{zd}^T \sigma_z S_{zd} \rightarrow -1 \end{matrix} \nonumber$
The combined measurement of both spins when the detectors have the same x-z orientation yields an expectation value of -1, because Ψ is a singlet state and the individual spins have opposite orientations.
$\begin{matrix} \Psi^T \text{kronecker(S, (0 deg), S(0 deg)} \Psi = -1 & \Psi^T \text{kronecker(S, (120 deg), S(120 deg)} \Psi = -1 & \Psi^T \text{kronecker(S, (240 deg), S(240 deg)} \Psi = -1 \end{matrix} \nonumber$
These results are in agreement with the local realist predictions in the right third of the table above. When the detectors are oriented at the same angle in the x-z plane they always register opposite spin values for their respective particles. At this point, quantum mechanics appears to support the realistic position.
However, overall (considering all nine detector settings) quantum mechanics predicts that opposite spins are recorded only 50% of the time, as opposed to 67% predicted by local realism. In other words, the overall expectation value for the joint spin measurements is 0. Quantum mechanics strongly disagrees with local realism in the experiments in which the detector settings differ. In these cases, quantum theory predicts the detectors register opposite spin values 25% of the time and the same value 75% of the time: 0.25(-1) + 0.75(+1) = 0.5 as shown by the off-diagonal vlaues in the results matrix below.
$\begin{matrix} i = 0.2 & j = 0.2 & R_{i,~j} = \Psi^T \text{kronecker[S[i (120 deg)], (120 deg)]]} \Psi & R = \begin{pmatrix} -1 & 0.5 & 0.5 \ 0.5 & -1& 0.5 \ 0.5 & 0.5 & -1 \end{pmatrix} \end{matrix} \nonumber$
Calculation of the overall spin value:
$\sum_i \sum_j \left[ \Psi^T \text{kronecker[S [i (120 deg)], S[j (120 deg)]]} \Psi \right] = 0 \nonumber$
By comparison the spin expectation value based on local realism as represented by the table above is:
$\frac{48}{72} (-1) + \frac{24}{72}(1) = -0.333 \nonumber$
The fact that local realism and quantum mechanics can lead to different predictions that might be adjudicated by experimental measurement was first pointed out by John S. Bell [Physics 1, 195 (1964)].
Sections 6.2, 6.3 and 11.3 of A. I. M. Rae's Quantum Mechanics, 3rd Ed. contain,in a clear and succinct fashion, the necessary mathematical background for this thought experiment. The key quantum spin correlation function of Rae's analysis (Fig 11.3) is calculated as follows in tensor format. The linear function in blue is the hidden-variable correlation function.
One more comment can be made about local realism and quantum mechanics in this particular example. If we look at the results for the first spin in the table above, its over-all expectation value is 0. There are as many +s as -s. Quantum mechanics agrees with local realism with this result in addition to the case in which the detectors have the same setting.
We can calculate the expectation values for the first spin independent of the second spin by replacing the spin operator of the second spin with the identity operator.
$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
We see that the expectation values for the three detector orientations of the first spin are 0. In other words, the measurement results on the first spin are perfectly random.
$\begin{matrix} \Psi^T \text{kronecker(S, (0 deg), I} \Psi = 0 & \Psi^T \text{kronecker(S, (120 deg), I} \Psi = 0 & \Psi^T \text{kronecker(S, (240 deg), I} \Psi = 0 \end{matrix} \nonumber$
The same holds for the second spin.
$\begin{matrix} \Psi^T \text{kronecker(I, S(0 deg)} \Psi = 0 & \Psi^T \text{kronecker(I, S(120 deg)} \Psi = 0 & \Psi^T \text{kronecker(I, S(240 deg)} \Psi = 0 \end{matrix} \nonumber$
And yet when the angles are the same the measurement results (as shown above) are correlated.
$\begin{matrix} \Psi^T \text{kronecker(S, (0 deg), S(0 deg)} \Psi = -1 & \Psi^T \text{kronecker(S, (120 deg), S(120 deg)} \Psi = -1 & \Psi^T \text{kronecker(S, (240 deg), S(240 deg)} \Psi = -1 \end{matrix} \nonumber$
"How is it possible that two events, each one objectively random, are always perfectly correlated." Anton Zeilinger, Nature, 8 December 2005.
All the calculations carried out above can also be done having the trace function operate on the product of the density matrix for Ψ and the matrices representing the measurement operators. Several examples are provided below.
Form the density matrix for Ψ.
$\rho \Psi = \Psi \Psi^T \nonumber$
$\begin{matrix} tr \left( \rho \Psi \text{kronecker(S (120 deg), S(120 deg))} \right) = -1 & tr \left( \rho \Psi \text{kronecker(S (0 deg), S(120 deg))} \right) = 0.5 \end{matrix} \nonumber$
$tr \left[ \sum_i \sum_j \left[ \rho \Psi \text{kronecker[S [i (120 deg)], S[j (120 deg)]} \right] \right] = 0 \nonumber$
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Tensor algebra is used to simplify the mathematical analysis of a gedanken experiment presented by John D. Norton in the February 2011 issue of the American Journal of Physics (pp 182-188). Norton's construction using spatial degrees of freedom is in my opinion highly contrived and actually not as simple as the original GHZ experiment involving polarized light [Nature 404, 515-519 (2000)] or Mermin's gedanken rendition involving spin-1/2 particles [Physics Today 43(6), 9-11 (1990)]. Besides using tensor algebra to avoid tedious algebraic manipulations, I adopt a more plausible model for the two-chambered Einstein boxes employed in Norton's approach.
The thought experiment involves three widely separated two-chamber boxes which I prefer to model as infinite potential wells with a finite internal barrier, as shown below. What I have in mind here is a simplified model for the ammonia molecule. See the Appendix for a more realistic double-well potential.
The orthonormal vectors |L> and |R> given below represent occupancy in the two chambers. Norton initially considers the consequences of placing the particle in two states, |P> and |M>, which are also orthonormal. I have no idea how these states would be created, and this is the main reason I find Norton's approach artificial.
$\begin{matrix} L = \begin{pmatrix} 1 \ 0 \end{pmatrix} & R = \begin{pmatrix} 0 \ 1 \end{pmatrix} & L^T L = 1 & R^T R = 1 & L^T = 0 \ P = \frac{1}{ \sqrt{2}} (L + iR) & M = \frac{1}{ \sqrt{2}} (L - iR) & \overline{(P^T)}P = 1 & \overline{(M^T)} M = 1 & \overline{(P^T)}M = 0 \end{matrix} \nonumber$
In the second part of my analysis I use the following symmetric and anti-symmetric states. These represent the ground and first excited superposition states of the ammonia molecule regarding the position of the nitrogen atom to the left or right of the plane created by the three hydrogen atoms.
$\begin{matrix} S = \frac{1}{ \sqrt{2}} (L+R) & A = \frac{1}{ \sqrt{2}} (L-R) & S^T S = 1 & A^T A = 1 & A^T S = 0 \end{matrix} \nonumber$
Construction of the initial three-box state highlighted below is facilitated by the use of three Mathcad commands: submatrix, kronecker, and augment. It is not difficult to do this by hand, however having a general expression for the wavefunction simplifies subsequent analysis.
$\Psi \text{a, b, c} = \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ a, \begin{pmatrix} 0 \ 0 \end{pmatrix} \right],~ \text{kronecker} \left[ \text{augment} \left[ b,~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right], ~ \text{augment} \left[ c,~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] \right] \right],~ 1,~ 8,~ 1,~ 1 \right] \nonumber$
Initial state:
$\begin{matrix} \Psi_{ABC} = \frac{1}{ \sqrt{2}} ( \Psi \text{ (P, P, P)} - \Psi \text{M, M, M))} & \Psi_{ABC}^T = \begin{pmatrix} 0 & 0.5i & 0.5i & 0 & 0.5i & 0 & 0 & -0.5i \end{pmatrix} \end{matrix} \nonumber$
For a three-box system this initial entangled states has eight possible position measurement outcomes, four of which are observed as shown in the probability calculations below.
$\begin{matrix} \left( \left| \Psi \text{(L, L, L)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(L, L, R)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(L, R, L)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi \text{(R, L, L)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(L, R, R)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(R, L, R)}^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi \text{(R, R, L)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(R, R, R)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \end{matrix} \nonumber$
These results show that position measurements on any two of the boxes enables one to determine which chamber the particle occupies in the third box without further measurement. If the measured particles are found to be on the same side (either both L or both R) the third particle is in the right chamber R. If they are different, the third particle is in the left chamber, L. Because we can predict with certainty the position of the particle in the third box, we say its position is an "element of reality." This leads to the conclusion that the permissible position eigenstates are: LLR, LRL, RLL, and RRR.
In a second set of experiments we shine light on two of the boxes with a frequency that just matches the energy difference between the ground state |S> and excited state |A>. If the light is absorbed the box is in the |S> state; if the box is in the |A> state stimulated emission occurs and a photon is released. This set of experiments measures the spectroscopic states of the boxes.
As the following probability calculations show, if the two boxes irradiated are found to be in the same spectroscopic state (SS or AA) a subsequent measurement of the position of the particle in the third box will yield L. Conversely, if they are in different spectroscopic state the position measurement on the third box will yield R. Thus, this set of measurements also indicate that position is an "element of reality."
$\begin{matrix} \left( \left| \Psi \text{(S, S, L)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(S, S, R)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(S, A, L)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi \text{(S, A, R)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(A, S, L)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(A, S, R)}^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi \text{(A, A, R)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(A, A, L)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi \text{(S, L, S)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(S, R, S)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(S, L, A)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi \text{(S, R, A)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(A, L, S)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(A, R, S)}^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi \text{(A, R, A)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(A, L, A)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi \text{(L, S, S)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(R, S, S)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(L, S, A)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi \text{(R, S, A)}^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi \text{(L, A, S)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(R, A, S)}^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi \text{(R, A, A)}^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi \text{(L, A, A)}^T \Psi_{ABC} \right| \right)^2 = 0.25 \end{matrix} \nonumber$
The allowed states for these measurements are listed in the following table. The top row showing that SS or AA implies L and that SA or AS implies R.
$\begin{pmatrix} \text{SSL} & \text{AAL} & \text{SLS} & \text{ALA} & \text{LSS} & \text{LAA} \ \text{SAR} & \text{ASR} & \text{SRA} & \text{ARS} & \text{RSA} & \text{RAS} \end{pmatrix} \nonumber$
The measurement outcomes can also be grouped in two different ways: (a) measurements in which three pairs have the same spectroscopic state and (b) measurements in which one pair has the same spectroscopic state and the remaining pairs have different spectroscopic states. The following table organizes the experimental results grouped in these categories. The first two rows satisfy criterion (a) and the remaining six rows satisfy criterion (b). The left column shows the implied total spectroscopic state, the middle three columns how it is achieved, and the right column the position state consistent with the spectroscopic results.
$\begin{pmatrix} \text{SSS} & ' & \text{SSL} & \text{LSS} & \text{SLS} & ' & \text{LLL} \ \text{AAA} & ' & \text{AAL} & \text{ALA} & \text{LAA} & ' & \text{LLL} \ \text{SSA} & ' & \text{SSL} & \text{SRA} & \text{RSA} & ' & \text{RRL} \ \text{AAS} & ' & \text{AAL} & \text{ARS} & \text{RAS} & ' & \text{RRL} \ \text{SAS} & ' & \text{SLS} & \text{SAR} & \text{RAS} & ' & \text{RLR} \ \text{ASA} & ' & \text{ALA} & \text{ASR} & \text{RSA} & ' & \text{RLR} \ \text{ASS} & ' & \text{LSS} & \text{ARS} & \text{ASR} & ' & \text{LRR} \ \text{SAA} & ' & \text{LAA} & \text{SRA} & \text{SAR} & ' & \text{LRR} \end{pmatrix} \nonumber$
As this table demonstrates, according to the spectroscopic measurements the permissible position states are RRL, RLR, LRR and LLL. This is in direct contradiction to the earlier position measurements which gave the result LLR, LRL, RLL, and RRR. In other words two measurement protocols lead to the notion that position is an element of reality, but they dramatically disagree on the actual values of the allowed positions. Norton concludes his paper with the following assessment.
Two principal assumptions were made in the arguments that generated this contradiction. One was that the empirical predictions of quantum theory are reliable. The other was that the EPR reality criterion, which in turn, depends on the assumptions of separability and locality. One of these assumptions must be given up. The continuing empirical success of quantum theory has led to a consensus that it is the second assumption, the EPR reality criterion, which is to be discarded.
Appendix
A more realistic potential well for this problem is shown below.
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This exercise explores the outcomes of measurements on the three spin-1/2 entangled state, ΨABC, highlighted below. It represents a lean GHZ protocol developed by H. J. Bernstein. Because much has been published on the GHZ protocol, I'm going to get right to the point without a lot of commentary.
The eigenstates for spin-up and spin-down in the z-, x- and y-directions in vector format are as follows.
$\begin{matrix} Z_u = \begin{pmatrix} 1 \ 0 \end{pmatrix} & Z_d = \begin{pmatrix} 0 \ 1 \end{pmatrix} & X_u = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & X_d = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} & Y_u = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} & Y_d = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \end{matrix} \nonumber$
These spin states are also shown on a Bloch sphere.
Tensor multiplication using Mathcad commands will be used to form the initial entangled spin state and subsequent measurement states. These results can be easily verified by hand calculation.
$\Psi \text{(a, b, c)} = \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \text{a, } \begin{pmatrix} 0 \ 0 \end{pmatrix} \right], ~ \text{kronecker} \left[ \text{augment} \left[ \text{b, } \begin{pmatrix} 0 \ 0 \end{pmatrix} \right], ~ \text{augment} \left[ \text{c, } \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] \right] \right], 1,~8,~1,~1 \right] \nonumber$
Initial state:
$\begin{matrix} \Psi_{ABC} = \frac{1}{ \sqrt{2}} \left( \Psi \left( Y_u,~Y_u,~Y_u \right) - \Psi \left( Y_d,~Y_d,~Y_d \right) \right) & \Psi_{ABC}^T = \begin{pmatrix} 0 & 0.5i & 0.5i & 0 & 0.5i 0 & 0 & -0.5i \end{pmatrix} \end{matrix} \nonumber$
Given this initial state the probabilities for the various measurement outcomes for spin in the z-direction are calculated.
$\begin{matrix} \left( \left| \Psi (Z_u,~ Z_u,~ Z_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (Z_u,~ Z_u,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (Z_u, ~Z_d, ~Z_u)^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi (Z_d,~ Z_u, ~Z_u)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (Z_u,~ Z_d,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (Z_d,~ Z_u,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi (Z_d,~ Z_d,~ Z_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (Z_d,~ Z_d,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 \end{matrix} \nonumber$
These calculations predict that the following spin states are observed: ZuZuZd, ZuZdZu, ZdZuZu and ZdZdZd. It is easily seen that the measurement of any two spin-states allows the prediction of the third without the need for a measurement. If two spins have the same value (uu or dd) then the third is d. If two spin states have different values (ud or du) then the third is u. Einstein, Podolsky and Rosen (EPR) wrote in their famous 1935 Physical Review paper,
"If, without in any way disturbing a system, we can predict with certainty (i.e. with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to this physical quantity."
According to this definition spin in the z-direction is an "element of reality." In other words, it has a definite value independent of measurement.
Next we calculate the quantum mechanical predictions for the measurement of spin in the x-direction on two of the spins and in the z-direction for the third.
$\begin{matrix} \left( \left| \Psi (Z_u,~X_u~Z_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (X_u,~X_u,~Z_d) ^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (X_u, X_d,~ X_u)^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi (X_u,~ X_d,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (X_d,~X_u,~Z_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (X_d,~ X_u,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi (X_d,~ X_d,~ Z_d)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (X_d,~ X_d,~ Z_u)^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi (X_u,~ Z_u,~ X_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (X_u,~ Z_d, X_u)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (X_u,~ Z_u,~ X_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi (X_u,~ Z_d,~ X_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (X_d,~ Z_u,~ X_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (X_d,~ Z_d,~ X_u)^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi (X_u,~ Z_d,~ X_d)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (X_d,~ Z_u,~ X_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi (Z_u,~ X_u,~X_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (Z_d,~X_u,~X_u)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (Z_u, X_u, X_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 \ \left( \left| \Psi (Z_d,~X_u,~X_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 & \left( \left| \Psi (Z_u, X_d, X_u)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (Z_d,~X_d,~X_u)^T \Psi_{ABC} \right| \right)^2 = 0 \ \left( \left| \Psi (Z_d,~X_d,~X_d)^T \Psi_{ABC} \right| \right)^2 = 0 & \left( \left| \Psi (Z_u,~X_d,~X_d)^T \Psi_{ABC} \right| \right)^2 = 0.25 \end{matrix} \nonumber$
The twelve possible outcomes are listed in two rows.
$\begin{pmatrix} X_uX_uZ_u & X_dX_dZ_u & X_uZ_uX_u & X_dZ_uX_d & Z_uX_uX_u & Z_uX_dX_d \ X_dX_dZ_d & X_dX_uZ_d & X_uZ_dX_d & X_dZ_dX_u & Z_dX_uX_d & Z_dX_dX_u \end{pmatrix} \nonumber$
These results are summarized as follows: if two spins have the same x-spin state then the z-spin state is u, but if they are different the z-spin state is d. Again according to the EPR criterion spin in the z-direction is an element of reality.
However, the calculations can be described in another way: (a) measurements in which three pairs have the same x spin state and (b) measurements in which one pair has the same x-spin state and the remaining pairs have different x-spin states. The following table organizes the experimental results grouped in these categories. The first two rows satisfy criterion (a) and the remaining six rows satisfy criterion (b). The left column shows the implied total x-spin state, the middle three columns how it is achieved, and the right column the z-spin state consistent with the x-spin measurement results.
$\begin{pmatrix} X_uX_uX_u & ' & X_uX_uZ_d & Z_uX_uX_u & X_uZ_uX_u & ' & Z_uZ_uZ_u \ X_dX_dX_d & ' & X_dX_dZ_u & X_dZ_uX_d & Z_uX_dX_d & ' & Z_uZ_uZ_u \ X_uX_uX_d & ' & X_uX_uZ_u & X_uZ_dX_d & Z_dX_uX_d & ' & Z_dZ_dZ_u \ X_dX_dX_u & ' & X_dX_dZ_u & X_dZ_dX_u & Z_dX_dX_u & ' & Z_dZ_dZ_u \ X_uX_dX_u & ' & X_uZ_uX_u & X_uX_dZ_d & Z_dX_dX_u & ' & Z_dZ_uZ_d \ X_dX_uX_d & ' & X_dZ_uX_d & X_dX_uZ_d & Z_dX_uX_d & ' & Z_dZ_uZ_d \ X_dX_uX_u & ' & Z_uX_uX_u & X_dZ_dX_u & X_dX_uZ_d & ' & Z_uZ_dZ_d \ X_uX_dX_d & ' & Z_uX_dX_d & X_uZ_dX_d & X_uX_dZ_d & ' & Z_uZ_dZ_d \end{pmatrix} \nonumber$
As this table demonstrates, according to the x-direction spin measurements the permissible z-direction states are ZuZuZu, ZdZdZu, ZdZuZd and ZuZdZd. This is in direct contradiction to the earlier z-direction measurements which gave the result ZdZdZd, ZuZuZd, ZuZdZu and ZdZdZu, using the same initial wavefunction ΨABC. In other words two measurement protocols lead to the notion that z-direction spin is an element of reality according to the EPR definition, but they dramatically disagree on the actual z-direction spin values allowed.
In the February 2001 issue of the American Journal of Physics on page 187, John D. Norton summarizes the situation as follows:
Two principal assumptions were made in the arguments that generated this contradiction. One was that the empirical predictions of quantum theory are reliable. The other was the EPR reality criterion, which in turn, depends on the assumptions of separability and locality. One of these assumptions must be given up. The continuing empirical success of quantum theory has led to a consensus that it is the second assumption, the EPR reality criterion, which is to be discarded.
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A tensor algebra approach is used to demonstrate the challenge to the local realistic position of reality that quantum mechanical entanglement creates. The example is drawn from Chapter 3 of David Z Albert's text, Quantum Mechanics and Experience.
A quon (any entity that exhibits both wave and particle aspects in the peculiar quantum manner - Nick Herbert, Quantum Reality, page 64) has a variety of properties each of which can take on two values. For example, it has the property of hardness and can be either hard or soft. It also has the property of color and can be either black or white.
In the matrix formulation of quantum mechanics these states are represented by the following vectors.
$\begin{matrix} \text{Hard} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{Soft} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \text{Black} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} \end{pmatrix} & \text{White} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{-1}{ \sqrt{2}} \end{pmatrix} \end{matrix} \nonumber$
Hard and Soft represent an orthonormal basis in the two-dimensional Hardness vector space.
$\begin{matrix} \text{Hard}^T \text{Hard} = 1 & \text{Soft}^T \text{Soft} = 1 & \text{Hard}^T \text{Hard} = 0 \end{matrix} \nonumber$
Likewise Black and White are an orthonormal in the two-dimensional Color vector space.
$\begin{matrix} \text{Black}^T \text{Black} = 1 & \text{White}^T \text{White} = 1 & \text{Black}^T \text{White} = 0 \end{matrix} \nonumber$
The relationship between the two bases is reflected in the following projection calculations.
$\begin{matrix} \text{Hard}^T \text{Black} = 0.707 & \text{Hard}^T \text{White} = 0.707 & \text{Soft}^T \text{Black} = 0.707 & \text{Soft}^T \text{White} = -0.707 & \frac{1}{ \sqrt{2}} = 0.707 \end{matrix} \nonumber$
Clearly Black and White can be written as superpositions of Hard and Soft, and vice versa.
$\begin{matrix} \frac{1}{ \sqrt{2}} \text{(Hard + Soft)} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & \frac{1}{ \sqrt{2}} \text{(Hard - Soft)} = \begin{pmatrix} 0.707 \ -0.707 \end{pmatrix} \ \frac{1}{ \sqrt{2}} \text{(Black + White)} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \text{(Black - White)} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Hard, Soft, Black and White are measurable properties and the vectors representing them are eigenstates of the Hardness and Color operators with eigenvalues +/- 1.
Operators
$\begin{matrix} \text{Hardness} = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \text{Color} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{Eigenvalue +1} & \text{Eigenvalue -1} \ \text{Hardness Hard} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{Hardness Soft} = \begin{pmatrix} 0 \ -1 \end{pmatrix} \ \text{Color Black} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & \text{Color White} = \begin{pmatrix} -0.707 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
Hard and Soft are not eigenfunctions of the Color operator, and Black and White are not eigenfunctions of the Hardness operator.
$\begin{matrix} \text{Hardness Black} = \begin{pmatrix} 0.707 \ -0.707 \end{pmatrix} & \text{Hardness White} = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} \ \text{Color Hard} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \text{Color Soft} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
As the Hardness-Color commutator shows, the Hardness and Color operators do not commute. They represent incompatible observables; observables that cannot simultaneously have well-defined values.
$\text{Hardness Color} - \text{Color Hardness} = \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix} \nonumber$
We now proceed with an analysis of the implications of the following two-quon entangled state, expressed in tensor format. A pair of quons is prepared in the following "singlet" state; one is hard and one is soft. (The Appendix shows how to set this state up using Mathcad.)
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \text{Hard} \rangle_1 | \text{Soft} \rangle_2 - | \text{Soft} \rangle_1 | \text{Hard} \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
Given |Ψ> the expectation value for measuring Hardness on the first quon is 0. The same is true for the second quon. In other words, it is equally likely for either quon to be Hard or Soft. (Kronecker is Mathcad's command for tensor multiplication of matrices. See the Appendix for more detail.)
$\begin{matrix} \Psi^T \text{kronecker(Hardness, I)} \Psi = 0 & \Psi^T \text{kronecker(I, Hardness)} \Psi = 0 \end{matrix} \nonumber$
However, if one quon is found to be Hard by measurement, the second will be measured Soft, and vice versa. In other words, there is perfect anti-correlation between the joint measurement of this property on the two quons.
$\Psi^T \text{kronecker(Hardness, Hardness)} \Psi = -1 \nonumber$
Given |Ψ> the expectation value for measuring Color on the first quon is 0. The same is true for the second quon. In other words, it is equally likely for either quon to be Black or White.
$\begin{matrix} \Psi^T \text{kronecker(Color, I)} \Psi = 0 & \Psi^T \text{kronecker(I, Color)} \Psi = 0 \end{matrix} \nonumber$
However, if one quon is found to be Black by measurement, the second will be measured White and vice versa. In other words, there is perfect anti-correlation between the joint measurement of this property on the two quons.
$\Psi^T \text{kronecker(Color, Color)} \Psi = -1 \nonumber$
Furthermore, as the following calculations show, there is no correlation between the measurement outcomes on Color and Hardness.
$\begin{matrix} \Psi^T \text{kronecker(Hardness, Color)} \Psi = 0 & \Psi^T \text{kronecker(Color, Hardness)} \Psi = 0 \end{matrix} \nonumber$
As the foundation for their belief in local realism, Einstein, Podolsky and Rosen (EPR) defined the concept of element of reality in their famous 1935 Physical Review paper,
"If, without in any way disturbing a system, we can predict with certainty (i.e. with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to this physical quantity."
It would seem from the above results, namely these,
$\begin{matrix} \Psi^T \text{kronecker(Color, Color)} \Psi = -1 & \Psi^T \text{kronecker(Hardness, Hardness)} \Psi = -1 \end{matrix} \nonumber$
that according to EPR both hardness and color are elements of reality. If the hardness of quon 1 is measured and found to be soft, we know without measurement (given the reliability of quantum mechanical predictions) that quon 2 is hard. Likewise, if the color of quon 2 is measured and found to be white, we know without measurement that quon 1 is black. On the basis of these calculations, the realist constructs the following table which assigns well-defined hardness and color states to both quons and is consistent with all the quantum calculations.
$\begin{pmatrix} \text{Quon 1} & \text{Quon 2} & \text{HardnessHardness} & \text{ColorColor} & \text{HardnessColor} \ \text{HB} & \text{SW} & -1 & -1 & -1 \ \text{HW} & \text{SB} & -1 & -1 & 1 \ \text{SB} & \text{HW} & -1 & -1 & 1 \ \text{SW} & \text{HB} & -1 & -1 & -1 \ \text{Realist} & \text{AverageValue} & -1 & -1 &0 \ \text{Quantum} & \text{AverageValue} & -1 & -1 & 0 \end{pmatrix} \nonumber$
The problem with this interpretation is that it has previously been shown that the Hardness and Color operators do not commute, meaning that they represent incompatible observables. Incompatible observables cannot be known (determined) simultaneously. A contradiction between the EPR reality criterion and quantum mechanics has thus been shown to exist.
Appendix
Tensor multiplication is used to construct the initial states using Mathcad commands submatrix, kronecker, and augment.
$\Psi = \frac{1}{ \sqrt{2}} \begin{bmatrix} \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix},~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right],~ \text{augment} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix},~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] \right], ~1,~4,~1,~1 \right] ~... \ +- \text{submatrix} \left[ \text{kronecker} \left[ \text{augment} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix},~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right],~ \text{augment} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix},~ \begin{pmatrix} 0 \ 0 \end{pmatrix} \right] \right], ~1,~4,~1,~1 \right] \end{bmatrix} \nonumber$
Kronecker is the Mathcad command that carries out the tensor multiplication of matrices. For example, consider the tensor multiplication of the Hardness and Color matrix operators.
$Hardness \otimes Color = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & 0 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \ 0 & \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & -1 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 \ 0 & 0 & -1 & 0 \end{pmatrix} \nonumber$
$\begin{matrix} \text{kronecker(Hardness, Color)} = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 \ 0 & 0 & -1 & 0 \end{pmatrix} & \text{kronecker} \begin{bmatrix} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix},~ \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \end{bmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 \ 0 & 0 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber$
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This tutorial is based on "Simulating Physics with Computers" by Richard Feynman, published in the International Journal of Theoretical Physics (volume 21, pages 481-485), and Julian Brown's Quest for the Quantum Computer (pages 91-100). Feynman used the experiment outlined below to establish that a local classical computer could not simulate quantum physics.
A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A |L \rangle_B + |R \rangle_A | R \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle of θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal.
In vector notation the left- and right-circular polarization states are expressed as follows:
Left circular polarization:
$\text{L} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
Right circular polarization:
$\text{R} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
In tensor notation the initial photon state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
However, as mentioned above, the photon polarization measurements will actually be made in the vertical/horizontal basis. These polarization states in vector representation are:
Vertical polarization:
$V = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Horizontal polarization:
$H = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
It is easy to show that the equivalent vertical/horizontal polarization state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle_A | V \rangle_B + |V \rangle_A | V \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B + \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
There are four measurement outcomes: both photons are vertically polarized, both are horizontally polarized, one is vertical and the other horizontal, and vice versa. The tensor representation of these measurement states are provided below.
$\begin{matrix} | VV \rangle = \begin{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & | VH \rangle = \begin{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & | HV \rangle = \begin{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & | HH \rangle = \begin{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \ \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & VV = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & VH = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & HV = \begin{pmatrix} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & HH = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Next, the operator representing the rotation of PAB by an angle θ relative to PAA is constructed using matrix tensor multiplication. Kronecker is Mathcad's command for tensor matrix multiplication.
$\text{RotOP}( \theta) = \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix},~ \begin{pmatrix} \cos \theta & \sin \theta \ - \sin \theta & \cos \theta \end{pmatrix} \right] \nonumber$
Now we can get on with some actual calculations. If the relative angle between PAA and PAB is zero degrees we observe perfect correlation. In other words, 50% of the time the analyzers agree that both photons are vertically polarized, and 50% of the time they agree that they are horizontally polarized.
Perfect correlation:
$\begin{matrix} \theta = 0 \text{deg} & \begin{bmatrix} \left( VV^T \text{RotOP}( \theta) \Psi \right)^2 & \left( VH^T \text{RotOP}( \theta) \Psi \right)^2 \ \left( HV^T \text{RotOP}( \theta) \Psi \right)^2 & \left( HH^T \text{RotOP}( \theta) \Psi \right)^2 \end{bmatrix} = \begin{pmatrix} 50 & 0 \ 0 & 50 \end{pmatrix}% \end{matrix} \nonumber$
However, if the relative angle between the two polarization analyzers is 90 degrees, perfect anti-correlation is observed - the photons are always detected with the opposite polarizations.
Perfect anti-correlation:
$\begin{matrix} \theta = 90 \text{deg} & \begin{bmatrix} \left( VV^T \text{RotOP}( \theta) \Psi \right)^2 & \left( VH^T \text{RotOp}( \theta) \Psi \right)^2 \ \left( HV^T \text{RotOP}( \theta) \Psi \right)^2 & \left( HH^T \text{RotOp}( \theta) \Psi \right)^2 \end{bmatrix} = \begin{pmatrix} 0 & 50 \ 50 & 0 \end{pmatrix} \% \end{matrix} \nonumber$
By comparison, if the relative angle between the analyzers is 30 degrees, the analysers behave the same way 75% of the time.
$\begin{matrix} \theta = 30 \text{deg} & \begin{bmatrix} \left( VV^T \text{RotOp}( \theta) \Psi \right)^2 & \left( VH^T \text{RotOp}( \theta) \Psi \right)^2 \ \left( HV^T \text{RotOp}( \theta) \Psi \right)^2 & \left( HH^T \text{RotOp}( \theta) \Psi \right)^2 \end{bmatrix} = \begin{pmatrix} 37.5 & 12.5 \ 12.5 & 37.5 \end{pmatrix} \% \end{matrix} \nonumber$
These quantum calculations are in agreement with experimental results. As we shall see they cannot be explained by a local realistic hidden-variable model of reality.
If you subscribe to the principle of local realism and believe that objects have well-defined properties independent of measurement or observation, the first two results (θ = 0 degrees, θ = 90 degrees) require that the photons carry the following instruction sets, where the hexagonal vertices refer to values of θ = 0, 30, 60, 90, 120, and 150 degrees. There are eight possible instruction sets, six of the type on the left and two of the type on the right. The white circles represent vertical polarization and the black circles represent horizontal polarization. In any given measurement, according to local realism, both photons (A and B) carry identical instruction sets, in other words the same one of the eight possible sets.
The problem is that while these instruction sets are in agreement with the 0 and 90 degree results, they can't explain the 30 degree data. The figure on the left shows that the same result should be obtained 2/3 of the time (4/6) and the figure on the right never. Thus, local realism predicts that the same result should be obtained 50% of the time, as opposed to the actual result of 75% of the time.
$\frac{6 \frac{2}{3} + 2 \times 0}{8} = 50 \% \nonumber$
When Feynman gets to this point in "Simulating Physics with Computers" he writes,
That's all. That's the difficulty. That's why quantum mechanics can't seem to be imitable by a local classical computer.
A local classical computer manipulates bits which are in well-defined states, 0s and 1s, shown above graphically in white and black. However, these classical states are incompatible with the quantum mechanical analysis which is consistent with experimental results. This two-photon experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s. Simulation of quantum physics requires a quantum computer!
Earlier in his paper Feynman sharpened the focus of his analysis by saying "But the physical world is quantum mechanical, and therefore the proper problem is the simulation of quantum physics ..."
He ends his presentation with the following remark.
And I'm not happy with all the analyses that go with just the classical theory, because nature isn't classical, dammit, and if you want to make a simulation of nature, you'd better make it quantum mechanical, ...
In conclusion, the quantum mechanical calculations are presented graphically for all θ values between 0 and 180 degrees.
$\begin{matrix} P_{same} ( \theta) = \left( VV^T \text{RotOp} ( \theta) \Psi \right)^2 + \left( HH^T \text{RotOp} ( \theta) \Psi \right)^2 \ P_{diff} ( \theta) = \left( VH^T \text{RotOp} ( \theta) \Psi \right)^2 + \left( HV^T \text{RotOp} ( \theta) \Psi \right)^2 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.42%3A_A_Summary_of_Feynman%27s_Simulating_Physics_with_Computers.txt
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This tutorial is based on "Simulating Physics with Computers" by Richard Feynman, published in the International Journal of Theoretical Physics (volume 21, pages 481-485), and Julian Brown's Quest for the Quantum Computer (pages 91-100). Feynman used the experiment outlined below to establish that a local classical computer could not simulate quantum physics.
A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A | \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal.
As mentioned above the photon polarization measurements are made in the V-H basis. The L and R polarization states are superpositions of the V and H polarization states, enabling the original state to be written in the measurement basis.
$\begin{matrix} | L \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle + i | H \rangle \right] & | R \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle - i | H \rangle \right] & | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle_A | V \rangle_B - | H \rangle_A |H \rangle_B \right] \end{matrix} \nonumber$
Rotating the right-hand polarizer counter-clockwise by θ yields the following polarization states for B.
$\begin{matrix} | V\ \rangle_B = \cos \theta | V \rangle_B - \sin \theta |H \rangle_B & | H' \rangle_B = \sin \theta | V \rangle_B + \cos \theta | H \rangle_B \end{matrix} \nonumber$
After rotation of B's analyzer the four measurement outcomes, both photons vertically polarized, both horizontally polarized, one vertical and the other horizontal, are expressed mathematically below.
$\begin{matrix} | \Psi_{VV'} \rangle = | V \rangle_A |V' \rangle_B & | \Psi_{VH'} \rangle = |V \rangle_A | H' \rangle_B & | \Psi_{HV'} \rangle = |H \rangle_A | V' \rangle_B & | \Psi_{HH'} \rangle = |H \rangle_A | H' \rangle_B \end{matrix} \nonumber$
Using the equations provided the probability the photons will behave the same way, both vertically polarized or both horizontally polarized is cos2(θ).
$\left| \left\langle \Psi_{VV'} | \Psi \right\rangle \right|^2 + \left| \right|^2 = \cos^2 \theta \nonumber$
Next the probability the photons will behave the same is calculated for three values of θ.
$P_{same} \theta = \cos ( \theta)^2 & P_{same} (0 \text{ deg}) = 1 & P_{same} (90 \text{ deg} = 0 & P_{same} (30 \text{ deg}) = 0.75 \nonumber$
These quantum calculations are in agreement with experimental results, but cannot be explained by a local realistic hidden-variable model of reality.
If objects have well-defined properties independent of measurement or observation, the first two results (θ = 0 degrees, θ = 90 degrees) require that the photons carry the following instruction sets, where the hexagonal vertices refer to θ values of 0, 30, 60, 90, 120, and 150 degrees. There are eight possible instruction sets, six of the type on the left and two of the type on the right. The white circles represent vertical polarization and the black circles represent horizontal polarization. In any given measurement, according to local realism, both photons (A and B) carry identical instruction sets, in other words the same one of the eight possible sets.
The problem is that while these instruction sets are in agreement with the 0 and 90 degree results, they can't explain the 30 degree data. The figure on the left shows that the same result should be obtained 2/3 of the time (4/6) and the figure on the right never. Thus, local realism predicts that the same result should be obtained 50% of the time, as opposed to the actual result of 75% agreement.
$\left( 6 \frac{2}{3} + 2(0) \right) \div 8 = 50 \% \nonumber$
When Feynman gets to this point in "Simulating Physics with Computers" he writes,
That's all. That's the difficulty. That's why quantum mechanics can't seem to be imitable by a local classical computer.
A local classical computer manipulates bits which are in well-defined states, 0s and 1s, shown above graphically in white and black. However, these classical states are incompatible with the quantum mechanical analysis which is in agreement with experimental results. This two-photon experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s. Simulation of quantum physics requires a quantum computer!
Earlier in his paper Feynman sharpened the focus of his analysis by saying "But the physical world is quantum mechanical, and therefore the proper problem is the simulation of quantum physics ..."
He ends his presentation with the following remark.
And I'm not happy with all the analyses that go with just the classical theory, because nature isn't classical, dammit, and if you want to make a simulation of nature, you'd better make it quantum mechanical, ...
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.43%3A_Another_Summary_of_Fenyman%27s_Stimulating_Physics_with_Computers.txt
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The purpose of this tutorial is to review Nick Herbert's "simple proof of Bell's theorem" as presented in Chapter 12 of Quantum Reality.
A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to the observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows that the polarization analyzers (PAs) can be oriented at different angles a and b. (The figure below is similar to the one on page 125 of Jim Baggott's The Meaning of Quantum Theory, which provides a thorough analysis of correlated two-photon experiments.)
The dramatize the quantum weirdness of this EPR experiment, Herbert places PA on Earth and PAB on Betelgeuse, 540 light years away. The source is a space ship midway between the PAs.
In the vertical/horizontal measurement basis the initial polarization state is (see the Appendix for a justification),
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle_A | V \rangle_B - |H \rangle_A | H \rangle_B \right] \nonumber$
There are four measurement outcomes: both photons are vertically polarized, both are horizontally polarized, one is vertical and the other horizontal, and vice versa. In other words, the PAs behave the same or differently. The probabilities for these events are given below (see "Simulating Physics with Computers" by Richard Feynman, published in the International Journal of Theoretical Physics, volume 21, pages 481-485, or The Meaning of Quantum Theory, by Jim Baggott, pages 125-127).
$\begin{matrix} P_{same} = P_{vv} + P_{hh} & P_{same} (a,~b) = \cos (a-b)^2 & P_{diff} = P_{vh} + P_{hv} & P_{diff} (a,~b) = \sin(a-b)^2 \end{matrix} \nonumber$
The following calculations show that if the PAs are oriented at the same angle they behave the same way 100% of the time. This is called perfect correlation.
$\begin{matrix} P_{same} \text{(0 deg, 0 deg)} = 100 \% & P_{same} \text{(30 deg, 30 deg)} = 100 \% & P_{same} \text{(90 deg, 90 deg)} = 100 \% \ P_{diff} \text{(0 deg, 0 deg)} = 0 \% & P_{diff} \text{(30 deg, 30 deg)} = 0 \% & P_{diff} \text{(90 deg, 90 deg)} = 0 \% \end{matrix} \nonumber$
These results appear to support the notion that the linear polarization states of the photons are "elements of reality." In other words, they are photon properties that exist independent of observation. This position is not supported by further calculation and experimentation.
Perfect anti-correlation occurs when the relative angle between the PAs is 90 degrees.
$\begin{matrix} P_{same} \text{(0 deg, 90 deg)} = 0 \% & P_{diff} \text{(0 deg, 90 deg)} = 100 \% \end{matrix} \nonumber$
At 45 degrees there is not correlation between the detectors.
$\begin{matrix} P_{same} \text{(0 deg, 45 deg)} = 50 \% & P_{diff} \text{(0 deg, 45 deg)} = 50 \% \end{matrix} \nonumber$
Using 0 degrees for both PAs at the benchmark, Herbert's analysis proceeds by moving PAB to 30 degrees and noting that this leads to a 25% (1 in 4) discrepancy between the analyzers.
$\begin{matrix} P_{same} \text{(0 deg, 30 deg)} = 75 \% & P_{diff} \text{(0 deg, 30 deg)} = 25 \% \end{matrix} \nonumber$
If instead PAA had been moved to -30 degrees the result is the same, the PAs disagree 25% of the time.
$\begin{matrix} P_{same} \text{(-30 deg, 0 deg)} = 75 \% & P_{diff} \text{(-30 deg, 0 deg)} = 25 \% \end{matrix} \nonumber$
Now the locality principal is invoked. The PAs are spatially separated so that according to conventional intuition, the change in the orientation of PAB has no effect on the results at PAA, and vice versa.
Now Herbert moves PAB back to 30 degrees with the following result.
$\begin{matrix} P_{same} \text{(-30 deg, 30 deg)} = 25 \% & P_{diff} \text{(-30 deg, 30 deg)} = 75 \% \end{matrix} \nonumber$
The angular difference is now 60 degrees, and the PAs disagree 75% of the time. On the basis of local realism one would expect a discrepancy of no more than 50% . If the measurements at the PAs are independent of each other, we should simply be able to add 25% and 25%.
The experiment was performed by John Clauser and Stuart Freedman at Berkeley in 1972 and confirmed the quantum predictions. The agreement between quantum theory and experiment requires that some element of local realism must be abandoned. The consensus is that nature allows non-local interactions for entangled systems such as the photons in this example. The results at PAA and PA2 (light years apart) are connected by a non-local interaction. This type of interaction is, in the words of Herbert, "unmediated, unmitigated and immediate."
Many other experiments besides those of Clauser and Freedman (most notably by Aspect and co-workers) have confirmed quantum mechanical predictions and refuted local realism. Anton Zeilinger described the current situation as follows:
By now, a number of experiments have confirmed quantum predictions to such an extent that a local-realistic world view can no longer be maintained.
It appears that, certainly at least for entangled quantum systems, it is wrong to assume that the features of the world which we observe, the measurement results, exist prior to and independently of our observation.
Appendix
In vector notation the left- and right-circular polarization states are expressed as follows:
Left circular polarization:
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
Right circular polarization:
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
In tensor notation the initial photon state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
Vertical polarization:
$V = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Horizontal polarization:
$H = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
It is easy to show that the equivalent vertical/horizontal polarization state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B + \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
Naturally there are other ways to do this. The most direct would be to write |L> and |R> as superpositions of |V> and |H>, and substitute them into the initial state involving the circular polarization states.
$\Psi = \frac{1}{ \sqrt{2}} \left( L_A L_B + R_A R_B \right) ~ \begin{array}{|l} \text{substitute,}~L_A = \frac{1}{ \sqrt{2}} (V_A + iH_A) \ \text{substitute,}~L_B = \frac{1}{ \sqrt{2}} (V_B + iH_B) \ \text{substitute,}~R_A = \frac{1}{ \sqrt{2}} (V_A - iH_A) \ \text{substitute,}~R_B = \frac{1}{ \sqrt{2}} (V_B - iH_B) \ \text{simplify} \end{array} \rightarrow ~ \Psi = \frac{ \sqrt{2} V_A V_B}{2} - \frac{ \sqrt{2} H_A H_B}{2} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.44%3A_Yet_Another_Assault_on_Local_Realism.txt
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The purpose of this tutorial is to review Nick Herbertʹs ʺsimple proof of Bellʹs theoremʺ as presented in Chapter 12 of Quantum Reality using matrix and tensor algebra.
A two‐stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to the observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows that the polarization analyzers (PAs) can be oriented at different angles a and b. (The figure below is similar to the one on page 125 of Jim Baggottʹs The Meaning of Quantum Theory, which provides a thorough analysis of correlated two‐photon experiments.)
To dramatize the quantum weirdness of this EPR experiment, Herbert places PAA on earth and PAB on Betelgeuse, 540 light years away. The source is a space ship midway between the PAs.
In the vertical/horizontal measurement basis the initial polarization state is (see Appendix A for a justification),
$\begin{matrix} | VV \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & | VH \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & | HV \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & | HH \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
We now write all states, Ψ and the measurement states, in Mathcad's vector format.
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & | VV \rangle = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & | VH \rangle = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & | HV \rangle = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & | HH \rangle = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Next, the operator representing the rotation of PAA by angle a clockwise and PAB by angle b counter‐clockwise (so that the PAs turn in the same direction) is constructed using matrix tensor multiplication. Kronecker is Mathcadʹs command for tensor matrix multiplication.
$\text{RotOp}(a,~b) = \text{kronecker} \begin{bmatrix} \begin{pmatrix} \cos a & - \sin a \ \sin a & \cos a \end{pmatrix}, ~ \begin{pmatrix} \cos b & \sin b \ - \sin b & \cos b \end{pmatrix} \end{bmatrix} \nonumber$
The probability that the detectors will behave the same or differently is calculated as follows.
$\begin{pmatrix} P_{same} (a,~b) = \left( VV^T \text{RotOp}(a,~b) \Psi \right)^2 + \left( HH^T \text{RotOp}(a,~b) \Psi \right)^2 \ P_{diff} (a,~b) = \left( VH^T \text{RotOp}(a,~b) \Psi \right)^2 + \left( HV^T \text{RotOp}(a,~b) \Psi \right)^2 \end{pmatrix} \nonumber$
Now we can get on with some actual calculations. The following calculations show that if the PAs are oriented at the same angle they behave the same way 100% of the time. This is called perfect correlation.
$\begin{matrix} P_{same} \text{(0 deg, 0 deg)} = 100 \% & P_{same} \text{(30 deg, 30 deg)} = 100 \% & P_{same} \text{(90 deg, 90 deg)} = 100 \% \ P_{diff} \text{(0 deg, 0 deg)} = 0 \% & P_{diff} \text{(30 deg, 30 deg)} = 0 \% & P_{diff} \text{(90 deg, 90 deg)} = 0 \% \end{matrix} \nonumber$
These results appear to support the notion that the linear polarization states of the photons are ʺelements of reality.ʺ In other words, they are photon properties that exist independent of observation. However, this position is not supported by further calculation and experimentation.
Perfect anti‐correlation occurs when the relative angle between the PAs is 90 degrees.
$\begin{matrix} P_{same} \text{(0 deg, 0 deg)} = 100 \% & P_{diff} \text{(0 deg, 0 deg)} = 0 \% \end{matrix} \nonumber$
At 45 degrees there is no correlation between the detectors.
$\begin{matrix} P_{same} \text{(0 deg, 45 deg)} = 50 \% & P_{diff} \text{(0 deg, 45 deg)} = 50 \% \end{matrix} \nonumber$
Using 0 degrees for both PAs as the bench mark, Herbertʹs analysis proceeds by moving PAB to 30 degrees and noting that this leads to a 25% (1 in 4) discrepancy between the analyzers.
$\begin{matrix} P_{same} \text{(0 deg, 30 deg)} = 75 \% & P_{diff} \text{(0 deg, 30 deg)} = 25 \% \end{matrix} \nonumber$
If instead PAA had been moved to ‐30 degrees the result is the same, the PAs disagree 25% of the time.
$\begin{matrix} P_{same} \text{(-30 deg, 0 deg)} = 75 \% & P_{diff} \text{(-30 deg, 0 deg)} = 25 \% \end{matrix} \nonumber$
Now the locality principal is invoked. The PAs are spatially separated so that according to conventional intuition, the change in the orientation of PAB has no effect on the results at PAA, and vice versa.
Now Herbert moves PAB back to 30 degrees with the following result.
$\begin{matrix} P_{same} \text{(-30 deg, 30 deg)} = 25 \% & P_{diff} \text{(-30 deg, 30 deg)} = 75 \% \end{matrix} \nonumber$
The angular difference is now 60 degrees, and the PAs disagree 75% of the time. On the basis of local realism one would expect a discrepancy of no more than 50% . If the measurements at the PAs are independent of each other, we should simply be able to add 25% and 25%. A more general version this approach to Bellʹs theorem can be found in the Appendix B.
The experiment was performed by John Clauser and Stuart Freedman at Berkeley in 1972 and confirmed the quantum predictions. The agreement between quantum theory and experiment requires that some element of local realism must be abandoned. The consensus is that nature allows non‐local interactions for entangled systems such as the photons in this example. The results at PAA and PAB (light years apart) are connected by a non‐local interaction. This type of interaction is, in the words of Herbert, ʺunmediated, unmitigated and immediate.ʺ
Many other experiments besides those of Clauser and Freedman (most notably by Aspect and co‐workers) have confirmed quantum mechanical predictions and refuted local realism. Anton Zeilinger described the current situation as follows:
By now, a number of experiments have confirmed quantum predictions to such an extent that a local‐realistic world view can no longer be maintained.
It appears that, certainly at least for entangled quantum systems, it is wrong to assume that the features of the world which we observe, the measurement results, exist prior to and independently of our observation.
Appendix A
In vector notation the left‐ and right‐circular polarization states are expressed as follows:
Left circular polarization:
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
Right circular polarization:
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
In tensor notation the initial photon state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
Vertical polarization:
$V = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Horizontal polarization:
$H = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
It is easy to show that the equivalent vertical/horizontal polarization state is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |V \rangle_A |V \rangle_B + |H \rangle_A |H \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B + \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
Naturally there are other ways to do this. The most direct would be to write |L> and |R> as superpositions of |V> and |H>, and substitute them into the initial state involving the circular polarization states.
$\psi = \frac{1}{ \sqrt{2}} (L_A L_B + R_A R_B ) ~ \begin{array}{|l} \text{substitute, } L_A = \frac{1}{ \sqrt{2}} (V_A + iH_A) \ \text{substitute, } L_B = \frac{1}{ \sqrt{2}} (V_B + iH_B) \ \text{substitute, } L_A = \frac{1}{ \sqrt{2}} (V_A + iH_A) \ \text{substitute, } L_A = \frac{1}{ \sqrt{2}} (V_A + iH_A) \ \text{simplify} \end{array} \rightarrow \psi = \frac{ \sqrt{2} V_A V_B}{2} - \frac{ \sqrt{2} H_A H_B}{2} \nonumber$
Figure 12.4 on page 223 in Herbert's Quantum Reality is reproduced.
Appendix B
It is not difficult to derive the following Bell inequality based on local realism involving three sets of polarization measurement angles for the experiment described above. See The Meaning of Quantum Theory by Jim Baggott, pages 133‐135.
$P_{diff} (a,~ b) + P_{diff} (a,~b) \geq P_{diff} (a,~c) \nonumber$
Below it is shown that the inequality is violated for 0, 22.5 and 45 degrees, as well as for 0, 30 and 60 degrees.
$\begin{matrix} P_{diff} \text{(0 deg, 22.5 deg)} + P_{diff} \text{(22.5 deg, 45 deg)} = 0.293 & P_{diff} \text{(0 deg, 45 deg)} = 0.5 \ P_{diff} \text{(0 deg, 30 deg)} + P_{diff} \text{(30 deg, 60 deg)} = 0.5 & P_{diff} \text{(0 deg, 60 deg)} = 0.75 \end{matrix} \nonumber$
The Bell inequality is also violated for other sets of angles. However, it is not violated for 0, 45 and 90 degrees.
$\begin{matrix} P_{diff} \text{(0 deg, 45 deg)} + P_{diff} \text{(45 deg, 90 deg)} = 1 & P_{diff} \text{(0 deg, 90 deg)} = 1 \end{matrix} \nonumber$
On page 135 Baggott summarizes the significance of Bellʹs inequality.
The most important assumption ... made in the reasoning which led to this inequality was that of ... the local reality of the photons. It is therefore an inequality that is quite independent of the nature of any local hidden variable theory that we could possibly devise. The conclusion is inescapable, quantum theory is incompatible with any local hidden variable theory and hence local reality.
On page 131 he writes that after Bellʹs work,
Questions about local hidden variables immediately changed character. From being rather academic questions about philosophy they became questions of profound importance for quantum theory. The choice between quantum theory and local hidden variable theories was no longer a question of taste, it was a matter of correctness.
This obviously suggests that clever experimentalist might be able to decide which view of reality is correct. To date experimental results have been consistent with the quantum mechanical predictions.
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The purpose of this tutorial is to review Jim Baggott's analysis of Bell's theorem as presented in Chapter 4 of The Meaning of Quantum Theory using matrix and tensor algebra.
A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to the observers in their path.
$\begin{matrix} | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis using calcite crystals, and allows that the polarization analyzers (PAs) can be oriented at different angles a and b. The figure below is taken from Chapter 4 of Jim Baggott's The Meaning of Quantum Theory.
The equivalent wave function in the vertical/horizontal basis is:
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle_A | V \rangle_B - |H \rangle_A | H \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B + \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
The matrix operator representing the calcite crystal, where θ is the angle of rotation of the crystal relative to the vertical is:
$\begin{matrix} \text{Calcite}( \theta) = \begin{pmatrix} \cos (2 \theta) & \sin (2 \theta) \ \sin (2 \theta) & - \cos (2 \theta) \end{pmatrix} & \text{Calcite(0)} = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \text{Calcite} \left( \frac{ \pi}{4} \right) = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{Calcite} \left( \frac{ \pi}{2} \right) = \begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The eigenvalues of the calcite operator for any orientation angle are +/- 1. Several examples are shown below.
$\begin{matrix} \text{eigenvals(Calcite(0))} = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvals} \left( \text{Calcite} \left( \frac{ \pi}{4} \right) \right) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvals} \left( \text{Calcite} \left( \frac{ \pi}{6} \right) \right) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The eigenvalue +1 is assigned to the vertical transmission channel and the -1 eigenvalue to the horizontal transmission channel.
Since only the relative angle between the two PAs is important, PA1 is held at the vertical orientation (θ = 0) and PA2 is rotated. The operator for this experimental set-up is as follows.
$Rot( \theta) \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos ( 2 \theta) & \sin (2 \theta) \ \sin (2 \theta) & - \cos (2 \theta) \end{pmatrix} = \begin{pmatrix} \cos (2 \theta) & \sin (2 \theta) & 0 & 0 \ \sin (2 \theta) & - \cos (2 \theta) & 0 & 0 \ 0 & 0 & - \cos(2 \theta) & - \sin (2 \theta) \ 0 & 0 & - \sin (2 \theta) & \cos (2 \theta) \end{pmatrix} \nonumber$
The expectation value of the Rot(θ) operator is a measure of the correlation between the joint measurements.
$E( \theta) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cos (2 \theta) & \sin (2 \theta) & 0 & 0 \ \sin (2 \theta) & - \cos (2 \theta) & 0 & 0 \ 0 & 0 & - \cos(2 \theta) & - \sin (2 \theta) \ 0 & 0 & - \sin (2 \theta) & \cos (2 \theta) \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} ~ \text{simplify} ~ \rightarrow \cos (2 \theta) \nonumber$
As shown above evaluation of E(θ) yields cos(2θ). For θ = 00 there is perfect correlation; for θ = 900 perfect anti-correlation; for θ = 450 no correlation.
$\begin{matrix} E \text{(0 deg)} = 1 & E \text{(90 deg)} = -1 & E \text{(45 deg)} = 0 \end{matrix} \nonumber$
Baggott derived a correlation function for this experiment based on a local hidden variable model of reality (pp. 110-113, 127-131). It (linear blue line) and the quantum mechanical correlation function, E(θ), are compared on the graph below. Quantum theory and local realism disagree at all angles except 0, 45 and 90 degrees.
Maximum disagreement between quantum theory and the local hidden variable model occurs at 20 and 70 degrees.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.46%3A_Jim_Baggott%27s_Bell_Theorem_Analysis.txt
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The purpose of this tutorial is to review Jim Baggott's analysis of Bell's theorem as presented in Chapter 4 of The Meaning of Quantum Theory using matrix and tensor algebra.
A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle of θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal.
In vector notation the left- and right-circular polarization states are expressed as follows:
Left circular polarization:
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
Right circular polarization:
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
In tensor notation the initial state is the following entangled superposition,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
However, as mentioned above, the photon polarization measurements will actually be made in the vertical/horizontal basis. These polarization measurement states for photons A and B in vector representation are given below. Θ is the angle through which the PA2 has been rotated.
Vertical polarization:
$\begin{matrix} V_A = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V_B = \begin{pmatrix} \cos \theta \ - \sin \theta \end{pmatrix} \end{matrix} \nonumber$
Horizontal polarization:
$\begin{matrix} V_A = \begin{pmatrix} 0 \ 1 \end{pmatrix} & V_B = \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} \end{matrix} \nonumber$
It is easy to show that |Ψ> in the vertical/horizontal basis is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle_A | V \rangle_B + |H \rangle_A | H \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B + \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
There are four possible measurement outcomes: both photons are vertically polarized, both are horizontally polarized, one is vertical and the other horizontal, and vice versa. The vector representations of the measurement states are obtained by tensor multiplication of the individual photon states.
$\begin{matrix} | V_A V_B \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta \ - \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \theta \ - \sin \theta \ 0 \ 0 \end{pmatrix} & | V_A H_B \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} = \begin{pmatrix} \sin \theta \ \cos \theta \ 0 \ 0 \end{pmatrix} \ | H_A V_B \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta \ - \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \cos \theta \ - \sin \theta \end{pmatrix} & | H_A H_B \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \sin \theta \ \cos \theta \end{pmatrix} \end{matrix} \nonumber$
The initial state and the measurement eigenstates are written in Mathcad syntax.
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & V_a V_b ( \theta) = \begin{pmatrix} \cos \theta \ - \sin \theta \ 0 \ 0 \end{pmatrix} & V_a H_b ( \theta) = \begin{pmatrix} \sin \theta \ \cos \theta \ 0 \ 0 \end{pmatrix} \ ~ & H_a V_b ( \theta) = \begin{pmatrix} 0 \ 0 \ \cos ( \theta) \ - \sin ( \theta) \end{pmatrix} & H_a H_b ( \theta) = \begin{pmatrix} 0 \ 0 \ \sin ( \theta) \ \cos ( \theta) \end{pmatrix} \end{matrix} \nonumber$
The projections of the initial state onto the four measurement states are,
Probability amplitude:
$\begin{pmatrix} V_a V_b ( \theta)^T \Psi \ V_a H_b ( \theta)^T \Psi \ H_a V_b ( \theta)^T \Psi \ H_a H_b ( \theta)^T \Psi \end{pmatrix} \rightarrow \begin{pmatrix} \frac{ \sqrt{2} \cos ( \theta)}{2} \ \frac{ \sqrt{2} \sin ( \theta)}{2} \ \frac{ \sqrt{2} \sin ( \theta)}{2} \ \frac{ \sqrt{2} \cos ( \theta)}{2} \end{pmatrix} \nonumber$
Probability:
$\begin{bmatrix} \left( V_a V_b ( \theta)^T \Psi \right)^2 \ \left( V_a H_b ( \theta)^T \Psi \right)^2 \ \left( H_a V_b ( \theta)^T \Psi \right)^2 \ \left( H_a H_b ( \theta)^T \Psi \right)^2 \end{bmatrix} \rightarrow \begin{pmatrix} \frac{ \cos ( \theta)^2}{2} \ \frac{ \sin ( \theta)^2}{2} \ \frac{ \sin ( \theta)^2}{2} \ \frac{ \cos ( \theta)^2}{2} \end{pmatrix} \nonumber$
Assigning an eigenvalue of +1 to a vertical polarization measurement and -1 to a horizontal polarization measurement allows the calculation of the expectation value for the joint polarization measurements, a function which quantifies the correlation between the joint measurements. The eigenvalues for the four joint measurement outcomes are: VaVb = 1; VaHb = -1; HaVb = -1; HaHb = 1. Weighting these by the probability of their occurence gives the expectation value or correlation function.
$E( \theta) = \left( V_a V_b ( \theta)^T \Psi \right)^2 - \left( V_a H_b ( \theta)^T \Psi \right)^2 - \left( H_a V_b ( \theta)^T \Psi \right)^2 + \left( H_a H_b ( \theta)^T \Psi \right)^2 ~ \text{simplify} \rightarrow \cos (2 \theta) \nonumber$
As shown above the evaluation of E(θ) yields cos(2θ). For θ = 00 there is perfect correlation; for θ = 900 perfect anti-correlation; for θ = 450 no correlation.
$\begin{matrix} E \text{(0 deg)} = 1 & E \text{(90 deg)} = -1 & E \text{(45 deg)} = 0 \end{matrix} \nonumber$
Baggott presented a correlation function for this experiment based on a local hidden variable model of reality (pp. 110-113, 127-131). It (linear blue line) and the quantum mechanical correlation function, E(θ), are compared on the graph below. Quantum theory and local realism disagree at all angles except 0, 45 and 90 degrees.
This example illustrates Bell's theorem: no local hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems. As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned.
Appendix
An equivalent computational approach creates a joint measurement operator from the A and B photon measurement eigenstates. This operator is then used to calculate the expectation value or correlation function.
$\left( V_A V_A^T - H_A H_A^T \right) \otimes \left( V_B V_B^T - H_B H_B^T \right) = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos (2 \theta) & - \sin ( 2 \theta) \ - \sin (2 \theta) & 2 \sin ( \theta)^2 - 1 \end{pmatrix} \nonumber$
Recalling that the vertical and horizontal measurement eigenvalues are +1 and -1, the mathematical structures of the A and B operators shown above are confirmed.
$\begin{matrix} V_A V_A^T - H_A H_A^T \rightarrow \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & V_B V_B^T - H_B H_B^T ~ \text{simplify} \rightarrow \begin{pmatrix} \cos (2 \theta) & - \sin ( 2 \theta) \ - \sin (2 \theta) & 2 \sin ( \theta)^2 - 1 \end{pmatrix} \end{matrix} \nonumber$
Tensor multiplication of the individual operators creates the joint measurement operator used in the calculation below.
$E ( \theta ) = \Psi^T \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin ( \theta)^2 - 1 & 0 & 0 \ 0 & 0 & - \cos(2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 1 - 2 \sin ( \theta)^2 \end{pmatrix} ~ \text{simplify} ~ \rightarrow \cos (2 \theta) \nonumber$
The expectation value calculations can also be performed using the trace function as shown below.
$\langle \Psi | \hat{O} | i \rangle \langle i | \Psi \rangle = \sum_i \langle \Psi | \hat{O} | i \rangle \langle i | \Psi \rangle = \sum_i \langle i | \Psi \rangle \langle \Psi | \hat{O} | i \rangle = Trace \left( | \Psi \rangle \langle \Psi | \hat{O} \right) ~ \text{where} ~ \sum_i |i \rangle \langle i | = \text{Identity} \nonumber$
$tr \begin{bmatrix} \Psi \Psi^T \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin ( \theta)^2 - 1 & 0 & 0 \ 0 & 0 & - \cos(2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 1 - 2 \sin ( \theta)^2 \end{pmatrix} \end{bmatrix} ~ \text{simplify} \rightarrow ~ \cos ( 2 \theta) \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.47%3A_Another_Bell_Theorem_Analysis.txt
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The purpose of this tutorial is to review Jim Baggott's analysis of Bell's theorem as presented in Chapter 4 of The Meaning of Quantum Theory using matrix and tensor algebra.
A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A | R \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle of θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal.
In vector notation the left- and right- circular polarization states are expressed as follows:
Left circular polarization:
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
Right circular polarization:
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
In tensor notation the initial state is the following entangled superposition,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A | L \rangle_B + |R \rangle_A | R \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
However, as mentioned above, the photon polarization measurements will actually be made in the vertical/horizontal basis. These polarization measurement states for photons A and B in vector representation are given below. Θ is the angle through which the PA2 has been rotated.
Vertical polarization:
$\begin{matrix} V_A = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V_B = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \end{matrix} \nonumber$
Horizontal polarization:
$\begin{matrix} V_A = \begin{pmatrix} 0 \ 1 \end{pmatrix} & V_B = \begin{pmatrix} \sin \theta \ \cos \theta \end{pmatrix} \end{matrix} \nonumber$
It is easy to show that |Ψ> in the vertical/horizontal basis is,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | V \rangle_A | V \rangle_B + |H \rangle_A |H \rangle_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B + \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
A joint measurement operator is created using the A and B photon measurement eigenstates, with vertical polarization assigned an eigenvalue of +1 and horizontal polarization an eigenvalue of -1. This operator is then used to calculate the expectation value or correlation function.
$\begin{matrix} V_A V_A^T - H_A H_A^T \rightarrow \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & V_B V_B^T - H_B H_B^T \text{simplify} \rightarrow \begin{pmatrix} \cos ( 2 \theta) & - \sin (2 \theta) \ - \sin (2 \theta) & 2 \sin ( \theta)^2 -1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} = \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 & 0 & 0 \ 0 & 0 & - \cos (2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} \nonumber$
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & E( \theta) = \Psi \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 & 0 & 0 \ 0 & 0 & - \cos (2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} \Psi \text{simplify} \rightarrow \cos (2 \theta) \end{matrix} \nonumber$
As shown above the evaluation of E(θ) yields cos(2θ). For θ = 00 there is perfect correlation; for θ = 900 perfect anti-correlation; for θ = 450 no correlation.
$\begin{matrix} E \text{(0 deg)} = 1 & E \text{(90 deg)} = -1 & E \text{(45 deg)} = 0 \end{matrix} \nonumber$
Baggott presented a correlation function for this experiment based on a local hidden variable model of reality (pp. 110-113, 127-131). It (linear blue line) and the quantum mechanical correlation function, E(θ), are compared on the graph below. Quantum theory and local realism disagree at all angles except 0, 45 and 90 degrees.
This example illustrates Bell's theorem: no local hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems. As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned.
In spite of the correlation shown in the joint polarization measurements, the individual measurements on photons 1 and 2 are totally random for all values of θ, i.e. E(θ) = 0. For example, the following operator is used to calculate the expectation value for measurements on photon 2 as a function of its analyzer's angle θ. The identity operator represents no measurement on spin 1.
$\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} = \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 & 0 & 0 \ 0 & 0 & - \cos (2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} \nonumber$
$E( \theta) = \Psi^T \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} = \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 & 0 & 0 \ 0 & 0 & - \cos (2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} \nonumber$
Naturally the same is true for photon 1.
$\begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) \ - \sin (2 \theta) & 2 \sin^2 ( \theta) - 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos (2 \theta) & 0 & - \sin (2 \theta) & 0 \ 0 & \cos (2 \theta) & 0 & - \sin (2 \theta) \ - \sin (2 \theta) & 0 & 2 \sin^2 ( \theta) -1 \ 0 & - \sin (2 \theta) & 0 & 2 \sin^2 ( \theta) - 1 \end{pmatrix} \Psi ~ \text{simplify} \rightarrow 0 \nonumber$
The expectation value calculations can also be performed using the trace function as shown below.
$\langle \Psi | \hat{O} | \Psi \rangle = \sum_i \langle \Psi | \hat{O} | i \rangle \langle i | \Psi \rangle = \sum_i \langle i | \Psi \rangle \langle \Psi | \hat{O} | i \rangle = Trace \left( | \Psi \rangle \langle \Psi | \hat{O} \right) ~ \text{where} ~ \sum_i |i \rangle \langle i | = \text{Identity} \nonumber$
$tr \begin{bmatrix} \Psi \Psi^T \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin ( \theta)^2 - 1 & 0 & 0 \ 0 & 0 & - \cos(2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 1 - 2 \sin ( \theta)^2 \end{pmatrix} \end{bmatrix} ~ \text{simplify} \rightarrow ~ \cos ( 2 \theta) \nonumber$
$tr \left[ \Psi ~ \Psi^T \begin{pmatrix} \cos (2 \theta) & - \sin (2 \theta) & 0 & 0 \ - \sin (2 \theta) & 2 \sin( \theta)^2 - 1 & 0 & 0 \ 0 & 0 & - \cos (2 \theta) & \sin (2 \theta) \ 0 & 0 & \sin (2 \theta) & 1 - 2 \sin ( \theta)^2 \end{pmatrix} \right] ~ \text{simplify}~ \rightarrow \cos (2 \theta) \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.48%3A_Another_Bell_Theorem_Analysis_-_Shorter_Version.txt
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A spin-1/2 pair is prepared in an entangled singlet state and the individual particles travel in opposite directions on the y-axis to a pair of Stern-Gerlach detectors which are set up to measure spin in the x-z plane. Particle 1's spin is measured along the z-axis, and particle 2's spin is measured at an angle θ with respect to the z-axis.
For the singlet state the arrows below indicate the spin orientation for any direction in the x-z plane.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 | \downarrow \rangle_2 - | \downarrow \rangle_1 | \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix}_1 \otimes \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix}_2 - \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix}_1 \otimes \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix}_2 \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The single particle spin operator in the x-z plane is constructed from the Pauli spin operators in the xand z-directions. θ is the angle of orientation of the measurement magnet with the z-axis.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma ( \theta) = \cos ( \theta) \sigma_z + \sin ( \theta) \sigma_x \rightarrow \begin{pmatrix} \cos ( \theta) & \sin ( \theta) \ \sin ( \theta) & - \cos ( \theta) \end{pmatrix} \end{matrix} \nonumber$
Tensor multiplication of σz and σ(θ) creates a joint spin measurement operator.
$\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & - \cos \theta & - \sin \theta \ 0 & 0 & - \sin \theta & \cos \theta \end{pmatrix} \nonumber$
The expectation value as a function of the measurement angle of particle 2 is calculated and the result is displayed graphically.
$E( \theta) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & - \cos \theta & - \sin \theta \ 0 & 0 & - \sin \theta & \cos \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \rightarrow - \cos \theta \nonumber$
The expectation value measures correlation. For θ = 00 there is perfect anti-correlation; for θ = 1800 perfect correlation (see the Appendix for a discussion of this case); for θ = 900 no correlation; for θ = 450 intermediate anti-correlation (-0.707). Calculations provided in the Appendix show that the individual spin measurements on particles 1 and 2 are totally random for all values of θ, i.e. E(θ) = 0.
$\begin{matrix} E \text{(0 deg)} = -1 & E \text{(180 deg)} = 1 & E \text{(90 deg)} = 0 & E \text{(45 deg)} = -0.707 \end{matrix} \nonumber$
If both observers measure their spins in the z-direction quantum mechanics predicts they will get opposite values due to the singlet nature of the spin state. In other words, the combined expectation value is -1 for these measurements. If spin 2 is measured in the x-direction (θ = 90o) quantum mechanics predicts, as noted above, an expectation value of 0.
A realist believes that objects have well-defined properties prior to and independent of observation. The following table provide a local realist's explanation of these results. Specific z- and x-spin states are assigned to the particles in the first two columns, with each particle in one of four equally probable spin orientations consistent with the composite singlet state.
$\begin{bmatrix} \text{Particle 1} & \text{Particle 2} & \hat{ \sigma}_z (1) \hat{ \sigma}_z (2) & \hat{ \sigma}_z (1) \hat{ \sigma}_x (2) \ | \uparrow_z \rangle | \uparrow_x \rangle & | \downarrow_z \rangle | \downarrow_x \rangle & -1 & -1 \ | \uparrow_z \rangle | \downarrow_x \rangle & | \downarrow_z \rangle | \uparrow_x \rangle & -1 & 1 \ | \downarrow_z \rangle | \uparrow_x \rangle & | \uparrow_z \rangle | \downarrow_x \rangle & -1 & 1 \ | \downarrow_z \rangle | \downarrow_x \rangle & | \uparrow_z \rangle | \uparrow_x \rangle & -1 & -1 \ \text{Expectation} & \text{Value} & -1 & 0 \end{bmatrix} \nonumber$
The quantum and classical pictures agree on the prediction of experimental results. The difficulty is that quantum mechanics does not accept the legitimacy of the states shown in the table on the left. One way to state the problem is to note that σx and σz are noncommuting operators.
$\sigma_z \sigma_x - \sigma_x \sigma_z \rightarrow \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix} \nonumber$
According to quantum mechanics spin in the z- and x-directions cannot simultaneously have well-defined values. The states in the table are not valid, in spite of their agreement with experimental results, because they give well-defined values to incompatible observables. Of course, the realist counters that these arguments simply indicate that quantum mechanics is not a complete theory because it cannot assign definite values to all elements of reality prior to and independent of measurement.
If the second spin is measured in the diagonal direction, θ = 45o, the realist again predicts an expectation value of 0, as shown in the following table.
$\begin{bmatrix} \text{Particle 1} & \text{Particle 2} & \hat{ \sigma}_z (1) \hat{ \sigma}_d (2) \ | \uparrow_z \rangle | \uparrow_d \rangle & | \downarrow_z \rangle | \downarrow_d \rangle & -1 \ | \uparrow_z \rangle | \downarrow_d \rangle & | \downarrow_z \rangle | \uparrow_d \rangle & 1 \ | \downarrow_z \rangle | \uparrow_d \rangle & | \uparrow_z \rangle | \downarrow_d \rangle & 1 \ | \downarrow_z \rangle | \downarrow_d \rangle & | \uparrow_z \rangle | \uparrow_d \rangle & -1 \ \text{Expectation} & \text{Value} & 0 \end{bmatrix} \nonumber$
The spin states in this table are also invalid according to quantum theory.
$\sigma_z \sigma \left( \frac{ \pi}{4} \right) - \sigma \left( \frac{ \pi}{4} \right) \sigma_z \rightarrow \begin{pmatrix} 0 & \sqrt{2} \ - \sqrt{2} & 0 \end{pmatrix} \nonumber$
However, as calculated earlier quantum mechanics predicts an expectation value of -0.707. This example illustrates Bell's theorem: no local realist hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems. As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned. Appendix If particle 2's detector is rotated by 180 degrees, spin-up in the z-direction (blue in the table below) has eigenvalue -1 and spin-down (red in the table below) has eigenvalue +1.
$\begin{matrix} \sigma \text{(180 deg)} = \begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix} & \text{eigenvec}( \sigma \text{(180 deg), -1)} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{eigenvec} ( \sigma \text{(180 deg), 1)} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
This leads to the local realist predictions in the right-hand column, which are in agreement with quantum mechanics.
$\begin{bmatrix} \text{Particle 1} & \text{Particle 2} & \hat{ \sigma}_z (1) \hat{ \sigma}_z ^{ \pi} (2) \ | \uparrow_z \rangle | \uparrow_x \rangle & | \downarrow_z \rangle | \downarrow_x \rangle & 1 \ | \uparrow_z \rangle | \downarrow_x \rangle & | \downarrow_z \rangle | \uparrow_x \rangle & 1 \ | \downarrow_z \rangle | \uparrow_x \rangle & | \uparrow_z \rangle | \downarrow_x \rangle & 1 \ | \downarrow_z \rangle | \downarrow_x \rangle & | \uparrow_z \rangle | \uparrow_x \rangle & 1 \ \text{Expectation} & \text{Value} & 1 \end{bmatrix} \nonumber$
The following operator is used to calculate the expectation value for measurements on spin 2 as a function of the detector angle θ. The identity operator represents no measurement on spin 1.
$\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & \cos \theta & \sin \theta \ 0 & 0 & \sin \theta & - \cos \theta \end{pmatrix} \nonumber$
The following calculation shows that the measurement results are totally random, yielding and an expectation value of 0 for all values of θ.
$E ( \theta) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & \cos \theta & \sin \theta \ 0 & 0 & \sin \theta & - \cos \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \rightarrow 0 \nonumber$
The same, of course, is true for the measurements on spin 1.
$\begin{pmatrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos \theta & 0 & \sin \theta & 0 \ 0 & \cos \theta & 0 & \sin \theta \ \sin \theta & 0 & - \cos \theta & 0 \ 0 & \sin \theta & 0 & - \cos \theta \end{pmatrix} \nonumber$
$E ( \theta) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta & 0 & \sin \theta & 0 \ 0 & \cos \theta & 0 & \sin \theta \ \sin \theta & 0 & - \cos \theta & 0 \ 0 & \sin \theta & 0 & - \cos \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \rightarrow 0 \nonumber$
The expectation value calculations can also be performed using the trace function as shown below.
$\langle \Psi | \hat{O} | \Psi \rangle = \sum_i \langle \Psi | \hat{O} | i \rangle \langle i | \Psi \rangle = \sum_i \langle i | \Psi \rangle \langle \Psi | \hat{O} | i \rangle = Trace \left( | \Psi \rangle \langle \Psi | \hat{O} \right) ~ \text{where} ~ \sum_i |i \rangle \langle i | = \text{Identity} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.49%3A_EPR_Analysis_for_a_Composite_Singlet_Spin_System.txt
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A spin-1/2 pair is prepared in an entangled singlet state and the individual particles travel in opposite directions on the y-axis to a pair of Stern-Gerlach detectors which are set up to measure spin in the x-z plane. Particle 1's spin is measured along the z-axis, and particle 2's spin is measured at an angle θ with respect to the z-axis.
For the singlet state the arrows below indicate the spin orientation for any direction in the x-z plane.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_1 | \downarrow \rangle_2 - | \downarrow \rangle_1 | \uparrow \rangle_2 \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix}_1 \otimes \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix}_2 - \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix}_1 \otimes \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix}_2 \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The single particle spin operator in the x-z plane is constructed from the Pauli spin operators in the x- and z-directions. φ is the angle of orientation of the measurement magnet with the z-axis.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & S( \theta) = \cos ( \theta) \sigma_z + \sin ( \theta) \sigma_x \rightarrow \begin{pmatrix} \cos ( \theta) & \sin ( \theta) \ \sin ( \theta) & - \cos ( \theta) \end{pmatrix} & S(0) = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
Tensor multiplication of S(0) and S(φ) creates a joint spin measurement operator.
$\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & - \cos \theta & - \sin \theta \ 0 & 0 - \sin \theta & \cos \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
The expectation value as a function of the measurement angle of particle 2 is calculated and the result displayed graphically.
$E( \theta) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & - \cos \theta & - \sin \theta \ 0 & 0 - \sin \theta & \cos \theta \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \rightarrow - \cos \theta \nonumber$
For θ = 00 there is perfect anti-correlation; for θ = 1800 perfect correlation; for θ = 900 no correlation. A correlation function based on a local-realistic, hidden-variable model (see Fig. 11.2 and related text in A. I. M. Rae's Quantum Mechanics, 2nd Ed.) and the quantum mechanical correlation function, E(θ), are compared on the graph below. Quantum theory and local realism disagree at all angles except 0, 90 and 180 degrees.
This example illustrates Bell's theorem: no local hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems. As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.50%3A_EPR_Analysis_for_a_Composite_Singlet_Spin_System_-_Short_Version.txt
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As will be demonstrated in this tutorial, the Stern-Gerlach experiment illustrates several key quantum concepts. The figure shown below is taken from Thomas Engel's text, Quantum Chemistry & Spectroscopy. The figure depicts the behavior of a beam of Na atoms as it interacts with a sequence of three Stern-Gerlach magnets.
We begin with a review of the quantum mechanics of electron spin.
Spin Eigenstates
Spin-up in the z-direction:
$\sigma_z = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Spin-down in the z-direction:
$\beta_z = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
Spin-up in the x-direction:
$\sigma_x = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \nonumber$
Spin-down in the x-direction:
$\beta_x = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \nonumber$
Operators
The matrix operators associated with the two Stern-Gerlach magnets are shown below.
SGZ operator:
$\text{SGZ} = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \nonumber$
SGX operator:
$\text{SGX} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
The αz and βz spin states are eigenfunctions of the SGZ operator with eigenvalues +1 and -1, respectively:
$\begin{matrix} \text{SGZ} \sigma_z = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \sigma_z^T \text{SGZ} \alpha_z^T = 1 & \text{SGZ} \beta_z = \begin{pmatrix} 0 \ -1 \end{pmatrix} & \beta_z^T \text{SGZ} \beta_z = -1 \end{matrix} \nonumber$
The αx and βx spin states are eigenfunctions of the SGX operator with eigenvalues +1 and -1, respectively:
$\begin{matrix} \text{SGX} \sigma_x = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} & \sigma_x^T \text{SGX} \alpha_x^T = 1 & \text{SGZ} \beta_x = \begin{pmatrix} -0.707 \ 0.707 \end{pmatrix} & \beta_z^T \text{SGX} \beta_x = -1 \end{matrix} \nonumber$
Analysis
Silver atoms are deflected by an inhomogeneous magnetic field because of the two-valued magnetic moment associated with their unpaired 5s electron ([Kr]5s14d10). The beam of silver atoms entering the Stern-Gerlach magnet oriented in the z-direction (SGZ) on the left is unpolarized. This means it is a mixture of randomly spin-polarized Ag atoms. As such, it is impossible to write a quantum mechanical wave function for this initial state. The density operator (or matrix), which is a more general quantum mechanical construct, can be used to represent both pure states and mixtures, as shown below.
$\begin{matrix} \hat{ \rho} = | \Psi \rangle \langle \Psi | & \hat{ \rho} = \sum p)i | \Psi_i \rangle \langle \Psi_i | \end{matrix} \nonumber$
In the equation on the right, pi is the fraction of the mixture in the state Ψi. The expectation value for a measurement on a pure or mixed state is written as follows in terms of the appropriate density operator.
$\langle A \rangle = Trace ( \hat{ \rho} \hat{A} ) \nonumber$
An unpolarized beam can be written as a 50-50 mixture of any of the orthogonal spin eigenstates - αz and βz, or αx and βx, or αy and βy. The density operator for the unpolarized spin beam entering the first SGZ is calculated using αz and βz.
$\rho_{mix} = \frac{1}{2} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \begin{pmatrix} 0.5 & 0 \ 0 & 0.5 \end{pmatrix} \nonumber$
The expectation value for passage through the SGZ magnet is 0, indicating equal amounts of Ag atoms in the spin-up and spin-down exit channels.
$\text{tr}( \rho_{mix} \text{SGZ}) = 0 \nonumber$
The αz beam emerging from SGZ is directed to the SGX magnet and the αx beam emerging from it is directed to another SGZ magnet. Before the second SGZ it might be assumed that the Ag atoms in the beam are in the electronic spin state |αz>|αx>, in other words after the SGZ and SGX magnets the Ag 5s electrons have well-defined values for spin in both the z- and x-directions. The vector representing this state is written in tensor format.
$\begin{matrix} I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \Psi^T \text{kronecker(SGZ, I)} \Psi = 1 & \text{kronecker(SGZ, I)} \Psi = \begin{pmatrix} 0.707 \ 0.707 \ 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
However, the actual experiment illustrated above shows that the expectation value is 0, with equal numbers of silver atoms emerging in αz and βz channels.
The correct quantum mechanical interpretation is that the SGZ and SGX operators do not commute, meaning that they cannot have simultaneous eigenstates.
$\text{SGX SGZ} - \text{SGZ SGX} = \begin{pmatrix} 0 & -2 \ 2 & 0 \end{pmatrix} \nonumber$
The Ag spin state entering the second SGZ magnet is αx, an eigenstate of the SGX operator, not simultaneously an eigenstate of SGZ and SGX. It can be written as a superposition of αz and βz.
$\begin{matrix} \alpha_x = \frac{1}{ \sqrt{2}} ( \alpha_z + \beta_z) & \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \begin{pmatrix} 0.707 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
Quantu theory predicts that the exit channels of the second SGZ magnet will be equally populated with Ag atoms.
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Hardy's paradox is based on analysis of the double Mach-Zehnder interferometer shown below. A positron enters one interferometer and an electron the other. One arm of each interferometer intersect allowing for electron-positron interaction.
In the subsequent quantum mechanical analysis it will be shown that the paradox in Hardy's thought experiment is that it shows that the electron and positron have some probability of interacting without annihilation occurring.
There are two 50-50 beam splitters in each interferometer. The probability amplitude for transmission is $\frac{1}{ \sqrt{2}}$ and for reflection it is $\frac{i}{ \sqrt{2}}$. By convention a 90 degree phase shift is assigned to reflection at the beam splitters.
Initially we ignore interaction at the intersection between the two interferometers. The initial state is |p>|e> represented by pe in Mathcad syntax - p represents the positron and e the electron. The rest of the terms are self-explanatory.
The evolution of the initial state is calculated below and we see that in the absence of interaction in the intersecting arms of the interferometers the electron and positron will be registered at their c-detectors. Recall that we calculate a probability amplitude and its absolute square gives the probability (|-1|2 = 1).
$p~e~ \begin{array}{|l} \text{substitute, e} = \frac{1}{ \sqrt{2}} (v_e + i w_e) \ \text{substitute, p} = \frac{1}{ \sqrt{2}} (v_p + i w_p) \ \text{substitute, } v_p = \frac{1}{ \sqrt{2}} (i c_p + d_p) \ \text{substitute, } w_p = \frac{1}{ \sqrt{2}} (c_p + i d_p) \ \text{substitute, } w_e = \frac{1}{ \sqrt{2}} (c_e + i d_e) \ \text{substitute, } v_e = \frac{1}{ \sqrt{2}} (i c_e + d_e) \end{array} \nonumber$
Now we look at the situation immediately after the first set of beam splitters and explore the implication of interaction, or in this case electron-positron annihilation.
$p~e~ \begin{array}{|l} \text{substitute, e} = \frac{1}{ \sqrt{2}} (v_e + i w_e) \ \text{substitute, p} = \frac{1}{ \sqrt{2}} (v_p + i w_p) \end{array} \rightarrow \frac{v_e v_p}{2} + \frac{v_p w_e i}{2} - \frac{w_e w_p}{2} + \frac{v_e w_p i}{2} \nonumber$
The term wewp gives the probability (25%) that the electron and positron will be in the intersecting arms of the interferometers and annihilate each other. Annihilation removes that amplitude term from further evolution. Replacing it with the symbol γ and calculating what happens at the second set of beam splitters yields the following result.
$p~e~ \begin{array}{|l} \text{substitute, } v_p = \frac{1}{ \sqrt{2}} (i c_p + d_p) \ \text{substitute, } w_p = \frac{1}{ \sqrt{2}} (c_p + i d_p) \ \text{substitute, } w_e = \frac{1}{ \sqrt{2}} (c_e + i d_e) \ \text{substitute, } v_e = \frac{1}{ \sqrt{2}} (i c_e + d_e) \end{array} \rightarrow - \frac{ \gamma}{2} - \frac{3 c_e c_p}{4} + \frac{c_p d_e i}{4} - \frac{d_e d_p}{4} + \frac{c_e d_p i}{4} \nonumber$
This calculation reveals the paradox. Earlier it was shown that in the absence of interaction ce and cp should fire simultaneously 100% of the time. Now, with interaction/annihilation, that drops to a probability of 9/16 and a photon detected 25% of the time. The bizarre part of the quantum mechanical calculation is in the next to the last term (dedp/4) above. For both d detectors to fire requires the amplitude before the second set of beam splitters to be wewp, as shown by the next to the last term on the right below.
$\frac{1}{2} w_ew_p~ \begin{array}{|l} \text{substitute, } w_e = \frac{1}{ \sqrt{2}} (c_e + i d_e) \ \text{substitute, } w_p = \frac{1}{ \sqrt{2}} (c_p + i d_p) \end{array} \rightarrow -\frac{c_e c_p}{4} + \frac{c_p d_e i}{4} - \frac{d_e d_p}{4} + \frac{c_e d_p i}{4} \nonumber$
But this amplitude should lead to annihilation. So quantum theory appears to permit a positron/electron interaction that does not lead to annihilation with a probability of 1/16.
The Appendix shows how to carry out the analysis using tensor algebra.
Appendix
The motion of the positron and electron to the left and right is represented by the following orthonormal vectors.
$\begin{matrix} L = \begin{pmatrix} 1 \ 0 \end{pmatrix} & R = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Initially the positron moves to the left and the electron to the right, giving the following initial state in tensor format.
$\text{p e} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \nonumber$
The beam splitters and mirrors are represented by 2x2 matrices. The mirrors are implied in the above figure by the change in direction between the two beam splitters.
Beam splitter:
$\text{BS} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i \ i & 1 \end{pmatrix} \nonumber$
Mirror:
$\text{M} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \nonumber$
The four output states are calculated in tensor format using motional basis vectors, |L> and |R>.
$\begin{matrix} c_p d_e = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & c_p c_e = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & d_p d_e = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & d_p c_e = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} & \end{matrix} \nonumber$
The operator representing the dual Mach-Zehnder interferometers is constructed by tensor multiplication of BS and M matrix operators.
$\text{DMZ} = \text{kronecker(BS, BS) kronecker(M, M) kronecker(BS, BS)} \nonumber$
The following calculations show that, in the absence of positron-electron interaction only the c-detectors fire simultaneously.
$\begin{matrix} \left[ \left| \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \text{DMZ} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \right| \right]^2 = 0 & \left[ \left| \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} \text{DMZ} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \right| \right]^2 = 1 \ \left[ \left| \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} \text{DMZ} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \right| \right]^2 = 0 & \left[ \left| \begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix} \text{DMZ} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \right| \right]^2 = 0 \end{matrix} \nonumber$
To complete the analysis in tensor format we look at the p-e state immediately after the first set of beam splitters.
$\text{kronecker(BS, BS)} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.5i \ 0.5 \ -0.5 \ 0.5i \end{pmatrix} \nonumber$
$\frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} i \ 0 \end{pmatrix} + \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ i \end{pmatrix} \begin{pmatrix} i \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ i \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{2} \left[ \begin{pmatrix} i \ 0 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ i \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} i \ 1 \ -1 \ i \end{pmatrix} \nonumber$
The third position on the product vector is the probability amplitude that the positron is in the right arm of its interferometer and that the electron is in the left arm of its interferometer - the w arms in the figure above. If the positron-electron pair interact annihilation occurs and a photon is produced. That term, therefore, is set to zero and the state used for the remaining evolution is,
$\frac{1}{2} \begin{pmatrix} i \ 1 \ 0 \ i \end{pmatrix} \nonumber$
This state interacts with the remaining mirrors and beam splitters.
$\text{BSM} = \text{kronecker(BS, BS) kronecker(M, M)} \nonumber$
The following output probability calculations are in exact agreement with the earlier calculations.
$\begin{matrix} \left[ \left| \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \text{BSM} \frac{1}{2} \begin{pmatrix} i \ 1 \ 0 \ i \end{pmatrix} \right| \right]^2 = 0.0625 & \left[ \left| \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} \text{BSM} \frac{1}{2} \begin{pmatrix} i \ 1 \ 0 \ i \end{pmatrix} \right| \right]^2 = 0.5625 \ \left[ \left| \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} \text{BSM} \frac{1}{2} \begin{pmatrix} i \ 1 \ 0 \ i \end{pmatrix} \right| \right]^2 = 0.0625 & \left[ \left| \begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix} \text{BSM} \frac{1}{2} \begin{pmatrix} i \ 1 \ 0 \ i \end{pmatrix} \right| \right]^2 = 0.0625 \end{matrix} \nonumber$
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The Bell states are maximally entangled superpositions of two-particle states. Consider two spin-1/2 particles created in the same event. There are four maximally entangled wave functions representing their collective spin states. Each particle has two possible spin orientations and therefore the composite state is represented by a 4-vector in a four-dimensional Hilbert space.
$\begin{matrix} | \Phi_p \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle + | \downarrow_1 \rangle | \downarrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} & \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \ | \Phi_m \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle - | \downarrow_1 \rangle | \downarrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \ | \Psi_p \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \downarrow_2 \rangle + | \downarrow_1 \rangle | \uparrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} & \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \ | \Psi_m \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \downarrow_2 \rangle - | \downarrow_1 \rangle | \uparrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The wave functions are not separable and consequently the entangled particles represented by these wave functions do not have separate identities or individual properties, they behave like a single entity. Individually the spins don't have a definite polarization, yet there is a definite spin orientation relationship between them. For example, if the spin orientation of particle 1 is learned through measurement, the spin orientation of particle 2 is also immediately known no matter how far away it may be. (The Appendix shows how such measurements destroy entanglement, forcing both particles into well-defined spin states.) Entanglement implies nonlocal phenomena which in the words of Nick Herbert are "unmediated, unmitigated and immediate."
The Bell states can be generated from two classical bits with the use of a quantum circuit involving a Hadamard (H) gate, the identity (I) and a controlled-not gate (CNOT) as shown below. The H gate operates on the top bit creating a superposition which controls the operation of the CNOT gate. The classical state on the left also serves as an index for the Bell state created from it: 0, 1, 2, 3. Kronecker, as discussed below, is Mathcad's command for matrix tensor multiplication.
$\begin{matrix} H = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \begin{matrix} \text{Bell} & | a \rangle & \triangleright & H & \cdot & \triangleright & \text{Bell} \ \text{State} ~ & ~ & ~ & | & ~ & \beta_{ab} \ \text{Index} & | \beta \rangle & \triangleright & \cdots & \oplus & \triangleright & \text{State} \end{matrix} \end{matrix} \nonumber$
Bell state generator:
$\text{BSG} = \text{CNOT kronecker(H, I)} \nonumber$
$\begin{matrix} \text{Index = 0} & \text{Index = 1} \ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} ~~ \text{BSG} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} ~~ \text{BSG} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} \ \text{Index = 2} & \text{Index = 3} \ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} ~~ \text{BSG} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} ~~ \text{BSG} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} \end{matrix} \nonumber$
It is not surprising that a Bell state measurement (BSM) is just the reverse of this process.
$\begin{matrix} \text{Bell} & \triangleright & \cdot & H & \triangleright & | a \rangle & \text{Bell} \ \beta_{ab} & ~ & | ~ & ~ & ~ & ~ & \text{State} \ \text{State} & \triangleright & \oplus & \cdots & \triangleright & | b \rangle & \text{Index} \end{matrix} \nonumber$
$\text{BSM = BSG}^{-1} \nonumber$
$\begin{matrix} \text{BSM} \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{BSM} \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \text{BSM} \begin{pmatrix} 0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{BSM} \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
We will now explore some of the characteristics of the Bell states through a variety of calculations. First we list the spin eigenfunctions in the Cartesian directions.
$\begin{matrix} \text{Szu} = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{Szd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \text{Sxu} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & \text{Sxd} = \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \end{pmatrix} & \text{Syu} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} & \text{Syd} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \end{matrix} \nonumber$
Working in units of h/4π we can use the Pauli matrices to represent the spin operators in the Cartesian directions. We will also need the identity operator from above.
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
Composite two-particle operators are constructed using matrix tensor multiplication implemented with Mathcad's kronecker command.
$\begin{matrix} \text{SxSx} = \text{kroncker}( \sigma_x,~ \sigma_x) & \text{SySy} = \text{kronecker}( \sigma_y,~ \sigma_y) & \text{SzSz} = \text{kroncker} ( \sigma_z,~ \sigma_z) \end{matrix} \nonumber$
The following matrix demonstrates that the Bell states are an orthonormal basis set.
$\begin{pmatrix} \Phi_p^T \Phi_p & \Phi_p^T \Phi_m & \Phi_p^T \Psi_p & \Phi_p^T \Psi_m \ \Phi_m^T \Phi_p & \Phi_m^T \Phi_m & \Phi_m^T \Psi_p & \Phi_m^T \Psi_m \ \Psi_p^T \Phi_p & \Psi_p^T \Phi_m & \Psi_p^T \Psi_p & \Psi_p^T \Psi_m \ \Psi_m^T \Phi_p & \Psi_m^T \Phi_m & \Psi_m^T \Psi_p & \Psi_m^T \Psi_m \ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \nonumber$
In the interest of brevity, the main quantum properties of the Bell states will be illustrated using the spin operator in the z-direction. Calculations involving the x- and y-spin operators can be found in the appendix.
The Bell states are eigenfunctions of the σzσz, σxσx and σyσy spin operators.
$\begin{array}{c & c} \text{eigenvalue} = 1 & \text{eigenvalue} = -1 \ \begin{matrix} SzSz \Phi_p = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} & SzSz \Phi_m = \begin{pmatrix} 0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} \ SxSx \Phi_p = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} & SxSx \Psi_p = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} \ SySy \Phi_m = \begin{pmatrix} 0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} & SySy \Psi_p = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} \end{matrix} & \begin{matrix} SzSz \Psi_p = \begin{pmatrix} 0 \ -0.707 \ -0.707 \ 0 \end{pmatrix} & SzSz \Psi_m = \begin{pmatrix} 0 \ - 0.707 \ 0.707 \ 0 \end{pmatrix} \ SxSx \Phi_m = \begin{pmatrix} - 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} & SxSx \Psi_m = \begin{pmatrix} 0 \ -0.707 \ 0.707 \ 0 \end{pmatrix} \ SySy \Phi_p = \begin{pmatrix} -0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} & SySy \Psi_m = \begin{pmatrix} 0 \ -0.707 \ 0.707 \ 0 \end{pmatrix} \end{matrix} \end{array} \nonumber$
Calculations of the expectation values yield consistent results.
$\begin{matrix} \Phi_p^T SzSz \Phi_p = 1 & \Phi_m^T SzSz \Phi_m = 1 & \Psi_p^T SzSz \Psi_p = -1 & \Psi_m^T SzSz \Psi_m = 1 \ \Phi_p^T SxSx \Phi_p = 1 & \Phi_m^T SxSx \Phi_m = -1 & \Psi_p^T SxSx \Psi_p = 1 & \Psi_m^T SxSx \Psi_m = -1 \ \Phi_p^T SySy \Phi_p = -1 & \Phi_m^T SySy \Phi_m = 1 & \Psi_p^T SySy \Psi_p = -1 & \Psi_m^T SySy \Psi_m = 1 \ \end{matrix} \nonumber$
We see that for the composite measurements in the z-basis the Φ Bell states exhibit perfect correlation, while the Ψ states show perfect anti-correlation. In the x-basis, the p states show perfect correlation and the m states show perfect anti-correlation.
This correlation is striking when you consider that the individual spin measurements on particles 1 and 2 are completely random giving expectation values of zero, as the following calculations show.
$\begin{matrix} \Phi_p^T \text{kronecker}( \sigma_z,~I) \Phi_p = 0 & \Phi_p^T \text{kronecker}(I,~ \sigma_z) \Phi_p = 0 \ \Phi_m^T \text{kronecker}( \sigma_z,~I) \Phi_m = 0 & \Phi_m^T \text{kronecker}(I,~ \sigma_z) \Phi_m = 0 \ \Psi_p^T \text{kronecker}( \sigma_z,~I) \Phi_p = 0 & \Psi_p^T \text{kronecker}(I,~ \sigma_z) \Psi_p = 0 \ \Psi_m^T \text{kronecker}( \sigma_z,~I) \Psi_m = 0 & \Psi_m^T \text{kronecker}(I,~ \sigma_z) \Psi_m = 0 \ \end{matrix} \nonumber$
Individually the particles appear to behave like classical mixtures, but collectively they exhibit quantum correlations. We will now look at this issue from another perspective. First the density operators (|Ψ><Ψ|) are calculated for each of the Bell states. (See the Appendix for a graphical representation of the density matrices.
$\begin{matrix} \Phi_p \Phi_p^T = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0.5 & 0 & 0 & 0.5 \end{pmatrix} & \Phi_m \Phi_m^T = \begin{pmatrix} 0.5 & 0 & 0 & -0.5 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ -0.5 & 0 & 0 & 0.5 \end{pmatrix} \ \Psi_p \Psi_p^T = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & 0.5 & 0 \ 0 & 0.5 & 0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} & \Psi_m \Psi_m^T = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & -0.5 & 0 \ 0 & -0.5 & 0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
We also demonstrate that the Bell states are pure states by showing that (|Ψ><Ψ|)2 = |Ψ><Ψ|.
$\begin{matrix} \left( \Phi_p \Phi_p^T \right)^2 = \begin{pmatrix} 0.5 & 0 & 0 & 0.5 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0.5 & 0 & 0 & 0.5 \end{pmatrix} & \left( \Phi_m \Phi_m^T \right)^2 = \begin{pmatrix} 0.5 & 0 & 0 & -0.5 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ -0.5 & 0 & 0 & 0.5 \end{pmatrix} \ \left( \Psi_p \Psi_p^T \right)^2 = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & 0.5 & 0 \ 0 & 0.5 & 0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} & \left( \Psi_m \Psi_m^T \right)^2 = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & -0.5 & 0 \ 0 & -0.5 & 0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
The existence of off-diagonal elements in these matrix operators is the quantum signature of an entangled superposition. We concentrate on the |Φm><Φm| density operator which is shown below. The following analysis is adapted from AJP 77, 244-252 (2009).
$| \Phi_m \rangle \langle \Phi_m | = \frac{1}{2} \left[ | \uparrow_1 \rangle \otimes | \uparrow_2 \rangle - | \downarrow_1 \rangle \otimes | \downarrow_2 \rangle \right] \left[ \langle \uparrow_2 | \otimes \langle \uparrow_1 | - \langle \downarrow_2 | \otimes \langle \downarrow_1 | \right] \nonumber$
In expanding this expression we will make use of identities like the following, which rearranges 1221 terms as 1122 terms.
$| \uparrow_1 \rangle \otimes | \uparrow_2 \rangle \langle \downarrow_2 | \otimes \langle \downarrow_1 | = | \uparrow_1 \rangle \langle \downarrow_1 | \otimes | \uparrow_2 \rangle \langle \downarrow_2 | \nonumber$
$| \Phi_m \rangle \langle \Phi_m | = \frac{1}{2} \left[ | \uparrow_1 \rangle \langle \uparrow_1 | \otimes | \uparrow_2 \rangle \langle \uparrow_2 | - | \uparrow_1 \langle \downarrow_1 | \otimes | \uparrow_2 \rangle \langle \downarrow_2 | - | \downarrow_1 \rangle \langle \uparrow_1 | \otimes | \downarrow_2 \rangle \langle \uparrow_2 | + | \downarrow_1 \rangle \langle \downarrow_1 | \otimes | \downarrow_2 \rangle \langle \downarrow_2 | \right] \nonumber$
To establish the validity of this arrangement we can recalculate |Φm><Φm|.
$\frac{1}{2} \begin{bmatrix} \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix},~ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] - \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix},~ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] ~ \cdots \ + \text{-kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix},~ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] + \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix},~ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \end{bmatrix} = \begin{pmatrix} 0.5 & 0 & 0 & -0.5 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ -0.5 & 0 & 0 & 0.5 \end{pmatrix} \nonumber$
The reason for writing |Φm><Φm| in this form is that it facilitates the calculation of the reduced density operator for particle 1 by "tracing" over the contribution of particle 2, or vice versa. (See the Appendix for a more transparent and thorough presentation of the partial trace.) Here we do the former and see that particle 1's "reduced or local" state operator is diagonal indicating a classical mixed state. This result is consistent with the random behavior observed earlier for the expectation values for the individual particles.
$\frac{1}{2} \begin{bmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] - \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \cdots \ + - \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \end{bmatrix} = \begin{pmatrix} 0.5 & 0 \ 0 & 0.5 \end{pmatrix} \nonumber$
Naturally the same is true if we trace over the contribution of particle 1 to obtain the local state operator of particle 2.
$\frac{1}{2} \begin{bmatrix} \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] - \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \cdots \ + - \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] + \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \end{bmatrix} = \begin{pmatrix} 0.5 & 0 \ 0 & 0.5 \end{pmatrix} \nonumber$
That these local density operators represent mixed states is further demonstrated by showing that for each the trace of the square of density operator is less than 1. The Bell states are pure (see above) because for them the trace of the square of the density operator equals 1.
$\begin{matrix} \text{tr} \left[ \begin{pmatrix} 0.5 & 0 \ 0 & 0.5 \end{pmatrix}^2 \right] = 0.5 & \begin{pmatrix} \text{tr} \left( \Phi_p \Phi_p^T \right)^2 & \text{tr} \left( \Phi_m \Phi_m^T \right)^2 & \text{tr} \left( \Psi_p \Psi_p^T \right)^2 & \text{tr} \left( \Phi_m \Psi_m^T \right)^2 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix} \end{matrix} \nonumber$
These calculations, like the previous ones, also establish that while the composite system is a coherent quantum superposition and pure state, the subsystems of the Bell state, spins 1 and 2, exhibit the classical statistical behavior of mixtures. The quantum correlations of the entangled superposition are only apparent when the measurement results on the individual spins are compared side by side.
Appendix
The following table summarizes the expectation value calculations for all three spin operators.
$\begin{pmatrix} \Phi_p^T SxSx \Phi_p & \Phi_p^T SySy \Phi_p & \Phi_p^T SzSz \Phi_p \ \Phi_m^T SxSx \Phi_m & \Phi_m^T SySy \Phi_m & \Phi_m^T SzSz \Phi_m \ \Psi_p^T SxSx \Psi_p & \Psi_p^T SySy \Psi_p & \Psi_p^T SzSz \Psi_p \ \Psi_m^T SxSx \Psi_m & \Psi_m^T SySy \Psi_m & \Psi_m^T SzSz \Psi_m \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1 \ -1 & 1 & 1 \ 1 & 1 & -1 \ -1 & -1 & -1 \end{pmatrix} \nonumber$
This calculation can also be performed by tracing over the product of the density operator and the appropriate spin operator.
$\langle \Psi | \hat{A} | \Psi \rangle = \sum_a \langle \Psi | \hat{A} | a \rangle \langle a | \Psi \rangle = \sum_a \langle a | \Psi \rangle \langle \Psi | \hat{A} | a \rangle = Tr \left( | \Psi \rangle \langle \Psi | \hat{A} \right) \nonumber$
$\begin{pmatrix} \text{tr} \left( \Phi_p \Phi_p^T SxSx \right) & \text{tr} \left( \Phi_p \Phi_p^T SySy \right) & \text{tr} \left( \Phi_p \Phi_p^T SzSz \right) \ \text{tr} \left( \Phi_m \Phi_m^T SxSx \right) & \text{tr} \left( \Phi_m \Phi_m^T SySy \right) & \text{tr} \left( \Phi_m \Phi_m^T SzSz \right) \ \text{tr} \left( \Psi_p \Psi_p^T SxSx \right) & \text{tr} \left( \Psi_p \Psi_p^T SySy \right) & \text{tr} \left( \Psi_p \Psi_p^T SzSz \right) \ \text{tr} \left( \Psi_m \Psi_m^T SxSx \right) & \text{tr} \left( \Psi_m \Psi_m^T SySy \right) & \text{tr} \left( \Psi_m \Psi_m^T SzSz \right) \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1 \ -1 & 1 & 1 \ 1 & 1 & -1 \ -1 & -1 & -1 \end{pmatrix} \nonumber$
It is clear from these results that the Bell states are eigenvectors of the total spin-spin operator.
$\begin{matrix} H = \text{kronecker} \left( \sigma_x,~ \sigma_x \right) + \text{kronecker} \left( \sigma_y,~ \sigma_y \right) + \text{kronecker} \left( \sigma_z,~ \sigma_z \right) & H = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 2 & 0 \ 0 & 2 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \Psi_m^T H \Psi_m = -3 & \Psi_p^T H \Psi_p = 1 & \Phi_p^T H \Phi_p = 1 & \Phi_m^T H \Phi_m = 1 \end{matrix} \nonumber$
$-3 \Psi_m \Psi_m^T + \Psi_p \Psi_p^T + \Phi_p \Phi_p^T + \Phi_m \Phi_m^T = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 2 & 0 \ 0 & 2 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \nonumber$
More about the partial trace:
The partial trace as given above is not particularly transparent in its matrix implementation. Here the partial trace over spin 1 is presented using conventional Dirac notation. First |Φm><Φm| is expanded.
$| \Phi_m \rangle \langle \Phi_m | = \frac{1}{2} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle \langle \uparrow_2 | \langle \uparrow_1 | - | \uparrow_1 \rangle | \uparrow_2 \rangle \langle \downarrow_2 | \langle \downarrow_1 | - | \downarrow_1 \rangle | \downarrow_2 \rangle \langle \uparrow_2 | \langle \uparrow_1 | + | \downarrow_1 \rangle | \downarrow_2 \rangle \langle \downarrow_2 | \downarrow_1 | \right] \nonumber$
Now the trace of spin 1 (bue) is generated showing that the cross terms in the middle of the above expression (red) vanish leaving the following result.
$\frac{1}{2} \left[ \langle \uparrow_1 | \Phi_m \rangle \langle \Phi_m | \uparrow_1 \rangle + \langle \downarrow_1 | \Phi_m \rangle \langle \Phi_m | \downarrow_1 \rangle \right] = \frac{1}{2} \left[ | \uparrow_2 \rangle \langle \uparrow_2 | + | \downarrow_2 \rangle \langle \downarrow_2 | \right] \nonumber$
Wave function collapse:
Previously it was noted that if the spin orientation of particle 1 is learned through measurement, the spin orientation of particle 2 is also immediately known no matter how far away it may be. To show this we first create projection (measurement) operators for spin up and down in the z- and x-directions.
$\begin{matrix} P_{zu} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} & P_{zd} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \ P_{zu} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \ \frac{1}{ \sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{ \sqrt{2}} & \frac{1}{ \sqrt{2}} \end{pmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{pmatrix} & P_{zd} = \begin{pmatrix} \frac{1}{ \sqrt{2}} \ - \frac{1}{ \sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{ \sqrt{2}} & - \frac{1}{ \sqrt{2}} \end{pmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} & - \frac{1}{2} \ - \frac{1}{2} & \frac{1}{2} \end{pmatrix} \end{matrix} \nonumber$
For the Bell state Ψp, measurement of the z-direction spin of the first particle has two equally likely (probability 1/2) outcomes - spin-up or spin-down. These measurement outcomes cause the second particle to have the opposite spin. This result can, of course, be easily predicted by inspection of Ψp.
$\Psi_p = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] ~~ \begin{matrix} \text{kronecker}(P_{zu},~ I) \Psi_p = \begin{pmatrix} 0 \ 0.707 \ 0 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \ \text{kronecker}(P_{zd},~ I) \Psi_p = \begin{pmatrix} 0 \ 0 \ 0.707 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
Measurement of the x-direction spin of the first particle has two equally likely (probability 1/2) outcomes - spin-up or spin-down. However, for Ψp the measurements cause the second particle to have the same spin. This result is not obtained by inspection because the Bell states have been written in the z-basis.
$\begin{matrix} \text{kronecker}(P_{xu},~I) \Psi_p = \begin{pmatrix} 0.354 & 0.354 & 0.354 & 0.354 \end{pmatrix} & \frac{1}{ \sqrt{2}} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \ \text{kronecker}(P_{xd},~I) \Psi_p = \begin{pmatrix} -0.354 & 0.354 & 0.354 & -0.354 \end{pmatrix} & \frac{-1}{ \sqrt{2}} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
This seemingly peculiar result occurs because in the x-basis Ψp has the following form.
$\Psi_p = \frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} - \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
Naturally, these results are consistent with the following expectation values.
$\begin{matrix} \Psi_p^T SzSz \Psi_p = -1 & \Psi_p^T SxSx \Psi_p = 1 \end{matrix} \nonumber$
If we repeat these calculations using Φm the following results are obtained.
$\Psi_p = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] ~~ \begin{matrix} \text{kronecker}(P_{zu},~ I) \Phi_m = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \ \text{kronecker}(P_{zd},~ I) \Phi_m = \begin{pmatrix} 0 \ 0 \ 0 \ -0.707 \end{pmatrix} & \frac{-1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The z-direction measurements can again be predicted by simple inspection of the entangled wave function.
$\begin{matrix} \text{kronecker}(P_{xu},~I) \Phi_m = \begin{pmatrix} 0.354 & -0.354 & 0.354 & -0.354 \end{pmatrix} & \frac{1}{ \sqrt{2}} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \ \text{kronecker}(P_{xd},~I) \Phi_m = \begin{pmatrix} 0.354 & 0.354 & -0.354 & -0.354 \end{pmatrix} & \frac{-1}{ \sqrt{2}} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The x-direction results are obvious if we write Φm in the x-basis.
$\Phi_m = \frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} + \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
The results are also consisten with the appropriate expectation values.
$\begin{matrix} \Phi_m^T SzSz \Phi_m = 1 & \Phi_m^T SxSx \Phi_m = -1 \end{matrix} \nonumber$
Graphical representations of the Bell state density matrices:
Algebraic analysis of Bell state generation (read left to right) and Bell state measurement (read right to left).
$\begin{matrix} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = | 0 \rangle | 0 \rangle \underleftrightarrow{H \otimes I} \frac{1}{ \sqrt{2}} ( | 0 \rangle + | 1 \rangle) | 0 \rangle = \frac{1}{ \sqrt{2}} (|00 \rangle + | 10 \rangle) \underleftrightarrow{CNOT} \frac{1}{ \sqrt{2}} ( | 00 \rangle + |11 \rangle ) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = | 0 \rangle | 1 \rangle \underleftrightarrow{H \otimes I} \frac{1}{ \sqrt{2}} ( | 0 \rangle + | 1 \rangle) | 1 \rangle = \frac{1}{ \sqrt{2}} (|01 \rangle + | 11 \rangle) \underleftrightarrow{CNOT} \frac{1}{ \sqrt{2}} ( | 01 \rangle + |10 \rangle ) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \ \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = | 1 \rangle | 0 \rangle \underleftrightarrow{H \otimes I} \frac{1}{ \sqrt{2}} ( | 0 \rangle - | 1 \rangle) | 0 \rangle = \frac{1}{ \sqrt{2}} (|00 \rangle - | 10 \rangle) \underleftrightarrow{CNOT} \frac{1}{ \sqrt{2}} ( | 00 \rangle - |11 \rangle ) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \ \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} = | 1 \rangle | 01\rangle \underleftrightarrow{H \otimes I} \frac{1}{ \sqrt{2}} ( | 0 \rangle - | 1 \rangle) | 1 \rangle = \frac{1}{ \sqrt{2}} (|01 \rangle - | 11 \rangle) \underleftrightarrow{CNOT} \frac{1}{ \sqrt{2}} ( | 00 \rangle + |11 \rangle ) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.53%3A_Bell_State_Exercises.txt
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Mathcad provides the kronecker command for matrix tensor multiplication. It requires square matrices for its arguments and therefore cannot be used directly for vector tensor multiplication. However, if a vector is augmented with a null vector (or matrix) to produce a square matrix, vector tensor multiplication can be carried out using kronecker and a submatrix command that discards everything except the first column of the product matrix. This technique is illustrated by putting the Bell and GHZ states in vector format.
The z- and x-direction spin eigenfunctions and the appropriate null vector are required.
$\begin{matrix} z_u = \begin{pmatrix} 1 \ 0 \end{pmatrix} & z_d = \begin{pmatrix} 0 \ 1 \end{pmatrix} & x_u = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & x_d = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} & n = \begin{pmatrix} 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The Mathcad syntax for tensor multiplication of two 2-dimensional vectors.
$\psi \text{(a, b)} = \text{submatrix(kronecker(augment(a, n), augment(b, n)), 1, 4, 1, 1)} \nonumber$
The four maximally entangled Bell states will be expressed in both the z- and x-basis.
$| \Phi_p \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle + | \downarrow_1 \rangle | \downarrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \nonumber$
$\begin{matrix} \Phi_p = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u) + \psi (z_d,~z_d) \right) & \Phi_p = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} & \Phi_p = \frac{1}{ \sqrt{2}} \left( \psi (x_u,~x_u) \right) + \left( \psi (x_d,~x_d) \right) & \Phi_p = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
$| \Phi_m \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle - | \downarrow_1 \rangle | \downarrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
$\begin{matrix} \Phi_m = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u) - \psi (z_d,~z_d) \right) & \Phi_m = \begin{pmatrix} 0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} & \Phi_m = \frac{1}{ \sqrt{2}} \left( \psi (x_u,~x_u) \right) + \left( \psi (x_d,~x_d) \right) & \Phi_m = \begin{pmatrix} 0.707 \ 0 \ 0 \ -0.707 \end{pmatrix} \end{matrix} \nonumber$
$| \Psi_p \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle + | \downarrow_1 \rangle | \downarrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \nonumber$
$\begin{matrix} \Psi_p = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u) + \psi (z_d,~z_d) \right) & \Phi_p = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} & \Psi_p = \frac{1}{ \sqrt{2}} \left( \psi (x_u,~x_u) \right) - \left( \psi (x_d,~x_d) \right) & \Psi_p = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} \end{matrix} \nonumber$
$| \Psi_m \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \uparrow_2 \rangle - | \downarrow_1 \rangle | \downarrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
$\begin{matrix} \Psi_m = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u) - \psi (z_d,~z_d) \right) & \Psi_m = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} & \Psi_m = \frac{1}{ \sqrt{2}} \left( \psi (x_u,~x_u) \right) - \left( \psi (x_d,~x_d) \right) & \Psi_m = \begin{pmatrix} 0 \ 0.707 \ -0.707 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The Mathcad syntax for tensor multiplication of three 2-dimensional vectors.
$\Psi \text{(a, b, c)} = \text{submatrix(kronecker(augment(a, n), kronecker(augment(b, n), augment(c, n))), 1, 8, 1, 1)} \nonumber$
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \pm \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & \pm 1 \end{pmatrix}^T \nonumber$
$\begin{matrix} \Psi_1 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u,~z_u) + \psi (z_d,~z_d,~z_d) \right) & \Psi_1^T = \begin{pmatrix} 0.707 & 0 & 0 & 0 & 0 & 0 & 0 & 0.707 \end{pmatrix} \ \Psi_2 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u,~z_u) - \psi (z_d,~z_d,~z_d) \right) & \Psi_2^T = \begin{pmatrix} 0.707 & 0 & 0 & 0 & 0 & 0 & 0 & -0.707 \end{pmatrix} \end{matrix} \nonumber$
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \pm \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \pm 1 & 0 \end{pmatrix}^T \nonumber$
$\begin{matrix} \Psi_3 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u,~z_d) + \psi (z_d,~z_d,~z_u) \right) & \Psi_3^T = \begin{pmatrix} 0 & 0.707 & 0 & 0 & 0 & 0 & 0.707 & 0 \end{pmatrix} \ \Psi_4 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_u,~z_d) - \psi (z_d,~z_d,~z_u) \right) & \Psi_4^T = \begin{pmatrix} 0 & 0.707 & 0 & 0 & 0 & 0 & -0.707 & 0 \end{pmatrix} \end{matrix} \nonumber$
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \pm \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & \pm 1 & 0 & 0 \end{pmatrix}^T \nonumber$
$\begin{matrix} \Psi_5 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_d,~z_u) + \psi (z_d,~z_u,~z_d) \right) & \Psi_5^T = \begin{pmatrix} 0 & 0 & 0.707 & 0 & 0 & 0.707 & 0 & 0 \end{pmatrix} \ \Psi_6 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_d,~z_u) - \psi (z_d,~z_u,~z_d) \right) & \Psi_6^T = \begin{pmatrix} 0 & 0 & 0.707 & 0 & 0 & -0.707 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \pm \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 0 & 0 & 1 & \pm 1 & 0 & 0 & 0 \end{pmatrix}^T \nonumber$
$\begin{matrix} \Psi_7 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_d,~z_d) + \psi (z_d,~z_u,~z_u) \right) & \Psi_7^T = \begin{pmatrix} 0 & 0 & 0 & 0.707 & 0.707 & 0 & 0 & 0 \end{pmatrix} \ \Psi_8 = \frac{1}{ \sqrt{2}} \left( \psi (z_u,~z_d,~z_d) - \psi (z_d,~z_u,~z_u) \right) & \Psi_8^T = \begin{pmatrix} 0 & 0 & 0 & 0.707 & -0.707 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.54%3A_Expressing_Bell_and_GHZ_States_in_Vector_Format_Using_Mathcad.txt
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The following circuit generates the eight GHZ maximally entangled states. A similar NMR circuit can be found on page 287 of The Quest for the Quantum Computer by Julian Brown.
$\begin{matrix} \text{Index} \ | a \rangle & \triangleright & \cdots & \oplus & \cdots & \triangleright & \text{G} \ ~ & ~ & ~ & | \ | b \rangle & \triangleright & \text{H} & \cdot & \cdot & \triangleright & \text{H} \ ~ & ~ & ~ & ~ & | \ | c \rangle & \triangleright & \cdots & \cdots & \oplus & \triangleright & \text{Z} \end{matrix} \nonumber$
Given the quantum gates in matrix form the quantum circuit is formed using Kronecker multiplication.
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \text{ICNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\text{GHZ} = \text{kronecker(I, CNOT) kronecker(ICNOT, I) kronecker(I, kronecker(H, I))} \nonumber$
Using the index as input the quantum circuit generates the corresponding GHZ state.
$\begin{matrix} G_0 = \text{GHZ} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0.707 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \ G_1 = \text{GHZ} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ 0 \ 0 \ 0 \ 0 \ 0.707 \ 0 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \ G_2 = \text{GHZ} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0.707 \ 0 \ 0 \ 0 \ 0 \ 0 \ \ 0 \ -0.707 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \ G_3 = \text{GHZ} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ 0 \ 0 \ 0 \ 0 \ \ -0.707 \ 0 \end{pmatrix} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \ G_4 = \text{GHZ} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0.707 \ 0.707 \ 0 \ \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \ G_5 = \text{GHZ} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0.707 \ 0 \ 0 \ 0.707 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \ G_6 = \text{GHZ} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ -0.707 \ 0.707 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \ G_7 = \text{GHZ} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ -0.707 \ 0 \ 0 \ 0.707 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \end{matrix} \nonumber$
Given the following truth tables the operation of the circuit is followed algebraically for G2 and G4.
$\begin{matrix} \text{Identity} & \begin{pmatrix} \text{0 to 0} \ \text{1 to 1} \end{pmatrix} & \text{Hadamard} & \text{H} = \begin{bmatrix} \text{0 to}~ \frac{(0 + 1)}{ \sqrt{2}} \text{ to 0} \ \text{1 to}~ \frac{(0 - 1)}{ \sqrt{2}} \text{ to 1} \end{bmatrix} \ \text{ICNOT} & \begin{pmatrix} \text{Decimal} & \text{Binary} & \text{to} & \text{Binary} & \text{Decimal} \ 0 & 00 & \text{to} & 00 & 0 \ 1 & 01 & \text{to} & 11 & 3 \ 2 & 10 & \text{to} & 10 & 2 \ 3 & 11 & \text{to} & 01 & 1 \end{pmatrix} & \text{CNOT} & \begin{pmatrix} \text{Decimal} & \text{Binary} & \text{to} & \text{Binary} & \text{Decimal} \ 0 & 00 & \text{to} & 00 & 0 \ 1 & 01 & \text{to} & 01 & 1 \ 2 & 10 & \text{to} & 11 & 3 \ 3 & 11 & \text{to} & 10 & 2 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} |010 \rangle & |100 \rangle \ \text{I} \otimes \text{H} \otimes \text{I} & \text{I} \otimes \text{H} \otimes \text{I} \ |0 \rangle \left( \frac{|0 \rangle - | 1 \rangle}{ \sqrt{2}} \right) |0 \rangle = \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |010 \rangle \right] & |1 \rangle \left( \frac{|0 \rangle + | 1 \rangle}{ \sqrt{2}} \right) |0 \rangle = \frac{1}{ \sqrt{2}} \left[ |100 \rangle + |110 \rangle \right] \ \text{ICNOT} \otimes \text{I} & \text{ICNOT} \otimes \text{I} \ \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |110 \rangle \right] & \frac{1}{ \sqrt{2}} \left[ |100 \rangle + |010 \rangle \right] \ \text{I} \otimes \text{CNOT} & \text{I} \otimes \text{CNOT} \ \frac{1}{ \sqrt{2}} \left[ |000 \rangle - |111 \rangle \right] & \frac{1}{ \sqrt{2}} \left[ |100 \rangle + |011 \rangle \right] \end{matrix} \nonumber$
Next the GHZ density matrices are presented graphically.
Using the GHZ states as input and running the circuit in reverse yields the indices.
$\text{IGHZ} = \text{GHZ}^-1 \nonumber$
$\begin{matrix} \text{IGHZ G}_0 = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{IGHZ G}_1 = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{IGHZ G}_2 = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{IGHZ G}_3 = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \ \text{IGHZ G}_4 = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{IGHZ G}_5 = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} & \text{IGHZ G}_6 = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{IGHZ G}_7 = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
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The purpose of this tutorial is restricted to a brief computational summary of the EPR experiment reported by Aspect, Grangier and Roger, ʺExperimental Realization of Einstein‐Podolsky‐Rosen‐Bohm Gedanken Experiment: A New Violation of Bellʹs Inequalities,ʺ in Phys. Rev. Lett. 49, 91 (1982). See Chapter 6 of The Quantum Challenge by Greenstein and Zajonc, Chapter 4 of Jim Baggottʹs The Meaning of Quantum Theory, and Chapter 12 of Quantum Reality by Nick Herbert for complete analyses of this historically important experiment.
A two‐stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to the observers in their path.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] \nonumber$
The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows that the polarization analyzers (PAs) can be oriented at different angles a and b. (The figure below is taken from Chapter 4 of Baggottʹs book.)
Naturally the bipartate photon wave function is identical in both the circular or linear polarization bases.
Left circular polarization:
$L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} \nonumber$
Right circular polarization:
$R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \nonumber$
Vertical polarization:
$V = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Horizontal polarization:
$H = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$| \Psi \rangle = \frac{1}{2 \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ i \end{pmatrix}_B + \begin{pmatrix} 1 \ -i \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ -i \end{pmatrix}_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ i \ i \ -1 \end{pmatrix} + \begin{pmatrix} 1 \ -i \ -i \ -1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
$| \Psi \rangle = \frac{1}{2 \sqrt{2}} \left[ |V \rangle_A |V \rangle_B + |H \rangle_A |H \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix}_A \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix}_B - \begin{pmatrix} 0 \ 1 \end{pmatrix}_A \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix}_B \right] = \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \nonumber$
There are four measurement outcomes: both photons are vertically polarized, both are horizontally polarized, one is vertical and the other horizontal, and vice versa. The tensor representation of these measurement states are provided below.
$\begin{matrix} |VV \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & |VH \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & |HV \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & |HH \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
We now write all states, Ψ and the measurement states, in Mathcad's vector format.
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & \text{VV} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{VH} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \text{HV} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{HH} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Next, the operator representing the rotation of PA1 by angle a clockwise and PA2 by angle b counter‐clockwise (so that the PAs turn in the same direction) is constructed using matrix tensor multiplication. Kronecker is Mathcadʹs command for tensor matrix multiplication.
$\text{RotOp(a, b)} = \text{kronecker} \left[ \begin{pmatrix} \cos a & \sin a \ - \sin a & \cos a \end{pmatrix},~ \begin{pmatrix} \cos b & - \sin b \ \sin b & \cos b \end{pmatrix} \right] \nonumber$
The probability that the detectors will behave the same or differently is calculated as follows.
$\begin{matrix} P_{ \text{same}} (a,~ b) = \left( VV^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HH^T \text{RotOp(a, b)} \Psi \right)^2 \ P_{ \text{diff}} (a,~ b) = \left( VH^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HV^T \text{RotOp(a, b)} \Psi \right)^2 \end{matrix} \nonumber$
The expectation value as a function of the relative orientation of the polarization detectors is the difference between these two expressions. This is generally called the correlation function. In other words, the composite eigenvalues are: ++ = ‐‐ = +1 (same) and +‐ = ‐+ = ‐1 (diff).
$\begin{matrix} \text{Corr(a, b)} = P_{same} \text{(a, b)} - P_{diff} \text{(a, b)} & \Theta = 0 \text{deg},~ 2 \text{deg}.. 90 \text{deg} \end{matrix} \nonumber$
These graphical representations of the Aspect experiment are in agreement with those presented in Aspect's paper and also in The Quantum Challenge, The Meaning of Quantum Theory, and Quantum Reality.
These calculations are now repeated for the three other Bell states.
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \end{pmatrix}^T \ P_{same} \text{(a, b)} = \left( VV^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HH^T \text{RotOp(a, b)} \Psi \right)^2 \ P_{diff} \text{(a, b)} = \left( VH^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HV^T \text{RotOp(a, b)} \Psi \right)^2 \ \text{Corr(a, b)} = P_{same} \text{(a, b)} - P_{diff} \text{(a, b)} \end{matrix} \nonumber$
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & 1 & 0 \end{pmatrix}^T \ P_{same} \text{(a, b)} = \left( VV^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HH^T \text{RotOp(a, b)} \Psi \right)^2 \ P_{diff} \text{(a, b)} = \left( VH^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HV^T \text{RotOp(a, b)} \Psi \right)^2 \ \text{Corr(a, b)} = P_{same} \text{(a, b)} - P_{diff} \text{(a, b)} \end{matrix} \nonumber$
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix}^T \ P_{same} \text{(a, b)} = \left( VV^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HH^T \text{RotOp(a, b)} \Psi \right)^2 \ P_{diff} \text{(a, b)} = \left( VH^T \text{RotOp(a, b)} \Psi \right)^2 + \left( HV^T \text{RotOp(a, b)} \Psi \right)^2 \ \text{Corr(a, b)} = P_{same} \text{(a, b)} - P_{diff} \text{(a, b)} \end{matrix} \nonumber$
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In the '90s N. David Mermin published two articles in the general physics literature (Physics Today, June 1990; American Journal of Physics, August 1990) on the Greenberger-Horne-Zeilinger (GHZ) gedanken experiment (American Journal of Physics, December 1990; Nature, 3 February 2000) involving three spin-1/2 particles that illustrated the clash between local realism and the quantum view of reality for the quantum nonspecialist. The purpose of this tutorial is to use the GHZ example to illustrate the Kochen-Specker (KS) theorem by stripping away the use of the three-spin wave function in the analysis of the thought experiment.
The KS theorem asserts that no noncontextual (NC) hidden variable (HV) model (NCHV) can agree with the measurement predictions of quantum theory for Hilbert space dimensions greater than 2. The problem dealt with here is, of course, three-dimensional.
The three spin-1/2 particles are created in a single event and move apart in the horizontal y-z plane. It will be shown that a consideration of spin measurements (in units of h/4π in the x- and y-directions reveals the impossibility of assigning values to the spin observables independent of measurement.
The x- and y-direction spin operators are the Pauli matrices:
$\begin{matrix} \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{matrix} \nonumber$
The eigenvalues of the Pauli matrices are +/- 1:
$\begin{matrix} \text{eigenvals}( \sigma_x) = \begin{pmatrix} 1 \ -1 \end{pmatrix} & \text{eigenvals}( \sigma_y) = \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
The following operators represent the measurement protocols for spins 1, 2, and 3.
$\begin{matrix} \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 & \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 & \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \end{matrix} \nonumber$
The tensor matrix product, also known as the Kronecker product, is available in Mathcad. The operators in tensor format are formed as follows.
$\begin{matrix} \sigma_{xyy} \sigma_{yxy} - \sigma_{yxy} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{yyx} - \sigma_{yyx} \sigma_{xyy} \rightarrow 0 & \sigma_{xyy} \sigma_{xxx} - \sigma_{xxx} \sigma_{xyy} \rightarrow 0 \ \sigma_{yxy} \sigma_{yyx} - \sigma_{yyx} \sigma_{yxy} \rightarrow 0 & \sigma_{yxy} \sigma_{xxx} - \sigma_{xxx} \sigma_{yxy} \rightarrow 0 & \sigma_{yyx} \sigma_{xxx} - \sigma_{xxx} \sigma_{yyx} \rightarrow 0 \end{matrix} \nonumber$
The next step is to compare the matrix for the product of the first three operators $(\sigma_{xyy} \sigma_{yxy} \sigma_{yyx})$ with that of the fourth $( \sigma_{xxx})$.
$\begin{matrix} \sigma_{xyy} \sigma_{yxy} \sigma_{yyx} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \ 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \ -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \sigma_{xxx} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
This indicates the following relationship between the four operators and leads quickly to a refutation of the concept of noncontextual, hidden values for quantum mechanical observables.
$\begin{matrix} \left( \sigma_x^1 \otimes \sigma_y^2 \otimes \sigma_y^3 \right) \left( \sigma_y^1 \otimes \sigma_x^2 \otimes \sigma_y^3 \right) \left( \sigma_y^1 \otimes \sigma_y^2 \otimes \sigma_x^3 \right) = - \left( \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \right) \end{matrix} \nonumber$
Local realism assumes that objects have definite properties independent of measurement. In this example it assumes that the x- and y-components of the spin have definite values prior to measurement. This position leads to a contradiction with the above result. There is no way to assign eigenvalues (+/-1) to the operators that is consistent with the above result.
Concentrating on the operator on the left side, we notice that there is a σy measurement on the first spin in the second and third term. If the spin state is well-defined before measurement those results have to be the same, either both +1 or both -1, so that the product of the two measurements is +1. There is a σy measurement on the second spin in terms one and three. By similar arguments those results will lead to a product of +1 also. Finally there is a σy measurement on the third spin in terms one and two. By similar arguments those results will lead to a product of +1 also. Incorporating these observations into the expression above leads to the following contradiction.
$\begin{matrix} \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 = - \sigma_x^1 \otimes \sigma_x^2 \otimes \sigma_x^3 \end{matrix} \nonumber$
A brute force method can be used to confirm this result by showing that the left and right sides of the equation are not equal for any legitimate set of values for the individual spins. This is shown for several such sets below.
$\begin{matrix} x1 = 1 & x2 = 1 & x3 = 1 & y1 = 1 & y2 = 1 & y3 = 1 \end{matrix} \nonumber$
$\begin{matrix} \text{(x1 y2 y3) (y1 x2 y3) (y1 y2 x3) = 1} & - \text{(x1 x2 x3)} = -1 \end{matrix} \nonumber$
$\begin{matrix} x1 = -1 & x2 = 1 & x3 = 1 & y1 = 1 & y2 = -1 & y3 = -1 \end{matrix} \nonumber$
$\begin{matrix} \text{(x1 y2 y3) (y1 x2 y3) (y1 y2 x3) = -1} & - \text{(x1 x2 x3)} = 1 \end{matrix} \nonumber$
$\begin{matrix} x1 = -1 & x2 = -1 & x3 = -1 & y1 = -1 & y2 = -1 & y3 = -1 \end{matrix} \nonumber$
$\begin{matrix} \text{(x1 y2 y3) (y1 x2 y3) (y1 y2 x3)} = -1 & - \text{(x1 x2 x3)} = 1 \end{matrix} \nonumber$
The Kochen-Specker theorem demonstrates that it is, in general, impossible to ascribe to an individual quantum system a definite value for each of a set of observables not all of which necessarily commute. N. David Mermin, Physical Review Letters, 65, 3373 (1990).
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In the field of quantum information interference, superpositions and entangled states are essential resources. Entanglement, a non-factorable superposition, is routinely achieved when two photons are emitted from the same source, say a parametric down converter (PDC). Entanglement swapping involves the transfer of entanglement to two photons that were produced independently and never previously interacted. The Bell states are the four maximally entangled two-qubit entangled basis for a four-dimensional Hilbert space and play an essential role in quantum information theory and technology, including teleportation and entanglement swapping. The Bell states are shown below.
$\begin{matrix} \Phi_p = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] & \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} & \Phi_m = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] & \Phi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \ \Psi_p = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] & \Psi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} & \Psi_m = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] & \Psi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
A four-qubit state is prepared in which photons 1 and 2 are entangled in Bell state Φp, and photons 3 and 4 are entangled in Bell state Ψm. The state multiplication below is understood to be tensor vector multiplication.
$\begin{matrix} \Psi = \Phi_p \Psi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \Psi = \frac{1}{2} \begin{pmatrix} 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \end{pmatrix}^T = & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Four Bell state measurements are now made on photons 2 and 3 which entangles photons 1 and 4. Projection of photons 2 and 3 onto Φp projects photons 1 and 4 onto Ψm.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Phi_p,~ \Phi_p^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0.25 & 0 & 0 & 0 & 0 & 0 & 0.25 & -0.25 & 0 & 0 & 0 & 0 & -0.25 & 0 \end{pmatrix} \ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Projection of photons 2 and 3 onto Φm projects photons 1 and 4 onto Ψp.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Phi_m,~ \Phi_m^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0.25 & 0 & 0 & 0 & 0 & 0 & 0.25 & -0.25 & 0 & 0 & 0 & 0 & -0.25 & 0 \end{pmatrix} \ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Projection of photons 2 and 3 onto Ψp projects photons 1 and 4 onto Φm.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Psi_p,~ \Psi_p^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0 & -0.25 & 0 & -0.25 & 0 & 0 & 0 & 0 & 0 & 0 & 0.25 & 0 & 0.25 & 0 & 0 \end{pmatrix} \ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 0 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
Finally, projection of photons 2 and 3 onto Ψm projects photons 1 and 4 onto Φp.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Psi_m,~ \Psi_m^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0 & -0.25 & 0 & 0.25 & 0 & 0 & 0 & 0 & 0 & 0 & -0.25 & 0 & 0.25 & 0 & 0 \end{pmatrix} \ \frac{-1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.58%3A_A_Brief_Introduction_to_Entanglement_Swapping.txt
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In the field of quantum information, interference, superpositions and entangled states are essential resources. Entanglement, a non-factorable superposition, is routinely achieved when two photons are emitted from the same source, perhaps a parametric down converter (PDC). Entanglement swapping involves the transfer (teleportation) of entanglement to two photons that were produced independently and never previously interacted. The Bell states are the four maximally entangled two-qubit entangled basis for a four-dimensional Hilbert space and play an essential role in quantum information theory and technology, including teleportation and entanglement swapping. This analysis attempts to provide the essential matrix math needed to understand parts of "Entangled delayed-choice entanglement swapping" (arXiv1203.4834) and "Delayed-choice gedanken experiments and their realizations" (arXiv1407.2930). The following analysis deals exclusively with entanglement swapping and does not consider the delayed-choice aspect of the research presented in these papers.
Bell states and the identity operator:
$\begin{matrix} \Phi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} & \Phi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & \Psi_p = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} & \Psi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
A four-qubit state is prepared in which photons 1 and 2 are entangled in Bell state Ψm, and photons 3 and 4 are also entangled in Bell state Ψm. The state multiplication below is understood to be tensor vector multiplication.
$\begin{matrix} \Psi_{1234} = \Psi_{m12} \Psi_{m34} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \Psi = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}^T \end{matrix} \nonumber$
The authors write this state as a superposition of Bell state products to suggest a way to transfer entanglement from 1&2-3&4 to 1&4-2&3: perform a Bell state measurement on photons 2 and 3.
$| \Psi \rangle_{1234} = \frac{1}{2} \left[ | \Psi_p \rangle_{14} \otimes | \Psi_P \rangle_{23} - | \Psi_m \rangle_{14} \otimes \Psi_m \rangle_{23} - | \Psi_m \rangle_{23} - | \Phi_p \rangle_{14} \otimes \Phi_p \rangle_{23} + | \Phi_m \rangle_{14} \otimes \Phi_m \rangle_{23} \right] \nonumber$
The following calculations agree with this product of entangled photon pairs 1&4 and 2&3. Projection of photons 2 and 3 onto Ψp projects photons 1 and 4 onto Ψp.
Projection of photons 2 and 3 onto Φp projects photons 1 and 4 onto Ψm.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Psi_p,~ \Psi_p^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0 & 0 & 0.25 & 0 & 0.25 & 0 & 0 & 0 & 0 & 0.25 & 0 & 0.25 & 0 & 0 & 0 \end{pmatrix} \ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
Projection of photons 2 and 3 onto Ψm projects photons 1 and 4 onto -Ψm.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Psi_m,~ \Psi_m^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0 & 0 & 0 & -0.25 & 0 & 0.25 & 0 & 0 & 0 & 0 & 0.25 & 0 & -0.25 & 0 & 0 & 0 \end{pmatrix} \ \frac{-1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 0 & 0 & 0 & -1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
Projection of photons 2 and 3 onto Φp projects photons 1 and 4 onto -Φp.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Phi_p,~ \Phi_p^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} -0.25 & 0 & 0 & 0 & 0 & 0 & -0.25 & 0 & 0 & 0 & 0 & 0 & -0.25 \end{pmatrix} \ \frac{-1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{-1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} -1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & -1 \end{pmatrix} \end{matrix} \nonumber$
Projection of photons 2 and 3 onto Φm projects photons 1 and 4 onto Φm.
$\begin{matrix} \left( \text{kronecker} \left( \text{I, kronecker} \left( \Phi_m,~ \Phi_m^T,~ \text{I} \right) \right) \Psi \right)^T = \begin{pmatrix} 0.25 & 0 & 0 & 0 & 0 & 0 & -0.25 & 0 & 0 & -0.25 & 0 & 0 & 0 & 0 & 0 & 0.25 \end{pmatrix} \ \frac{1}{ 2 \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right]^T = \frac{1}{4} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The initial four-particle state can be written in the H/V, A/D and R/L polarization bases. See the Appendix for details.
$\begin{matrix} \Psi_{1234} = \frac{1}{2} (H_1 V_2 - V_1 H_2) = (H_3 V_4 - V_3 H_4) = \frac{1}{2} (H_1 V_2 V_3 V_4 - H_1 V_2 V_3 H_4 - V_1 H_2 H_3 V_4 + V_1 H_2 V_3 H_4 + V_1 H_2 V_3 H_4) \ \Psi_{1234} = \frac{1}{2} (A_1 D_2 - D_1 A_2) = (A_3 D_4 - D_3 A_4) = \frac{1}{2} (A_1 D_2 A_3 D_4 - A_1 D_2 A_3 D_4 - A_1 D_2 D_3 A_4 + D_1 A_2 A_3 D_4 + D_1 A_2 D_3 A_4) \ \Psi_{1234} = \frac{1}{2} (L_1 R_2 - R_1 L_2) = (L_3 R_4 - R_3 L_4) = \frac{1}{2} (L_1 R_2 R_3 R_4 - L_1 R_2 R_3 L_4 - R_1 L_2 L_3 R_4 + R_1 L_2 R_3 L_4) \end{matrix} \nonumber$
Where,
$\begin{matrix} H = \begin{pmatrix} 1 \ 0 \end{pmatrix} & V = \begin{pmatrix} 0 \ 1 \end{pmatrix} & R = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} & L = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & A = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} H = \frac{1}{ \sqrt{2}} (R+L) & V = \frac{i}{ \sqrt{2}} (L-R) & H = \frac{1}{ \sqrt{2}} (D+A) & V = \frac{1}{ \sqrt{2}} (D-A) \end{matrix} \nonumber$
$\begin{matrix} R = \frac{1}{ \sqrt{2}} (H + iV) & L = \frac{1}{ \sqrt{2}} (H - iV) & R = \frac{1}{ \sqrt{2}} (D+A) \end{matrix} \nonumber$
After production, photon 1 is sent to Alice and photon 4 is sent to Bob. Photons 2 and 3 are sent to Victor. Alice and Bob can measure their photons in either the H/V, R/L or A/D bases. Victor can measure his photons separately in the H/V basis or he can carry out one of the four Bell state measurements on his photon pairs. This later choice as shown earlier projects photons 1 and 4 into the corresponding entangled Bell state.
The following table shows the possible measurement results that Alice and Bob will obtain, depending on the type of measurement Victor makes on his photons. In experiments 1-4 Victor measures his photons separately in the H/V basis and Alice and Bob do the same. In the remaining experiments Victor does a Bell state measurement on his photons and Alice and Bob measure in any of the three bases.
$\begin{pmatrix} \text{Experiment} & 1 & 2 & 3 & 4 & ' & 5 & 6 & 7 & 8 & ' & 9 & 10 & 11 & 12 & ' & 13 & 14 & 15 & 16 \ \text{Alice1} & H & H & V & V ' & H & V & H & V & ' & L & R & L & R & ' & D & A & A & D \ \text{Bob4} & V & H & V & H & ' & V & H & H & V & ' & L & L & R & R & ' & D & D & A & A \ \text{Victor23} & VH & VV & HH & HV & ' & \Psi_p & \Psi_m & \Phi_p & \Phi_m & ' & \Psi_p & \Psi_m & \Phi_p & \Phi_m ' & \Psi_p & \Psi_m & \Phi_p & \Phi_m \end{pmatrix} \nonumber$
Results 1-4 are consistent with the original state expressed in the H/V basis.
$\Psi_{1234} = \frac{1}{2} (H_1 V_2 H_3 V_4 - H_1 V_2 V_3 H_4 - V_1 H_2 H_3 V_4 + V_1 H_2 V_3 H_4) \nonumber$
In the remaining experiments Victor makes a Bell state measurement on photons 2 and 3, and projects photons 1 and 4 into the following Bell states. The table shows measurement results that Alice and Bob could make on their photons given these states. See the Appendix for more detail.
$\begin{matrix} \Psi_{p14} = \frac{1}{ \sqrt{2}} (H_1 V_4 + V_1 H_4) = \frac{i}{ \sqrt{2}} (L_1 L_4 - R_1 R_4) = \frac{1}{ \sqrt{2}} (D_1 D_4 - A_1 A_4) \ \Psi_{m14} = \frac{1}{ \sqrt{2}} (H_1 V_4 - V_1 H_4) = \frac{i}{ \sqrt{2}} (L_1 L_4 - R_1 R_4) = \frac{1}{ \sqrt{2}} (A_1 D_4 - D_1 A_4) \ \Phi_{p14} = \frac{1}{ \sqrt{2}} (H_1 H_4 + V_1 V_4) = \frac{i}{ \sqrt{2}} (L_1 R_4 + R_1 L_4) = \frac{1}{ \sqrt{2}} (A_1 A_4 - D_1 D_4) \ \Phi_{m14} = \frac{1}{ \sqrt{2}} (H_1 H_4 - V_1 V_4) = \frac{i}{ \sqrt{2}} (L_1 L_4 + R_1 R_4) = \frac{1}{ \sqrt{2}} (A_1 D_4 + D_1 A_4) \end{matrix} \nonumber$
Appendix
The Bell states are written in the H/V, R/L, and A/D bases.
$\begin{matrix} \Psi_p = \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i V_j + V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (R_j - L_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( - \frac{ R_i R_j i}{2} + \frac{L_i L_j i}{2} \right) \ ~ & \frac{1}{ \sqrt{2}} (H_i V_j + V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i A_j i}{2} - \frac{D_i D_j i}{2} \right) \ \Psi_m = \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i V_j - V_i H_j) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (L_j - R_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( - \frac{ L_i R_j i}{2} + \frac{L_i R_j i}{2} \right) \ ~ & \frac{1}{ \sqrt{2}} (H_i V_j - V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i D_j i}{2} - \frac{A_i D_j i}{2} \right) \ \Phi_p = \begin{pmatrix} 1 \ 0 \ 0 \ 1 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i H_j + V_i V_j) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (L_j - R_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ L_i R_j i}{2} + \frac{L_i R_j i}{2} \right) \ ~ & \frac{1}{ \sqrt{2}} (H_i V_j + V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i A_j i}{2} - \frac{D_i D_j i}{2} \right) \ \Phi_m = \begin{pmatrix} 1 \ 0 \ 0 \ -1 \end{pmatrix} & \frac{1}{ \sqrt{2}} (H_i H_j - V_i V_j) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (R_i + L_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (L_j - R_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (L_i - R_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (R_j + L_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ L_i L_j i}{2} + \frac{R_i R_j i}{2} \right) \ ~ & \frac{1}{ \sqrt{2}} (H_i V_j - V_i H_j ) & \begin{array}{|l} \text{substitute, } H_i = \frac{1}{ \sqrt{2}} (D_i + A_i) \ \text{substitute, } V_j = \frac{i}{ \sqrt{2}} (D_j - A_j) \ \text{substitute, } V_i = \frac{i}{ \sqrt{2}} (D_i - A_i) \ \text{substitute, } H_j = \frac{1}{ \sqrt{2}} (D_j + A_j) \end{array} \rightarrow \sqrt{2} \left( \frac{ A_i D_j i}{2} - \frac{A_j D_i i}{2} \right) \end{matrix} \nonumber$
The initial state is written in the H/V, R/L, and A/D bases.
$\begin{matrix} \frac{1}{2} (H_1 V_2 - V_1 H_2) (H_3 V_4 - V_3 H_4 ) & \begin{array}{|l} \text{substitute, H}_1 = \frac{1}{ \sqrt{2}} (D_1 + A_1 ) \ \text{substitute, V}_2 = \frac{1}{ \sqrt{2}} (D_2 - A_2 ) \ \text{substitute, V}_1 = \frac{1}{ \sqrt{2}} (D_1 - A_1 ) \ \text{substitute, H}_2 = \frac{1}{ \sqrt{2}} (D_2 + A_2 ) \ \text{substitute, H}_3 = \frac{1}{ \sqrt{2}} (D_3 + A_3) \ \text{substitute, V}_3 = \frac{1}{ \sqrt{2}} (D_3 - A_3) \ \text{substitute, H}_4 = \frac{1}{ \sqrt{2}} (D_4 + A_4) \ \text{substitute, V}_4 = \frac{1}{ \sqrt{2}} (D_4 - A_4) \end{array} \rightarrow \frac{A_1 D_2 - A_2 D_1)(A_3D_4-A_4D_3)}{2} \ \frac{1}{2} (H_1 V_2 - V_1 H_2) (H_3 V_4 - V_3 H_4) & \begin{array}{|l} \text{substitute, H}_1 = \frac{1}{ \sqrt{2}} (R_1 + L_1 ) \ \text{substitute, V}_2 = \frac{i}{ \sqrt{2}} (L_2 - R_2) \ \text{substitute, V}_1 = \frac{i}{ \sqrt{2}} (L_1 - R_1) \ \text{substitute, H}_2 = \frac{1}{ \sqrt{2}} (R_2 + L_2) \ \text{substitute, H}_3 = \frac{1}{ \sqrt{2}} (R_3 + L_3) \ \text{substitute, V}_3 = \frac{i}{ \sqrt{2}} (L_3 - R_3) \ \text{substitute, H}_4 = \frac{1}{ \sqrt{2}} (R_4 + L_4) \ \text{substitute, V}_4 = \frac{1}{ \sqrt{2}} (L_4 - R_4) \end{array} \rightarrow \frac{L_1 R_2 - L_2 R_1)(L_3R_4-L_4R_3)}{2} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.59%3A_An_Entanglement_Swapping_Protocol.txt
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In order to explore the conflict between quantum mechanics and local realism a spin-1/2 pair is prepared in an entangled singlet state and the individual particles travel in opposite directions on the y-axis to a pair of Stern-Gerlach detectors which are set up to measure spin in the x-z plane. Particle A's spin is measured along the z-axis, and particle B's spin is measured at any angle θ with respect to the z-axis. The experimental setup is shown below.
The entangled singlet spin state is written in both the z- and θ-direction spin eigenstates.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow \rangle_A | \downarrow \rangle_B - | \downarrow \rangle_A | \uparrow \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix}_A \otimes \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix}_B - \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix}_A \otimes \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix}_B \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
$\begin{matrix} \text{Spin-up Eigenvalue +1} & \varphi_u ( \theta) = \begin{pmatrix} \cos \left( \frac{ \theta}{2} \right) \ \sin \left( \frac{ \theta}{2} \right) \end{pmatrix} & \text{Spin-down Eigenvalue -1} & \varphi_d ( \theta) = \begin{pmatrix} - \sin \left( \frac{ \theta}{2} \right) \ \cos \left( \frac{ \theta}{2} \right) \end{pmatrix} \end{matrix} \nonumber$
If particle A is observed to be spin-up in the z-direction (eigenvalue +1), particle B is spin-down in the z-direction due to the singlet nature of the entangled state.
The probability B will be found on measurement to be spin-up in the θ-direction yielding a composite eigenvalue of +1 is:
$\left( \varphi_u ( \theta)^T \varphi_d (0) \right)^2 \text{ simplify } \rightarrow \frac{1}{2} - \frac{ \cos \theta}{2} \nonumber$
The probability B will be found on measurement to be spin-down in the θ-direction yielding a composite eigenvalue of -1 is:
$\left( \varphi_d ( \theta)^T \varphi_d (0) \right)^2 \text{ simplify } \rightarrow \frac{ \cos \theta}{2} + \frac{1}{2} \nonumber$
Therefore the overall quantum correlation or expectation value is:
$E( \theta) = \left( \varphi_u ( \theta)^T \varphi_d (0) \right)^2 - \left( \varphi_d ( \theta)^T \varphi_d (0) \right)^2 \text{ simplify } \rightarrow - \cos \theta \nonumber$
The expectation value as a function of the measurement angle difference is displayed below. In what follows we will concentrate on the data for only 0 degrees and 45 degrees, and show that a local realistic model is consistent with the 0-degree result but not the 45-degree result.
If the observers measure their spins in the same direction (both θ = 0 deg or both θ = 45 deg) quantum mechanics predicts they will get opposite values due to the singlet nature of the spin state. In other words, the combined expectation value is -1 for these measurements as shown in the figure above. However, if they measure their spins at 0 and 45 degrees, the expectation value is -0.707.
Realists believe that objects have well-defined properties prior to and independent of observation. Specific 0- and 45-deg spin states are assigned to the particles in the first two columns, with each particle in one of four equally probable spin orientations consistent with the composite singlet state. The next two columns show that these assignments agree with the quantum predictions when both spins are measured at the same angle. The last column shows that these spin assignments disagree with the quantum prediction when one spin is measured at 0 degrees and the other at 45 degrees.
$\begin{bmatrix} \text{Particle A} & \text{Particle B} & \hat{S}_0 \text{(A)} \hat{S}_0 \text{(B)} & \hat{S}_{45} \text{(A)} \hat{S}_{45} \text{(B)} & \hat{S}_0 \text{(A)} \hat{S}_{45} \text{(B)} \ | \uparrow \rangle | \nearrow \rangle & | \downarrow \rangle | \swarrow \rangle & -1 & -1 & -1 \ | \uparrow \rangle | \swarrow \rangle & | \downarrow \rangle | \nearrow \rangle & -1 & -1 & 1 \ | \downarrow \rangle | \nearrow \rangle & | \uparrow \rangle | \swarrow \rangle & -1 & -1 & 1 \ | \downarrow \rangle | \swarrow \rangle & | \uparrow \rangle | \nearrow \rangle & -1 & -1 & -1 \ \text{Realist} & \text{Value} & -1 & -1 & 0 \ \text{Quantum} & \text{Value} & -1 & -1 & -0.707 \end{bmatrix} \nonumber$
This brief analysis demonstrates that there are conceptually simple, Stern-Gerlach like, experiments on spin-1/2 systems which can adjudicate the conflict between local realism and quantum mechanics.
In addition to the disagreement shown in the last column of the table, quantum theory asserts that the realist's spin states are invalid. The spin operator at an angel θ to the vertical in the xz-plane is
$\text{Op} ( \theta) = \varphi_u ( \theta)^T \varphi_d ( \theta) - \varphi_d ( \theta) \varphi_d ( \theta)^T \text{ simplify } \rightarrow \begin{pmatrix} \cos ( \theta) & \sin ( \theta) \ \sin ( \theta) & - \cos ( \theta) \end{pmatrix} \nonumber$
The operators for spin measurements at 0 and 45 degrees in the xz-plane do not commute.
$\text{Op (0 deg) Op(45 deg)} - \text{Op(45 deg) Op(0 deg)} = \begin{pmatrix} 0 & 1.414 \ -1.414 & 0 \end{pmatrix} \nonumber$
Therefore, according to quantum theory a particle's spin cannot be simultaneously well-defined for both 0 and 45 degrees.
Addendum
According to Richard Feynman it takes a quantum computer to simulate quantum pheonomenon. The following quantum circuit produces results that are in agreement with experiment as summarized in the graph above. The Hadamard and CNOT gates create the singlet state from the |11> input. Rz(θ) is the rotation of the measuring device of the second spin. The final Hadamard gates prepare the system for measurement in the x-basis. See arXiv:1712.05642v2 for further detail.
$\begin{matrix} |1 \rangle & \triangleright & \text{H} & \cdot & \cdots & \text{H} & \triangleright & \text{Measure 0 or 1} \ ~ & ~ & ~ & | \ |1 \rangle & \triangleright & \cdots & \oplus & \text{R}_z( \theta) & \text{H} & \triangleright & \text{Measure 0 or 1} \end{matrix} \nonumber$
The quantum operators required to execute this circuit are:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{R}_z ( \theta) = \begin{pmatrix} 1 & 0 \ 0 & e^{i \theta} \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\text{BellCircuit} ( \theta) = \text{kronecker(H, H) kronecker} \left( \text{(I, R)}_z ( \theta) \right) \text{CNOT kronecker(H, I)} \nonumber$
The circuit is run for θ = π/4 to demonstrate that it produces the result highlighted in the graph above. In addition, by varying θ it can be shown that the circuit reproduces the entire plot of E(θ). There are four output states shown below. If the spins are measured in the same state, |00> or |11>, the eigenvalue is +1, if they are different, |01> or |10>, the eigenvalue is -1. The probability for each output state is calculated on the right.
$\begin{matrix} \text{Output state} & \text{Eigenvalue} & \text{Probability} \ | 00 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & 1 & \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.0732 \ | 01 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & -1 & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.4268 \ | 10 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & -1 & \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.4268 \ |11 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} & 1 & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.0732 \end{matrix} \nonumber$
Expectation value or correlation coeficient:
$0.0732 - 0.4268 - 0.4268 + 0.0732 = -0.707 \nonumber$
A classical computer manipulates bits which are in well-defined states consisting of 0s and 1s.This entangled two-spin experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s. Simulation of quantum physics requires a quantum computer, and the circuit shown above is a quantum computer.
An alternative computational method using projection operators:
$\begin{matrix} \text{Output state} & \text{Eigenvalue} & \text{Probability} \ | 00 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & 1 & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.0732 \ | 01 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & -1 & \left[ \left|\text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.4268 \ | 10 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & -1 & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.4268 \ |11 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} & 1 & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.0732 \end{matrix} \nonumber$
Measuring only one spin using a projection operator and the identity:
$\begin{matrix} \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix},~ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.5 & \left[ \left|\text{kronecker} \left[ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix},~ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.5 \ \left[ \left|\text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix},~ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.5 & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix},~ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \right] \text{BellCircuit} \left( \frac{ \pi}{4} \right) \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.5 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.60%3A_Quantum_Correlations_Simplified.txt
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According to Richard Feynman it takes a quantum computer to simulate quantum phenomena. In this tutorial we begin with a traditional quantum analysis of a well-know thought experiment involving correlated spin-1/2 particles. After that the operation of a quantum circuit designed to simulate the thought experiment is analyzed. It will be shown that the quantum analysis and the simulation lead to the same result for the expectation value for the experiment.
The Thought Experiment
A spin-1/2 pair is prepared in an entangled singlet state and the individual particles travel in opposite directions on the y-axis to a pair of Stern-Gerlach (SG) detectors which are set up to measure spin in the x-z plane.
$| \Psi_m \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \downarrow_2 \rangle - | \downarrow_1 \rangle | \uparrow_2 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ \right] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} \nonumber$
Particle A's spin is measured along with z-axis, and particle B's spin is measured at any angle θ with respect to the z-axis in the x-z plane. The experimental apparatus is shown below.
The single particle spin operator in the x-z plane is constructed from the Pauli spin operators in the x- and z-directions. θ is the angle of orientation of the measurement magnet with the z-axis.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & \sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & S( \theta) = \cos ( \theta) \sigma_z + \sin ( \theta) \sigma_x \rightarrow \begin{pmatrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{pmatrix} \end{matrix} \nonumber$
Tensor multiplication of SA(0) and SB(θ) creates a joint measurement operator for spins A and B.
$S_A (0) \otimes S_B ( \theta) = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta & \sin \theta \ \sin \theta & - \cos \theta \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & - \cos \theta & - \sin \theta \ 0 & 0 & - \sin \theta & \cos \theta \end{pmatrix} \nonumber$
The expectation value as a function of the measurement angle of spin B is calculated and the result displayed graphically.
$\begin{matrix} \Psi_m = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & E( \theta) = \Psi_m^T \begin{pmatrix} \cos \theta & \sin \theta & 0 & 0 \ \sin \theta & - \cos \theta & 0 & 0 \ 0 & 0 & - \cos \theta & - \sin \theta \ 0 & 0 & - \sin \theta & \cos \theta \end{pmatrix} \Psi_m \rightarrow - \cos ( \theta) \end{matrix} \nonumber$
The Quantum Simulation
A classical computer can't simulate this thought experiment because it manipulates bits which are in well-defined states, 0s and 1s. These classical states are incompatible with the quantum mechanical state above which involves an entangled superposition. This two-spin thought experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s. As Feynman asserted, the simulation of quantum physics requires a quantum computer!
The following quantum circuit produces results that are in agreement with the thought experiment as summarized in the graph above. The initial Hadamard and CNOT gates create the singlet state from the |11> input. R z(θ) rotates the spin of B. The final Hadamard gates prepare the system for measurement. See arXiv:1712.05642v2 for further detail.
$\begin{matrix} |1 \rangle & \triangleright & \text{H} & \cdot & \cdots & \text{H} & \triangleright & \text{Measure 0 or 1} \ ~ & ~ & ~ & | \ |1 \rangle & \triangleright & \cdots & \oplus & \text{R}_z ( \theta) & \text{H} & \triangleright & \text{Measure 0 or 1} \end{matrix} \nonumber$
The quantum gates required to execute this circuit and their corresponding truth tables:
$\begin{matrix} \text{Identity} & \text{Hadamard gate} & \text{R}_z \text{ rotation} & \text{Controlled NOT} \ \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{R}_z ( \theta) = \begin{pmatrix} 1 & 0 \ 0 & e^{i \theta} \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \ \begin{pmatrix} \text{0 to 0} \ \text{1 to 1} \end{pmatrix} & \begin{bmatrix} \text{0 to } \frac{1}{ \sqrt{2}} (0 + 1) \ \text{1 to } \frac{1}{ \sqrt{2}} (0 - 1) \end{bmatrix} & \begin{pmatrix} \text{0 to 0} \ \text{1 to e}^{i \theta} \end{pmatrix} & \begin{pmatrix} \text{00 to 00} \ \text{01 to 01} \ \text{10 to 11} \ \text{11 to 10} \end{pmatrix} \end{matrix} \nonumber$
A flow diagram for the operation of the simulation circuit is prepared using the truth tables above. It clearly shows the formation of superpositions and an entangled superposition, the singlet spin state highlighted in red below, as the operation of the circuit proceeds.
$\begin{matrix} |0 \rangle = | \uparrow \rangle ~ \text{eigenvalue +1} ~ |1 \rangle = | \downarrow \rangle ~ \text{eigenvalue -1} \ |1 \rangle |1 \rangle = |11 \rangle \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] |1 \rangle = \frac{1}{ \sqrt{2}} \left[ |01 \rangle - |11 \rangle \right] \ \text{CNOT} \ \frac{1}{ \sqrt{2}} \left[ |01 \rangle - |10 \rangle \right] \ \text{I} \otimes \text{R}_z ( \theta) \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle | e^{i \theta} \rangle - |1 \rangle |0 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ |0 \rangle e^{i \theta} |1 \rangle - |1 \rangle |0 \rangle \right] \ \text{H} \otimes \text{H} \ \frac{1}{ \sqrt{2}} \left[ \frac{1}{ \sqrt{2}} (|0 \rangle + |1 \rangle ) e^{i \theta} \frac{1}{ \sqrt{2}} (|0 \rangle - |1 \rangle) - \frac{1}{ \sqrt{2}} ( |0 \rangle - |1 \rangle ) \frac{1}{ \sqrt{2}} ( |0 \rangle + |1 \rangle ) \right] \ \Downarrow \ \frac{1}{ 2 \sqrt{2}} \left[ |00 \rangle (e^{i \theta} - 1) - |01 \rangle (e^{i \theta} +1 ) + |10 \rangle (e^{i \theta} +1) - |11 \rangle (e^{i \theta} - 1) \right] \end{matrix} \nonumber$
There are four output states highlighted in blue above. If the spins are measured in the same state, |00> or |11>, the eigenvalue is +1, if they are different, |01> or |10>, the eigenvalue is -1. The probabilities for each of the two types of output states are now calculated.
$\begin{matrix} |00 \rangle \text{ or } |11 \rangle \rightarrow \left| \pm \frac{1}{ 2 \sqrt{2}} (e^{i \theta} - 1) \right|^2 = \frac{1 - \cos \theta}{4} & |01 \rangle \text{ or } \rightarrow \left| \pm \frac{1}{2 \sqrt{2}} (e^{i \theta} + 1) \right|^2 = \frac{ \cos \theta + 1}{4} \end{matrix} \nonumber$
Thus we see that the expectation value (or correlation coefficient) generated experimentally by the operation of this circuit is identical to the one calculated by the initial quantum mechanical analysis of the two-spin thought experiment. A quantum circuit has simulated an unperformed quantum experiment.
$E( \theta) = 2 \left( \frac{1- \cos \theta}{4} \right) - 2 \left( \frac{\cos \theta + 1}{4} \right) \rightarrow - \cos \theta \nonumber$
Another Look at the Simulation
As a companion to this algebraic analysis, the same result will now be demonstrated numerically. The probabilities of observing the four output states (|00>, |01>, |10> and |11>) are calculated for 0, 45, 60 and 90 degrees and shown to be in agreement with the following expectation values.
$\begin{matrix} \text{E(0 deg) = -1} & \text{E(45 deg) = -0.707} & \text{E(60 deg) = -0.5} & \text{E(90 deg) = 0} \end{matrix} \nonumber$
The operator representing the circuit is constructed from the matrix operators provided alone.
$\text{Op}( \theta) = \text{kronecker(H, H) kronecker(I, R}_z ( \theta)) \text{CNOT kronecker(H, I)} \nonumber$
$\begin{matrix} |00 \rangle ~ \text{eigenvalue +1} & \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0 & |01 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.5 \ |10 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.5 & |11 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{Op(0 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0 \end{matrix} \nonumber$
Expectation value: $0-0.5-0.5+0 = -1$
$\begin{matrix} |00 \rangle ~ \text{eigenvalue +1} & \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.073 & |01 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.427 \ |10 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.427 & |11 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{Op(45 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.073 \end{matrix} \nonumber$
Expectation value: $0.073-0.427-0.427+0.073 = -0.708$
$\begin{matrix} |00 \rangle ~ \text{eigenvalue +1} & \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{Op(60 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.125 & |01 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{Op(60 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.375 \ |10 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{Op(60 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.375 & |11 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{Op(60 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.125 \end{matrix} \nonumber$
Expectation value: $0.125-0.375-0.375+0.125 = -0.5$
$\begin{matrix} |00 \rangle ~ \text{eigenvalue +1} & \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{Op(90 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.25 & |01 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{Op(90 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.25 \ |10 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{Op(90 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.25 & |11 \rangle ~ \text{eigenvalue -1} & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{Op(90 deg)} \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right| \right]^2 = 0.25 \end{matrix} \nonumber$
Expectation value: $0.25-0.25-0.25+0.25 = 0.5$
Conclusion
"Quantum simulation is a process in which a quantum computer simulates another quantum system. Because of the various types of quantum weirdness, classical computers can simulate quantum systems only in a clunky, inefficient way. But because a quantum computer is itself a quantum system, capable of exhibiting the full repertoire of quantum weirdness, it can efficiently simulate other quantum systems. The resulting simulation can be so accurate that the behavior the computer will be indistinguishable from the behavior of the simulated system itself ." (Seth Lloyd, Programming the Universe, page 149.)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.61%3A_Simulating_Quantum_Correlations_with_a_Quantum_Computer.txt
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A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path. The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal.
The entangled two-photon polarization state is written in the circular and linear polarization states,
$\Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ |V \rangle_A |V \rangle_B - |H \rangle_A |H \rangle_B \right] \nonumber$
using
$\begin{matrix} | L \rangle = \frac{1}{ \sqrt{2}} \left[ |V \rangle + i| H \rangle \right] & |R \rangle = \frac{1}{ \sqrt{2}} \left[ |V \rangle - i| H \rangle \right] \end{matrix} \nonumber$
The vertical (eigenvalue +1) and horizontal (eigenvalue -1) polarization states for the photons in the measurement plane are given below. Θ is the angle of PAB.
$\begin{matrix} V( \theta) = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} & H ( \theta) = \begin{pmatrix} - \sin \theta \ \cos \theta \end{pmatrix} & V(0) = \begin{pmatrix} 1 \ 0 \end{pmatrix} & H(0) = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
If photon A has vertical polarization photon B also has vertical polarization, the probability that photon B has vertical polarization when measured at an angle θ giving a composite eigenvalue of +1 is,
$\left( V( \theta)^T V(0) \right)^2 \rightarrow \cos ( \theta)^2 \nonumber$
If photon A has vertical polarization photon B also has vertical polarization, the probability that photon B has horizontal polarization when measured at an angle θ giving a composite eigenvalue of -1 is,
$\left( H( \theta)^T V(0) \right)^2 \rightarrow \sin ( \theta)^2 \nonumber$
Therefore the overall quantum correlation coefficient or expectation value is:
$E( \theta) = \left( V( \theta)^T V(0) \right)^2 - \left( H( \theta)^T V(0) \right)^2 ~ \text{simplify}~ \rightarrow \cos (2 \theta) \nonumber$
Now it will be shown that a local-realistic, hidden-variable model can be constructed which is in agreement with the quantum calculations for 0 and 90 degrees, but not for 30 and 60 degrees (highlighted).
$\begin{matrix} \text{E(0 deg) = 1} & \text{E(30 deg) = 0.5} & \text{E(60 deg) = -0.5} & \text{E(90 deg) = -1} \end{matrix} \nonumber$
If objects have well-defined properties independent of measurement, the results for θ = 0 degrees and θ = 90 degrees require that the photons carry the following instruction sets, where the hexagonal vertices refer to θ values of 0, 30, 60, 90, 120, and 150 degrees.
There are eight possible instruction sets, six of the type on the left and two of the type on the right. The white circles represent vertical polarization with eigenvalue +1 and the black circles represent horizontal polarization with eigenvalue -1. In any given measurement, according to local realism, both photons (A and B) carry identical instruction sets, in other words the same one of the eight possible sets.
The problem is that while these instruction sets are in agreement with the 0 and 90 degree quantum calculations, with expectation values of +1 and -1 respectively, they can't explain the 30 degree predictions of quantum mechanics. The figure on the left shows that the same result should be obtained 4 times with joint eigenvalue +1, and the opposite result twice with joint eigenvalue of -1. For the figure on the right the opposite polarization is always observed giving a joint eigenvalue of -1. Thus, local realism predicts an expectation value of 0 in disagreement with the quantum result of 0.5.
$\frac{6(1-1+1+1-1+1)+2(-1-1-1-1-1-1)}{8} = 0 \nonumber$
This exercise illustrates Bell's theorem: no local hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems . As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned.
This analysis is based on "Simulating Physics with Computers" by Richard Feynman, published in the International Journal of Theoretical Physics (volume 21, pages 481-485), and Julian Brown's Quest for the Quantum Computer (pages 91-100). Feynman used the experiment outlined above to establish that a local classical computer could not simulate quantum physics.
A local classical computer manipulates bits which are in well-defined states, 0s and 1s, shown above graphically in white and black. However, these classical states are incompatible with the quantum mechanical analysis which is consistent with experimental results. This two-photon experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s.
Simulation of quantum physics requires a quantum computer. The following quantum circuit simulates this experiment exactly. The Hadamard and CNOT gates transform the input, |10>, into the required entangled Bell state. R(θ) rotates the polarization of photon B clockwise through an angle θ. Finally measurement yields one of the four possible output states: |00>, |01>, |10> or |11>.
$\begin{matrix} |1 \rangle & \triangleright & \text{H} & \cdot & \cdots & \triangleright & \text{Measure 0 or 1} \ ~ & ~ & ~ & | \ |0 \rangle & \triangleright & \cdots & \oplus & \text{R}( \theta) & \triangleright & \text{Measure 0 or 1} \end{matrix} \nonumber$
The matrix operators required to build this circuit are as follows:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{R}( \theta) = \begin{pmatrix} \cos \theta & - \sin \theta \ \sin \theta & \cos \theta \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\text{BellCircuit}( \theta) = \text{kronecker(I, R} ( \theta)) \text{CNOT kronecker(H, I)} \nonumber$
Calculating the probability of observing the four output states ( |00>, |01>, |10> and |11>) shows that the quantum circuit correctly simulates the experiment for the 30 and 60 degree rotations.
$\begin{matrix} |00 \rangle = |VV \rangle ~ \text{eigenvalue = +1} & |01 \rangle = |VH \rangle ~ \text{eigenvalue = -1} \ \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{6} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.375 & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{6} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.125 \ |10 \rangle = |HV \rangle ~ \text{eigenvalue = -1} & |11 \rangle = |HH \rangle ~ \text{eigenvalue = +1} \ \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{6} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.125 & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{6} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.375 \end{matrix} \nonumber$
Expectation value: $0.375 - 0.125 + 0.375 - 0.125 = 0.5$
$\begin{matrix} |00 \rangle = |VV \rangle ~ \text{eigenvalue = +1} & |01 \rangle = |VH \rangle ~ \text{eigenvalue = -1} \ \left[ \left| \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{3} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.125 & \left[ \left| \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{3} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.375 \ |10 \rangle = |HV \rangle ~ \text{eigenvalue = -1} & |11 \rangle = |HH \rangle ~ \text{eigenvalue = +1} \ \left[ \left| \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{3} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.375 & \left[ \left| \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix}^T \text{BellCircuit} \left( \frac{ \pi}{3} \right) \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.125 \end{matrix} \nonumber$
Expectation value: $0.125 - 0.375 - 0.375 + 0.125 = -0.5$
Next an algebraic analysis of the quantum circuit shows that it yields the correct expectation value for all values of θ. This analysis requires the truth tables for the matrix operators.
$\begin{matrix} \text{Identity} & \text{Hadamard gate} & \text{R}( \theta) \text{ rotation} & \text{Controlled NOT} \ \begin{pmatrix} \text{0 to 0} \ \text{1 to 1} \end{pmatrix} & \begin{bmatrix} \text{0 to } \frac{1}{ \sqrt{2}} (0 +1) \ \text{1 to } \frac{1}{ \sqrt{2}} (0-1) \end{bmatrix} & \begin{matrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \xrightarrow{R ( \theta)} \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} \ \begin{pmatrix} 0 \ 1 \end{pmatrix} \xrightarrow{R( \theta)} \begin{pmatrix} - \sin \theta \ \cos \theta \end{pmatrix} \end{matrix} & \begin{pmatrix} \text{00 to 00} \ \text{01 to 01} \ \text{10 to 11} \ \text{11 to 10} \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} |0 \rangle = |V \rangle ~ \text{eigenvalue +1} & |1 \rangle = |H \rangle ~ \text{eigenvalue -1} \end{matrix} \nonumber$
$\begin{matrix} |1 \rangle |0 \rangle = |10 \rangle \ \text{H} \otimes \text{I} \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] |0 \rangle = \frac{1}{ \sqrt{2}} \left[ |00 \rangle - |10 \rangle \right] \ \text{CNOT} \ \frac{1}{ \sqrt{2}} \left[ |00 \rangle - |11 \rangle \right] \ \text{I} \otimes \text{R} ( \theta) \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle ( \cos \theta |0 \rangle + \sin \theta |1 \rangle ) -|1 \rangle ( - \sin \theta | 0 \rangle + \cos \theta |1 \rangle ) \right] \ \Downarrow \frac{1}{ \sqrt{2}} \left[ \cos \theta | 00 \rangle + \sin \theta | 01 \rangle + \sin \theta |10 \rangle - \cos \theta |11 \rangle \right] \ \text{Probabilities} \ \Downarrow \ \frac{ \cos^2 \theta}{2} |00 \rangle + \frac{ \sin^2 \theta}{2} |01 \rangle + \frac{ \sin^2 \theta}{2} |10 \rangle + \frac{ \cos^2 \theta}{2} |11 \rangle \end{matrix} \nonumber$
|00> = |VV> and |11> = |HH> have composite eigenvalues of +1. |01> = |VH> and |10> = |HV> have composites eigenvalue of -1. Therefore,
$\text{E} ( \theta) = \cos \theta^2 - \sin \theta^2 \text{ simplify } \rightarrow \cos (2 \theta) \nonumber$
Summary
"Quantum simulation is a process in which a quantum computer simulates another quantum system. Because of the various types of quantum weirdness, classical computers can simulate quantum systems only in a clunky, inefficient way. But because a quantum computer is itself a quantum system, capable of exhibiting the full repertoire of quantum weirdness, it can efficiently simulate other quantum systems. The resulting simulation can be so accurate that the behavior the computer will be indistinguishable from the behavior of the simulated system itself. " (Seth Lloyd, Programming the Universe, page 149.)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.62%3A_Quantum_Computer_Simulation_of_Photon_Correlations.txt
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A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path. The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal.
The entangled two-photon polarization state is written in the circular and linear polarization bases,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | L \rangle_A |L \rangle_B + |R \rangle_A |R \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ |V \rangle_A |V \rangle_B - |H \rangle_A |H \rangle_B \right] \nonumber$
The vertical (eigenvalue +1) and horizontal (eigenvalue -1) polarization states for the photons in the measurement plane are given below. Θ is the angle of the measuring PA.
$\begin{matrix} V( \theta) = \begin{pmatrix} \cos \theta \ \sin \theta \end{pmatrix} & H( \theta) = \begin{pmatrix} - \sin \theta \ \cos \theta \end{pmatrix} & V(0) = \begin{pmatrix} 1 \ 0 \end{pmatrix} & H(0) = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
If photon A has vertical polarization photon B also has vertical polarization, the probability that photon B has vertical polarization when measured at an angle θ giving a composite eigenvalue of +1 is,
$\left( V( \theta)^T V(0) \right)^2 \rightarrow \cos ( \theta)^2 \nonumber$
If photon A has vertical polarization photon B also has vertical polarization, the probability that photon B has horizontal polarization when measured at an angle θ giving a composite eigenvalue of -1 is,
$\left( H( \theta)^T V(0) \right)^2 \rightarrow \sin ( \theta)^2 \nonumber$
Therefore the overall quantum correlation coefficient or expectation value is:
$E( \theta) = \left( V( \theta)^T V(0) \right)^2 - \left( H( \theta)^T V(0) \right)^2 \text{ simplify } \rightarrow \cos (2 \theta) \nonumber$
Now it will be shown that a local-realistic, hidden-variable model can be constructed which is in agreement with the quantum calculations for 0, 45 and 90 degrees, but not for 30 and 60 degrees (highlighted).
$\begin{matrix} \text{E(0 deg) = 1} & \text{E(30 deg) = 0.5} & \text{E(45 deg) = 0} & \text{E(60 deg) = -0.5} & \text{E(90 deg) = -1} \end{matrix} \nonumber$
If objects have well-defined properties independent of measurement, the results for θ = 0 degrees and θ = 90 degrees require that the photons carry the following instruction sets, where the hexagonal vertices refer to θ values of 0, 30, 60, 90, 120, and 150 degrees.
There are eight possible instruction sets, six of the type on the left and two of the type on the right. The white circles represent vertical polarization with eigenvalue +1 and the black circles represent horizontal polarization with eigenvalue -1. In any given measurement, according to local realism, both photons (A and B) carry identical instruction sets, in other words the same one of the eight possible sets.
The problem is that while these instruction sets are in agreement with the 0 and 90 degree quantum calculations, with expectation values of +1 and -1 respectively, they can't explain the 30 degree predictions of quantum mechanics. The figure on the left shows that the same result should be obtained 4 times with joint eigenvalue +1, and the opposite result twice with joint eigenvalue of -1. For the figure on the right the opposite polarization is always observed giving a joint eigenvalue of -1. Thus, local realism predicts an expectation value of 0 in disagreement with the quantum result of 0.5.
$\frac{6(1-1+1+1-1+1)+2(-1-1-1-1-1-1)}{8} = 0 \nonumber$
If we look at 60 degrees, E(60 deg) = 0,we reach the same conclusion: local realism disagrees with quantum result of -0.5.
$\frac{^(-1-1+1-1-1+1)+2(1+1+1+1+1+1)}{8} = 0 \nonumber$
Instruction sets for 45 degrees, E(45 deg) = 0, are shown below. The vertices are 0, 45, 90 and 135 degrees. The instruction sets are constructed so that they satisfy the 0 and 90 degree quantum expectation values of 1 and -1, respectively. It is clear that they also satisfy the 45 degree result yielding an expectation value of zero: (1-1+1-1) = (-1+1-1+1) = (1-1+1-1) = (-1+1-1+1) = 0.
This exercise illustrates Bell's theorem: no local hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems . As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned.
This analysis is based on "Simulating Physics with Computers" by Richard Feynman, published in the International Journal of Theoretical Physics (volume 21, pages 481-485), and Julian Brown's Quest for the Quantum Computer (pages 91-100). Feynman used the experiment outlined above to establish that a local classical computer could not simulate quantum physics.
A local classical computer manipulates bits which are in well-defined states, 0s and 1s, shown above graphically in white and black. However, these classical states are incompatible with the quantum mechanical analysis which is consistent with experimental results. This two-photon experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s. Simulation of quantum physics requires a quantum computer!
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.63%3A_Quantum_Correlations_Illustrated_with_Photons.txt
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Art Hobson recently posted "Implications of bipartite interferometry for the measurement problem" at arXiv:1301.1673. In this post he provides the following interpretation (modified slightly by F. Rioux) of entangled bipartite spin systems. The mathematics below is provided in support of Hobson's analysis.
When a bipartite system is in an entangled superposition, its subsystems are not in superpositions but are instead mixed states with each subsystem in a definite, but unknown state. An entangled spin is always in its local state, the state described by its reduced density operator, because this is the state actually detected by an observer of the spin. The entangled state is a global superposition of spin correlations, not a superposition of local spin states.
The reduced density operator of an entangled spin is obtained by tracing (averaging) the total density operator over the other spin. This procedure shows that a spin's reduced density operator is diagonal indicating a classical mixed state - it has been stripped of its off-diagonal interference terms.
An entangled bipartite superposition:
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ | \uparrow_1 \rangle | \downarrow_2 \rangle - | \downarrow_1 \rangle | \uparrow_2 \rangle \right] \nonumber$
where
$\begin{matrix} | \uparrow \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & | \downarrow \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Its total density operator:
$\hat{ \rho_{12}} = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left[ | \uparrow_1 \rangle | \downarrow_2 \rangle - | \downarrow_1 \rangle | \uparrow_2 \rangle \right] \left[ \langle \downarrow_2| \langle \uparrow_1 | - \langle \uparrow_2 | \langle \downarrow_1 | \right] = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 1 & -1 & 0 \ 0 & -1 & 1 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
The total density operator written in two equivalent forms:
$\begin{matrix} \hat{ \rho_{12}} = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left[ | \uparrow_1 \rangle \otimes \langle \downarrow_2 \rangle \langle \downarrow_2 | \otimes \langle \uparrow_1 | - | \uparrow_1 \rangle \otimes | \downarrow_2 \rangle \langle \uparrow_2 \otimes \langle \downarrow_1 | - | \uparrow_1 \rangle \otimes | \uparrow_2 \rangle \langle \downarrow_2 | \otimes \langle \uparrow_1 | + | \downarrow_1 \rangle \otimes | \uparrow_2 \rangle \langle \uparrow_2 | \otimes \langle \downarrow_1 | \right] \ \rho_{12} = \frac{1}{2} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} \right] = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & -0.5 & 0 \ 0 & -0.5 & 0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \ \hat{ \rho_{12}} = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left[ | \uparrow_1 \rangle \langle \uparrow_1 | \otimes | \downarrow_2 \rangle \langle \downarrow_2 | - | \uparrow_1 \rangle \langle \downarrow_1 | \otimes | \downarrow_2 \rangle \langle \uparrow_2 | - | \downarrow_1 \rangle \langle \uparrow_1 | \otimes | \uparrow_2 \rangle \langle \downarrow_2 | + | \downarrow_1 \rangle \langle \downarrow_1 | \otimes | \uparrow_2 \rangle \langle \uparrow_2 | \right] \ \rho_{12} = \frac{1}{2} \begin{bmatrix} \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] - \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] ... \ +- \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] + \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \end{bmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & -0.5 & 0 \ 0 & -0.5 & 0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
Calculation of the reduced density operator of spin 1 by tracing the total density operator over spin 2:
$\hat{ \rho_1} = \frac{1}{2} \left[ \langle \uparrow_2 | \Psi \rangle \langle \Psi | \uparrow_2 \rangle + \langle \downarrow_2 | \Psi \rangle \langle \Psi | \downarrow_2 \rangle \right] = \frac{1}{2} \left[ | \downarrow_1 \rangle \langle \downarrow_1 | + | \uparrow_1 \rangle \langle \uparrow_1 | \right] = \frac{1}{2} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$rho_1 = \frac{1}{2} \begin{bmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} ~ \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] - \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} ~ \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] ... \ +- \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} ~ \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} ~ \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \end{bmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{pmatrix} \nonumber$
Calculation of the reduced density operator of spin 2 by tracing the total density operator over spin 1:
$\hat{ \rho}_2 = \frac{1}{2} \left[ \langle \uparrow_1 | \Psi \rangle \langle \Psi | \uparrow_1 \rangle + \langle \downarrow_1 | \Psi \rangle \langle \Psi | \downarrow_1 \rangle \right] = \frac{1}{2} \left[ | \downarrow_2 \rangle \langle \downarrow_2 | + | \uparrow_2 \rangle \langle \uparrow_2 | \right] = \frac{1}{2} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \nonumber$
$rho_2 = \frac{1}{2} \begin{bmatrix} \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} - \text{tr} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} ... \ +- \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \right] \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} + \text{tr} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} \right] \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix} \end{bmatrix} \rightarrow \begin{pmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{pmatrix} \nonumber$
Now let's flesh this out with specific calculations. Below we have the entangled singlet spin state, its total density operator, and the reduced density operators for the individual spins.
$\begin{matrix} \Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ -1 \ 0 \end{pmatrix} & \rho_{12} = \Psi \Psi^T = \begin{pmatrix} 0 & 0 & 0 & 0 \ 0 & 0.5 & -0.5 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix} & \rho_1 = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \rho_2 = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Next we need the operator required for spin measurements in the z-direction and the identity operator, do nothing.
$\begin{matrix} \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Now we calculate expectation values for some spin measurements in the z-direction. As shown below this can be accomplished in two ways, the latter on the right, uses the trace function.
$\begin{matrix} \langle \Psi | \hat{O} | \Psi \rangle = \sum_{i} \langle i | \Psi \rangle \langle \Psi | \hat{O} | i \rangle = Trace \left( | \Psi \rangle \langle \Psi | \hat{O} \right) \text{ where } \sum_{i} | i \rangle \langle i | = Identity \end{matrix} \nonumber$
If we measure the z-component of both spins we find perfect anti-correlation which is expected for an entangled singlet state.
$\begin{matrix} \Psi^T \text{kronecker} \left( \sigma_z,~ \sigma_z \right) \Psi = -1 & \text{tr} \left( \Psi \Psi^T \text{kronecker} \left( \sigma_z, ~ \sigma_z \right) \right) = 1 \end{matrix} \nonumber$
However, if only one spin is measured the expectation value is 0.
$\begin{matrix} \Psi^T \text{kronecker} \left( \sigma_z,~ \text{I} \right) \Psi = 0 & \text{tr} \left( \Psi \Psi^T \text{kronecker} \left( \sigma_z,~ \text{I} \right) \right) = 0 & \Psi^T \text{kronecker} \left( \text{I},~ \sigma_z \right) \Psi = 0 & \text{tr} \left( \Psi \Psi^T \text{kronecker} \left( \text{I},~ \sigma_z \right) \right) = 0 \end{matrix} \nonumber$
These results are consistent with the following calculations, which use the reduced density operators for the individual spins. In other words it assumes the spins are in local mixed states, as was asserted previously. The trace function is required for the calculation of expectation values of mixed states.
$\begin{matrix} \text{tr} \left( \rho_1,~ \sigma_z \right) = 0 & \text{tr} \left( \rho_2,~ \sigma_z \right) = 0 \end{matrix} \nonumber$
To repeat, the entangled state Ψ is a global superposition of spin correlations, not a superposition of local spin states.
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A quantum computer exploits quantum mechanical effects such as superpositions, entanglement and interference to perform new types of calculations that are impossible on a classical computer. Quantum computation is therefore nothing less than a distinctly new way of harnessing nature. (Adapted from David Deutsch, The Fabric of Reality, page 195.)
Whereas classical computers perform operations on classical bits, which can be in one of two discrete states, 0 or 1, quantum computers perform operations on quantum bits, or qubits, which can be put into any superposition of two quantum states, |0> and |1>. Peter Pfeifer, McGraw-Hill Encyclopedia of Science and Industry.
The following example demonstrates how a quantum circuit can function as an algorithm for the evaluation of a mathematical function f(x), and how the same algorithm is capable of parallel evaluations of that function.
$\begin{matrix} \begin{pmatrix} \text{x} & \text{f(x)} \ 0 & 1 \ 1 & 0 \end{pmatrix} & \begin{matrix} |x \rangle & \cdots & \cdot & \cdots & \cdots & | x \rangle \ ~ & ~ & | \ |0 \rangle & \cdots & \oplus & \fbox{NOT} & \cdots & | f(x) \rangle \end{matrix} & \hat{U}_f | x \rangle |0 \rangle = |x \rangle f(x) \rangle \end{matrix} \nonumber$
As shown below, when |x> is |0> or |1> the circuit behaves like a classical computer yielding the value of f(x). When |x> is a superposition of |0> and |1> the circuit is a quantum computer, operating on both input values simultaneously in a single pass through the circuit, yielding both values of f(x). Note that in the latter case, the intermediate and final states are entangled Bell superpositions. The Bell states are an essential resource in many quantum information applications.
$\begin{matrix} \text{Input} & \text{Operation} & \text{Intermediate} & \text{Operation} & \text{Output} \ |0 \rangle |0 \rangle & ~ & |0 \rangle |0 \rangle & ~ & |0 \rangle |1 \rangle \ |1 \rangle |0 \rangle & \xrightarrow{ \text{CNOT}} & |1 \rangle |1 \rangle & \xrightarrow{ \text{I} \otimes \text{NOT}} & |1 \rangle |0 \rangle \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] = \frac{1}{ \sqrt{2}} \left[ |0 \rangle |0 \rangle + |1 \rangle |0 \rangle \right] & ~ & \frac{1}{ \sqrt{2}} \left[ |0 \rangle |0 \rangle + |1 \rangle |1 \rangle \right] & ~ & \frac{1}{ \sqrt{2}} \left[ |0 \rangle |1 \rangle + |1 \rangle |0 \rangle \right] \end{matrix} \nonumber$
Haroche and Raimond (pages 94-95 of Exploring the Quantum) describe the latter process as follows: "By superposing the inputs of a computation, one operates the machine 'in parallel', making it compute simultaneously all the values of a function and keeping its state, before any final bit detection is performed, suspended in a coherent superposition of all the possible outcomes." However, as Haroche and Raimond note, on a practical level only one result can be realized for each operation of the circuit because on measurement the superposition created by the circuit collapses to one of the states forming the superposition. Therefore, the exploitation of quantum parallelism for practical purposes such as searches and factorization requires more elaborate quantum circuits than the one presented here.
Truth tables for the quantum circuit:
$\begin{matrix} \text{NOT} & \begin{pmatrix} 0 \text{ to } 1 \ 1 \text{ to } 0 \end{pmatrix} & \text{CNOT} \begin{pmatrix} \text{Decimal} & \text{Binary} & \text{to} & \text{Binary} & \text{Decimal} \ 0 & 00 & \text{to} & 00 & 0 \ 1 & 01 & \text{to} & 01 & 1 \ 2 & 10 & \text{to} & 11 & 3 \ 3 & 11 & \text{to} & 10 & 2 \end{pmatrix} \end{matrix} \nonumber$
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This tutorial deals with quantum function evaluation and parallel computation. The example is taken from pages 94-95 of Exploring the Quantum by Haroche and Raimond. A certain function of x yields the following table of results.
$\begin{pmatrix} \text{x} & 0 & 1 \ \text{f(x)} & 1 & 0 \end{pmatrix} \nonumber$
First we establish that the circuit shown below yields the results given in the table, and then demonstrate that it also carries out a parallel calculation in one step using both input values of x.
$\begin{matrix} \begin{matrix} |x \rangle & \cdots & \cdot & \cdots & \cdots & |x \rangle \ ~ & ~ & | \ |0 \rangle & \cdots & \oplus & \fbox{ NOT} & \cdots & |f(x) \rangle \end{matrix} & \text{ where, for example } & |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The top wire carries the value of x and the bottom wire is initially set to |0 >. After operation of the controlled-NOT and NOT gates, x remains on the top wire while the bottom wire carries the value of the function, f(x). In other words,
$\hat{U}_f |x \rangle |0 \rangle = |x \rangle | f(x) \rangle \nonumber$
The quantum gates in matrix form are:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{NOT} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Uf (controlled-NOT, followed by a NOT operation on the lower wire) is a reversible operator. Doing it twice in succession on the initial two-qubit state is equivalent to the identity operation.
Kronecker is Mathcad's command for carrying out matrix tensor multiplication. Note that the identity operator is required when a wire is not involved in an operation. In what follows the quantum circuit is constructed, displayed and its reversibility demonstrated. In other words, repeating the circuit is equivalent to the identity operation. Reversibility is a crucial property in quantum computer circuitry.
$\text{QuantumCircuit = kronecker(I, NOT) CNOT} \nonumber$
$\begin{matrix} \text{QuantumCircuit} = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} & \text{QuantumCircuit}^2 = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Given the simplicity of the matrix representing the circuit, the following calculations can easily be done by hand.
$\begin{matrix} ~ & \text{Input} & \text{Calculation} & \text{Output} \ \text{f(0) = 1} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \ \text{f(1) = 0} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
These calculations demonstrate that the quantum circuit is a valid algorithm for the calculation of f(x). We now demonstrate parallel computation by putting |x> in a balanced superposition of |0> and |1>. As shown below, the operation of the circuit yields a superposition of the previous results. The function has been evaluated for both values of x in a single pass through the circuit.
$\begin{matrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} & \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ 0.707 \ 0 \end{pmatrix} \end{matrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \nonumber$
Haroche and Raimond describe this process as follows: "By superposing the inputs of a computation, one operates the machine 'in parallel', making it compute simultaneously all the values of a function and keeping its state, before any final bit detection is performed, suspended in a coherent superposition of all the possible outcomes."
In summary, simple calculations have demonstrated how a quantum circuit can function as an algorithm for the evaluation of a mathematical function, and how the same circuit is capable of parallel evaluations of that function.
$\begin{matrix} \text{Input} & \text{Operation} & \text{Intermediate} & \text{Operation} & \text{Output} \ |00 \rangle & ~ & |00 \rangle & ~ & |01 \rangle \ |10 \rangle & \xrightarrow{ \text{CNOT}} & |11 \rangle & \xrightarrow{ \text{I} \otimes \text{NOT}} & |10 \rangle \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] |0 \rangle = \frac{1}{ \sqrt{2}} \left[ |00 \rangle + |10 \rangle \right] & ~ & \frac{1}{ \sqrt{2}} \left[ |00 \rangle + |11 \rangle \right] & ~ & \frac{1}{ \sqrt{2}} \left[ |01 \rangle + |10 \rangle \right] \end{matrix} \nonumber$
However, as Haroche and Raimond note, on a practical level only one result can be realized for each operation of the circuit because on measurement the superposition created by the circuit collapses to one of the states forming the superposition. This is simulated with projection operators (|0><1|) on both registers for the four possible measurement outcomes for each value of x.
$\begin{matrix} \text{f(0) = 0} & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T,~ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T \right] ~ \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0 \ \text{f(0) = 1} & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T,~ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T \right] ~ \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.5 \ \text{f(1) = 0} & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T,~ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T \right] ~ \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0.5 \ \text{f(1) = 1} & \left[ \left| \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T,~ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T \right] ~ \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} \right| \right]^2 = 0 \end{matrix} \nonumber$
As Haroche and Raimond write, "It is, however, one thing to compute potentially at once all the values of f(x) and quite another to be able to exploit this quantum parallelism and extract from it more information than from a mundane classical computation. The final stage of information acquisition must always be a measurement." Therefore, the exploitation of quantum parallelism for practical purposes such as searches and factorization requires more elaborate quantum circuits than the one presented here.
Truth tables for quantum circuit elements:
$\begin{matrix} \text{Identity} & \begin{pmatrix} \text{0 to 0} \ \text{1 to 1} \end{pmatrix} & \text{NOT} & \begin{pmatrix} \text{0 to 1} \ \text{1 to 0} \end{pmatrix} & \text{CNOT} & \begin{pmatrix} \text{Decimal} & \text{Binary} & \text{to} & \text{Binary} & \text{Decimal} \ 0 & 00 & \text{to} & 00 & 0 \ 1 & 01 & \text{to} & 01 & 1 \ 2 & 10 & \text{to} & \text{11} & 3 \ 3 & 11 & \text{to} & 10 & 2 \end{pmatrix} \end{matrix} \nonumber$
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This tutorial deals with quantum function evaluation and parallel computation. A certain function of x yields the following table of results.
$\begin{pmatrix} \text{x} & 0 & 1 \ \text{f(x)} & 1 & 0 \end{pmatrix} \nonumber$
First we establish that the circuit shown below yields the results given in the table, and then demonstrate that it also carries out a parallel calculation in one step using both input values of x.
$\begin{matrix} \begin{matrix} |x \rangle & \cdots & \cdot & \cdots & \cdots & |x \rangle \ ~ & ~ & | \ |0 \rangle & \cdots & \oplus & \fbox{ NOT} & \cdots & |f(x) \rangle \end{matrix} & \text{ where, for example } & |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The top wire carries the value of x and the bottom wire is initially set to |0 >. After operation of the controlled-NOT and NOT gates, x remains on the top wire while the bottom wire carries the value of the function, f(x). In other words,
$\hat{U}_f |x \rangle |0 \rangle = |x \rangle | f(x) \rangle \nonumber$
The quantum gates in matrix form are:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{NOT} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{ICNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{QuantumCircuit = kronecker(I, NOT) ICNOT} & \text{QuantumCircuit} = \begin{pmatrix} 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Given the simplicity of the matrix representing the circuit, the following calculations can easily be done by hand.
$\begin{matrix} ~ & \text{Input} & \text{Calculation} & \text{Output} \ \text{f(0) = 1} & \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \ \text{f(1) = 0} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
These calculations demonstrate that the quantum circuit is a valid algorithm for the calculation of f(x). We now demonstrate parallel computation by putting |x> in a balanced superposition of |0> and |1>. As shown below, the operation of the circuit yields a superposition of the previous results. The function has been evaluated for both values of x in a single pass through the circuit.
$\begin{matrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} & \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ 0 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \nonumber$
In summary, simple calculations have demonstrated how a quantum circuit can function as an algorithm for the evaluation of a mathematical function, and how the same circuit is capable of parallel evaluations of that function.
$\begin{matrix} \text{Input} & \text{Operation} & \text{Intermediate} & \text{Operation} & \text{Output} \ |00 \rangle & ~ & |00 \rangle & ~ & |01 \rangle \ |10 \rangle & \xrightarrow{ \text{ICNOT}} & |11 \rangle & \xrightarrow{ \text{I} \otimes \text{NOT}} & |10 \rangle \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] |0 \rangle = \frac{1}{ \sqrt{2}} \left[ |00 \rangle + |10 \rangle \right] & ~ & \frac{1}{ \sqrt{2}} \left[ |00 \rangle + |10 \rangle \right] & ~ & \frac{1}{ \sqrt{2}} \left[ |01 \rangle + |11 \rangle \right] \end{matrix} \nonumber$
However, on a practical level only one result can be realized for each operation of the circuit because on measurement the superposition created by the circuit collapses to one of the states forming the superposition.
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This tutorial is closely related to the preceding one. A certain function of x maps {0,1} to {0,1}. The four possible outcomes of the evaluation of f(x) are given in tabular form.
$\begin{pmatrix} \text{x} & ' & 0 & 1 & ' & 0 & 1 & ' & 0 & 1 & ' & 0 & 1 \ \text{f(x)} & ' & 0 & 0 & ' & 1 & 1 & ' & 0 & 1 & ' & 1 & 0 \end{pmatrix} \nonumber$
In the previous tutorial we established that the circuit shown below yields the result given in the right most section of the table. In other words, f(x) is a balanced function, because $f(0) \neq f(1)$, as is the result immediately to its left. The results in the first two sections are labelled constant because $f(0) = f(1)$.
where, for example
$\begin{matrix} |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
This circuit carries out,
$\hat{U}_f |x \rangle |0 \rangle = |x \rangle | f(x) \rangle \nonumber$
where Uf (controlled-NOT, followed by a NOT operation on the lower wire) is a unitary operator that accepts input |x> on the top wire and places f(x) on the bottom wire.
From the classical perspective, if the question (as asked by Deutsch) is whether f(x) is constant or balanced then one must calculate both f(0) and f(1) to answer the question. Deutsch pointed out that quantum superpositions and the interference effects between them allow the answer to be given with one pass through a modified version of the circuit as shown here.
The input |0>|1> is followed by a Hadamard gate on each wire. As is well known the Hadamard operation creates the following superposition states.
$\begin{matrix} \text{H} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & \text{H} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
Therefore the Hadamard operations transform the input state to the following two-qubit state which is fed to Uf.
$|x \rangle |y \rangle = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \end{pmatrix} \nonumber$
Uf processes its input to generate the following output.
$\hat{U}_f |x \rangle |y \rangle = |x \rangle | \text{mod}_2 (y + f(x)) \rangle = |x \rangle |y \oplus f(x) \rangle \nonumber$
To facilitate an algebraic analysis of the circuit operation the input state is written as,
$|x \rangle |y \rangle = \frac{1}{ \sqrt{2}} \left[ |0 \rangle +|1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] = \frac{1}{2} \left[ |0 \rangle |0 \rangle - |0 \rangle |1 \rangle + |1 \rangle |0 \rangle - |1 \rangle |1 \rangle \right] \nonumber$
In this format Uf creates the following output state.
$\Psi_{out} = \frac{1}{2} \left[ |0 \rangle | f(0) \rangle - |0 \rangle |1 \oplus f(0) \rangle + |1 \rangle | f(1) \rangle - |1 \rangle |1 \oplus f(1) \rangle \right] \nonumber$
As the table shows there are four possible outcomes depending on whether the function is constant (the first two) or balanced (the second two).
$\begin{matrix} \Psi_{out} \xrightarrow[f(0) = 0]{f(0) = f(1)} \frac{1}{2} \left[ |0 \rangle |0 \rangle - |0 \rangle |1 \rangle + |1 \rangle |0 \rangle - |1 \rangle |1 \rangle \right] = \frac{1}{2} \left( |0 \rangle + |1 \rangle \right) \left( |0 \rangle - |1 \rangle \right) \ \Psi_{out} \xrightarrow[f(0) = 1]{f(0) = f(1)} \frac{1}{2} \left[ |0 \rangle |1 \rangle - |0 \rangle |0 \rangle + |1 \rangle |1 \rangle - |1 \rangle |0 \rangle \right] = - \frac{1}{2} \left( |0 \rangle + |1 \rangle \right) \left( |0 \rangle - |1 \rangle \right) \ \Psi_{out} \xrightarrow[f(0) = 0]{f(0) \neq f(1)} \frac{1}{2} \left[ |0 \rangle |0 \rangle - |0 \rangle |1 \rangle + |1 \rangle |1 \rangle - |1 \rangle |0 \rangle \right] = \frac{1}{2} \left( |0 \rangle + |1 \rangle \right) \left( |0 \rangle - |1 \rangle \right) \ \Psi_{out} \xrightarrow[f(0) = 1]{f(0) \neq f(1)} \frac{1}{2} \left[ |0 \rangle |1 \rangle - |0 \rangle |0 \rangle + |1 \rangle |0 \rangle - |1 \rangle |1 \rangle \right] = - \frac{1}{2} \left( |0 \rangle - |1 \rangle \right) \left( |0 \rangle - |1 \rangle \right) \end{matrix} \nonumber$
The Hadamard operation (see matrix below) on the first qubit brings about the following transformations.
$\begin{matrix} H \frac{1}{2} \left( |0 \rangle + |1 \rangle \right) = \frac{1}{ \sqrt{2}} |0 \rangle & H \frac{1}{2} \left( |0 \rangle - |1 \rangle \right) = \frac{1}{ \sqrt{2}} |1 \rangle \end{matrix} \nonumber$
The four possible output states are now,
$\begin{matrix} \Psi_{out} \xrightarrow[f(0) = 0]{f(0) = f(1)} \frac{1}{ \sqrt{2}} |0 \rangle \left( |0 \rangle - |1 \rangle \right) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 0 \ 0 \end{pmatrix} & \Psi_{out} \xrightarrow[f(0) = 1]{f(0) = f(1)} - \frac{1}{ \sqrt{2}} |0 \rangle \left( |0 \rangle - |1 \rangle \right) = \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \ 0 \ 0 \end{pmatrix} \ \Psi_{out} \xrightarrow[f(0) = 0]{f(0) \neq f(1)} \frac{1}{ \sqrt{2}} |1 \rangle \left( |0 \rangle - |1 \rangle \right) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 0 \ 1 \ -1 \end{pmatrix} & \Psi_{out} \xrightarrow[f(0) = 1]{f(0) \neq f(1)} - \frac{1}{ \sqrt{2}} |1 \rangle \left( |0 \rangle - |1 \rangle \right) = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 0 \ -1 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Quantum mechanics answers Deutsch's question with a single measurement. A measurement on the first qubit reveals whether the function is constant (|0>) or balanced (|1>).
We now look at the same calculation using matrix algebra. The required quantum operators in matrix form are:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{NOT} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Kronecker is Mathcad's command for carrying out matrix tensor multiplication. Note that the identity operator is required when a wire is not involved in an operation.
$\begin{matrix} \text{U}_f = \text{kronecker(I, NOT) CNOT} & \text{QuantumCircuit} = \text{kronecker(H, I) U}_f ~ \text{kronecker(H, H)} \end{matrix} \nonumber$
Input state:
$\begin{matrix} |0 \rangle |1 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ -0.707 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
Comparing this with the previous algebraic analysis, we see that the quantum circuit produces the result $f(0) \neq f(1)$ with f(0) = 1, which we already knew from previous work.
The measurement on the first qubit is implemented with projection operators |0><1|, and confirms that the function is not constant but belongs to the balanced category.
$\begin{matrix} \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T, ~ \text{I} \right] \text{QuantumCircuit} \begin{pmatrix} 0 \ . \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{Top qubit is not |0 >.} \ \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T,~\text{I} \right] \text{QuantumCircuit} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ -0.707 \ 0.707 \end{pmatrix} & \text{Top qubit is |1 >.} \end{matrix} \nonumber$
This could have also been easily determined by inspection:
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 0 \ -1 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \nonumber$
The following provides an algebraic analysis of the Deutsch algorithm.
$\begin{matrix} |0 \rangle & \triangleright & \fbox{H} & \cdots & \cdot & \cdots & \fbox{H} & \triangleright & \text{measure} \frac{ |0 \rangle \text{ constant}}{|1 \rangle \text{ balance}} \ ~ & ~ & ~ & ~ | \ |1 \rangle & \triangleright & \fbox{H} & \cdots & \oplus & \fbox{NOT} & \cdots \end{matrix} \nonumber$
Hadamard operation:
$\begin{matrix} \text{H} |0 \rangle \rightarrow \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] & \text{H} |1 \rangle \rightarrow \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] \end{matrix} \nonumber$
$\begin{matrix} \text{NOT} & \begin{pmatrix} \text{0 to 1} \ \text{1 to 0} \end{pmatrix} & \text{CNOT} & \begin{pmatrix} \text{Decimal} & \text{Binary} & \text{to} & \text{Binary} & \text{Decimal} \ 0 & 00 & \text{to} & 00 & 0 \ 1 & 01 & \text{to} & 01 & 1 \ 2 & 10 & \text{to} & 11 & 3 \ 3 & 11 & \text{to} & 10 & 2 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} |01 \rangle \ \text{H} \otimes \text{H} \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] = \frac{1}{2} \left[ |00 \rangle - |01 \rangle + |10 \rangle - |11 \rangle \right] \ \text{CNOT} \ \frac{1}{2} \left[ |00 \rangle - |01 \rangle + 11 \rangle - |10 \rangle \right] = \frac{1}{2} \left( |0 \rangle - |1 \rangle \right) \left( |0 \rangle - |1 \rangle \right) \ \text{I} \otimes \text{NOT} \ \frac{1}{2} \left[ \left( |0 \rangle - |1 \rangle \right) \left( |1 \rangle - |1 \rangle \right) \right] \ \text{H} \otimes \text{I} \ |1 \rangle \frac{1}{ \sqrt{2}} \left( |1 \rangle - |0 \rangle \right) \end{matrix} \nonumber$
The top wire contains |1 > indicating the function is balanced.
The Hadamard operation is actually a simple example of a Fourier transform. In other words, the final step of Deutsch's algorithm is to carry out a Fourier transform on the input wire. This also occurs on the input wires in Grover's search algorithm, Simon's query algorithm and Shor's factorization algorithm.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.68%3A_A_Simple_Solution_to_Deutsch%27s_Problem.txt
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See the previous tutorial "A Simple Solution to Deutsch's Problem" in this section for further detail on Deutsch's algorithm. This tutorial presents a similar example, so only a brief outline will be provided.
A certain function of x maps {0,1} to {0,1}. The four possible outcomes of the evaluation of f(x) are given in tabular form.
$\begin{pmatrix} \text{x} & ' & 0 & 1 & ' & 0 & 1 & ' & 0 & 1 & ' & 0 & 1 \ \text{f(x)} & ' & 0 & 0 & ' & 1 & 1 & ' & 0 & 1 & ' & 1 & 0 \end{pmatrix} \nonumber$
From the classical perspective, if the question (as asked by Deutsch) is whether f(x) is constant $(f(0) = f(1))$ or balanced $(f(0) \neq f(1))$ then one must calculate both f(0) and f(1) to answer the question as shown below.
The following circuit carries out,
$\hat{U}_f |x \rangle |0 \rangle = |x \rangle | f(x) \rangle \nonumber$
where Uf (inverse controlled-NOT, followed by a NOT operation on the lower wire) is a unitary operator that accepts input |x> on the top wire and places f(x) on the bottom wire.
$\begin{matrix} |x \rangle & \cdots & \oplus & \cdots & \cdots & |x \rangle \ ~ & ~ & | \ |0 \rangle & \cdots & \cdot & \fbox{NOT} & \cdots & | f(x) \rangle \end{matrix} \nonumber$
The required circuit elements in matrix format are:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} & \text{NOT} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{ICNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{QuantumCircuit} = \text{kronecker(I, NOT) ICNOT} & \text{QuantumCircuit} = \begin{pmatrix} 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} ~ & \text{Input} & \text{Calculation} & \text{Output} \ f(0) = 1 & \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \ f(1) = 1 & \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
We see by these operations of the circuit that f(x) is constant:
$\begin{pmatrix} \text{x} & 0 & 1 \ \text{f(x)} & 1 & 1 \end{pmatrix} \nonumber$
A quantum computer can also process a superposition of inputs, in this case a superposition of 0 and 1. The circuit produces a superposition of the outputs calculated above.
$\begin{matrix} \text{Input state}: & \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \end{pmatrix} ~~ \text{QuantumCircuit} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.707 \ 0 \ 0.707 \end{pmatrix} \ \text{Output state:} & \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 \ 1 \ 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] = \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \end{matrix} \nonumber$
However, this does not answer the balanced/constant question because observation of the output causes it to collapse to one value of the other, either |0>|1> or |1>|1>. Running the circuit just once will not answer the question.
Deutsch pointed out that quantum superpositions and the interference effects between them allow the answer to be given with one pass through a modified version of the circuit shown here.
$\begin{matrix} |0 \rangle & \triangleright & \fbox{H} & \cdots & \oplus & \cdots & \fbox{H} & \triangleright & \text{measure} \frac{|0 \rangle \text{ constant}}{|1 \rangle \text{ balanced}}\end{matrix} \nonumber$
$\text{QuantumCircuit} = \text{kronecker(H, I) kronecker(I, NOT) ICNOT kronecker(H, H)} \nonumber$
$\text{QuantumCircuit} = \begin{pmatrix} 0.707 & -0.707 & 0 & 0 \ 0.707 & 0.707 & 0 & 0 \ 0 & 0 & -0.707 & 0.707 \ 0 & 0 & 0.707 & 0.707 \end{pmatrix} \nonumber$
$\begin{matrix} \text{QuantumCircuit} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} -0.707 \ 0.707 \ 0 \ 0 \end{pmatrix} & \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \frac{1}{ \sqrt{2}} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} - \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \end{matrix} \nonumber$
The appearance of |0 > on the top wire confirms that the function is constant with one pass through the circuit.
It is also possible using the following truth tables to trace the evolution of the input quibits through the quantum circuit.
Hadamard operation:
$\begin{bmatrix} 0 & ' & \text{H} & ' & \frac{1}{ \sqrt{2}} (0 + 1) & ' & \text{H} & ' & 0 \ 1 & ' & \text{H} & ' & \frac{1}{ \sqrt{2}} (0-1) & ' & \text{H} & ' & 1 \end{bmatrix} \nonumber$
$\begin{matrix} \text{Identity} & \begin{pmatrix} 0 & \text{to} & 1 \ 1 & \text{to} & 0 \end{pmatrix} & \text{ICNOT} & \begin{pmatrix} \text{Decimal} & \text{Binary} & \text{to} & \text{Binary} & \text{Decimal} \ 0 & 00 & \text{to} & 00 & 0 \ 1 & 01 & \text{to} & 11 & 3 \ 2 & 10 & \text{10} & 10 & 2 \ 3 & 11 & \text{to} & 01 & 1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} |01 \rangle \ \text{H} \otimes \text{H} \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] = \frac{1}{2} \left[ |00 \rangle - |01 \rangle + |10 \rangle - |11 \rangle \right] \ \text{ICNOT} \ \frac{1}{2} \left[ |00 \rangle - |11 \rangle + |10 \rangle - |01 \rangle \right] = \frac{1}{2} \left( |0 \rangle +|1 \rangle \right) \left( |0 \rangle -|1 \rangle \right) \ \text{I} \otimes \text{NOT} \ \frac{1}{2} \left[ \left( |0 \rangle + |1 \rangle \right) \left( |1 \rangle - |0 \rangle \right) \right] \ \text{H} \otimes \text{I} \ |0 \rangle \frac{1}{ \sqrt{2}} \left( |1 \rangle - |0 \rangle \right) \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.69%3A_Another_Example_of_Deutsch%27s_Algorithm.txt
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This tutorial deals with quantum function evaluation and parallel computation. The example is taken from pages 94-95 of Exploring the Quantum by Haroche and Raimond. A certain function of x yields the following table of results.
$\begin{pmatrix} \text{x} & 0 & 1 & 2 & 3 \ \text{f(x)} & 1 & 0 & 0 & 1 \end{pmatrix} \nonumber$
First we establish that the circuit given below yields the results shown in the table, and then demonstrate that it also carries out a parallel calculation in one step using all four input values of x.
In Figure 2.22(b) Haroche and Raimond show a three wire quantum circuit which evaluates f(x). The top two wires are coded for the values of x (2 bits, 4 values) and the third wire is initially set to |0 >. After operation of the circuit on the initial state, the value of the function appears on the third wire. The qubits on the top wires are not changed by the circuit. The circuit consists of a CnNOT (controlled-controlled NOT) gate employing all three wires, followed by a CNOT gate involving the second and third wires, and finally a NOT gate operating on the third wire. A sketch is provided below.
$\begin{matrix} \begin{matrix} |a \rangle & \triangleright & \cdot & \cdots & \cdots & \cdots & \cdots & \triangleright & |a \rangle \ ~ & ~ & | \ |b \rangle & \triangleright & | & \cdots & \cdot & \cdots & \cdots & \triangleright & |b \rangle \ ~ & ~ & | & ~ & | \ |0 \rangle & \triangleright & \oplus & \cdots & \oplus & \cdots & \fbox{NOT} & \triangleright & | f(x) \rangle \end{matrix} \text{ where } |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The required quantum gates in matrix operator form are as follows:
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{NOT} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Kronecker is Mathcad's command for carrying out the tensor multiplication of matrices. Note that the identity operator is required when a wire is not involved in an operation. In the Appendix the matrix form of the circuit is displayed and its reversibility is demonstrated.
$\text{QuantumCircuit = kronecker(I, kronecker(I, NOT)) kronecker(I, CNOT) CnNOT} \nonumber$
The operation of the quantum circuit is now illustrated.
$\begin{matrix} ~ & \text{Input} & \text{Calculation} & \text{Output} \ \text{f(0) = 1} & \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \ \text{f(1) = 0} & \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \ \text{f(2) = 0} & \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} \ \text{f(3) = 0} & \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \end{pmatrix} & \text{QuantumCircuit} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Now we come to the demonstration of parallel calculation that is possible in a quantum gate circuit such as the one shown above. If all the values of x (0, 1, 2, 3) are present simultaneously as a superposition, the |a>|b> part of the input register is (1/2 is the normalization constant),
$\begin{matrix} \frac{1}{2} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} & \text{which leads to this total input register} & \frac{1}{2} \begin{pmatrix} 1 \ 1 \ 1 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
Operating on this register with the quantum circuit yields the following result, which by inspection is clearly a balanced superposition of the earlier results. The function has been evaluated for all values of x in one pass through the circuit.
$\begin{matrix} ~ & \text{f(0) = 1} & \text{f(1) = 0} & \text{f(2) = 0} & \text{f(3) = 1} \ \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.5 \ 0.5 \ 0 \ 0.5 \ 0 \ 0 \ 0.5 \end{pmatrix} & \frac{1}{2} \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & + \frac{1}{2} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & + \frac{1}{2} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} & + \frac{1}{2} \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0.5 \ 0.5 \ 0 \ 0.5 \ 0 \ 0 \ 0.5 \end{pmatrix} \end{matrix} \nonumber$
The amplitude of each contribution to the superposition is 1/2, so the probability of being observed on measurement is 0.25. The contributions to the superposition are rewritten to emphasize that the top two wires carry the input value of x, while the bottom wire has the value of f(x).
$\begin{matrix} \text{f(0) = 1} & \text{f(1) = 0} & \text{f(2) = 0} & \text{f(3) = 1} \ \begin{pmatrix} 0 \ 1 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 1\ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} & \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Haroche and Raimond describe this process as follows: "By superposing the inputs of a computation, one operates the machine 'in parallel', making it compute simultaneously all the values of a function and keeping its state, before any final bit detection is performed, suspended in a coherent superposition of all the possible outcomes."
This seems absolutely marvelous, four calculations in a single operation of the circuit. However, there's a catch, and that is that an inquiry (measurement) into the results of the calculation can yield only one of the results. On measurement the quantum superposition collapses, as it always does, to one value and the others are irretrievably lost. You can only learn one thing from a quantum state measurement. This is illustrated with projection operators on the top two wires to learn the value of x by measurment. The identity operator leaves the lower wire alone which now carriers the value of f(x).
$\begin{matrix} ~ & \text{f(0) = 1} \ \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T, ~ \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T, ~ \text{I} \right] \right] \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.5 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \frac{1}{2} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \ ~ & \text{f(1) = 0} \ \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T, ~ \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T, ~ \text{I} \right] \right] \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0.5 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & \frac{1}{2} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \ ~ & \text{f(2) = 0} \ \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T, ~ \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T, ~ \text{I} \right] \right] \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0.5 \ 0 \ 0 \ 0 \end{pmatrix} & \frac{1}{2} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \ ~ & \text{f(2) = 0} \ \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T, ~ \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T, ~ \text{I} \right] \right] \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0.5 \end{pmatrix} & \frac{1}{2} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{matrix} \nonumber$
As Haroche and Raimond write, "It is, however, one thing to compute potentially at once all the values of f and quite another to be able to exploit this quantum parallelism and extract from it more information than from a mundane classical computation. The final stage of information acquisition must always be a measurement."
In summary, this tutorial has demonstrated how a quantum circuit can function as an algorithm for the evaluation of a mathematical function, and how the same algorithm is capable of parallel evaluations of that function. The exploitation of quantum parallelism for practical outcomes such as factorization requires more elaborate quantum circuits than the one presented here.
Appendix
As promised earlier the matrix form of the quantum circuit is displayed here. As can be seen it is sparse!
$\text{QuantumCircuit} = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \nonumber$
An important property of circuits in quantum computation is reversibility. The reversibility of the quantum circuit is demonstrated by showing that using it twice is equivalent to the identity operator.
$\text{QuantumCircuit}^2 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \nonumber$
An alternative summary of the calculation uses the truth tables for the NOT, CNOT and CnNOT directly.
$\begin{matrix} \text{NOT} & \text{CNOT} & \text{CnNOT} \ \begin{pmatrix} 0 & ' & 1 \ 1 & ' & 0 \end{pmatrix} & \begin{pmatrix} \text{Decimal} & \text{Binary} & ' & \text{Binary} & \text{Decimal} \ 0 & 00 & ' & 00 & 0 \ 1 & 01 & ' & 01 & 1 \ 2 & 10 & ' & 11 & 3 \ 3 & 11 & ' & 10 & 2 \end{pmatrix} & \begin{pmatrix} \text{Decimal} & \text{Binary} & ' & \text{Binary} & \text{Decimal} \ 0 & 000 & ' & 000 & 0 \ 1 & 001 & ' & 001 & 1 \ 2 & 010 & ' & 010 & 2 \ 3 & 011 & ' & 011 & 3 \ 4 & 100 & ' & 101 & 5 \ 5 & 101 & ' & 100 & 4 \ 6 & 110 & ' & 111 & 7 \ 7 & 111 & ' & 110 & 6 \end{pmatrix} \end{matrix} \nonumber$
The calculations are summarized in table format as follows. The CnNOT is a three-qubit gate, but only the four input states in which the third qubit is |0> are needed. The CNOT operates on the second and third qubits, while the NOT gate operates only on the third qubit.
$\begin{matrix} |000 \rangle & \xrightarrow{CnNOT} & |0 \rangle |00 \rangle & \xrightarrow{I \otimes CNOT} & |00 \rangle |0 \rangle & \xrightarrow{I \otimes I \otimes NOT} & |00 \rangle |1 \rangle & f(0) = 1 \ |010 \rangle & ~ & |0 \rangle |10 \rangle & ~ & |01 \rangle |1 \rangle & ~ & |01 \rangle |1 \rangle & f(1) = 0 \ |100 \rangle & ~ & |1 \rangle |01 \rangle & ~ & |10 \rangle |1 \rangle & ~ & |10 \rangle |0 \rangle & f(2) = 0 \ |110 \rangle & ~ & |1 \rangle |11 \rangle & ~ & |11 \rangle |0 \rangle & ~ & |11 \rangle |1 \rangle & f(3) = 1 \end{matrix} \nonumber$
A superposition of all four input states would behave as follows under the operation of the quantum circuit.
$\frac{1}{2} \begin{pmatrix} |00 \rangle |0 \rangle \ + \ |01 \rangle |0 \rangle \ + \ |10 \rangle |0 \rangle \ + \ |11 \rangle |0 \rangle \end{pmatrix} \xrightarrow{CnNOT} \frac{1}{2} \begin{pmatrix} |0 \rangle |00 \rangle \ + \ |0 \rangle |10 \rangle \ + \ |1 \rangle |01 \rangle \ + \ |1 \rangle |11 \rangle \end{pmatrix} \xrightarrow{I \otimes CNOT} \frac{1}{2} \begin{pmatrix} |0 \rangle |00 \rangle \ + \ |0 \rangle |11 \rangle \ + \ |1 \rangle |01 \rangle \ + \ |1 \rangle |10 \rangle \end{pmatrix} \xrightarrow{I \otimes I \otimes NOT} \frac{1}{2} \begin{pmatrix} |00 \rangle |1 \rangle \ + \ |01 \rangle |0 \rangle \ + \ |10 \rangle |0 \rangle \ + \ |11 \rangle |1 \rangle \end{pmatrix} \xrightarrow{Decimal} \frac{1}{2} \begin{pmatrix} |0 \rangle |1 \rangle \ + \ |1 \rangle |0 \rangle \ + \ |2 \rangle |0 \rangle \ + \ |3 \rangle |1 \rangle \end{pmatrix} \nonumber$
Another feature of this circuit is that if the bottom wire is measured, it projects the top two wires into a superposition of inputs that give the bottom wire result.
$\text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 0 \ 1 \ 1 \ 0 \ 1 \ 0 \ 0 \ 1 \end{pmatrix} = \frac{1}{2} \left[ \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} \right] \begin{pmatrix} 1 \ 0 \end{pmatrix} + \frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \right] \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
$\begin{matrix} ~ & \text{f(1) = f(2) = 0} \ \text{kronecker} \left[ \text{I, kronecker} \left[ \text{I}, ~ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \right] \right] \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0.5 \ 0 \ 0.5 \ 0 \ 0 \ 0 \end{pmatrix} & \frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \right] \begin{pmatrix} 1 \ 0 \end{pmatrix} \ ~ & \text{f(0) = f(3) = 1} \ \text{kronecker} \left[ \text{I, kronecker} \left[ \text{I}, ~ \begin{pmatrix} 0 & 0 \ 0 & 1 \end{pmatrix} \right] \right] \text{QuantumCircuit} \frac{1}{2} \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0.5 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0.5 \end{pmatrix} & \frac{1}{2} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Here's an alternative analysis of the circuit operation.
$\begin{matrix} \frac{1}{2} \left[ |00 \rangle + |01 \rangle + |10 \rangle + |11 \rangle \right] |0 \rangle = \frac{1}{2} \left[ |000 \rangle + |010 \rangle + |100 \rangle + |110 \rangle \right] \ \text{CnNOT} \ \frac{1}{2} \left[ |000 \rangle + |010 \rangle + |101 \rangle + |111 \rangle \right] = \frac{1}{2} \left[ |0 \rangle |00 \rangle + |0 \rangle |10 \rangle + |1 \rangle |01 \rangle + |1 \rangle |11 \rangle \right] \ \text{I} \otimes \text{CNOT} \ \frac{1}{2} \left[ |0 \rangle |00 \rangle + |0 \rangle |11 \rangle + |1 \rangle |01 \rangle + |1 \rangle |10 \rangle \right] = \frac{1}{2} \left[ |00 \rangle |0 \rangle + |01 \rangle |1 \rangle + |10 \rangle |1 \rangle + |11 \rangle |0 \rangle \right] \ \text{I} \otimes \text{I} \otimes \text{NOT} \ \frac{1}{2} \left[ |00 \rangle |1 \rangle + |01 \rangle |0 \rangle + |10 \rangle |0 \rangle + |11 \rangle |1 \rangle \right] \xrightarrow{ \text{Decimal}} \frac{1}{2} \left[ |0 \rangle |1 \rangle + |1 \rangle |0 \rangle + |2 \rangle |0 \rangle + |3 \rangle |1 \rangle \right] \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.70%3A_Evaluating_a_Function_Using_a_Quantum_Circuit_and_a_Demonstration_of_Parallel_Computation.txt
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This tutorial deals with quantum function evaluation and parallel computation. The example is taken from pages 94-95 of Exploring the Quantum by Haroche and Raimond. A certain function of x yields the following table of results.
$\begin{pmatrix} \text{x} & 0 & 1 & 2 & 3 \ \text{f(x)} & 1 & 0 & 0 & 1 \end{pmatrix} \nonumber$
In the quantum circuit provided by Haroche and Raimond the top two wires carry the values of x in binary code (2 bits, 4 values) and the third wire is initially set to |0 >. After operation of the circuit on the initial state, the value of the function appears on the third wire. The qubits on the top wires are unchanged by the circuit. The circuit consists of a CnNOT gate employing all three wires, followed by a CNOT gate involving the second and third wires, and finally a NOT gate operating on the third wire.
$\begin{matrix} \begin{matrix} |a \rangle & \triangleright & \cdot & \cdots & \cdots & \cdots & \cdots & \triangleright & |a \rangle \ ~ & ~ & | \ |b \rangle & \triangleright & | & \cdots & \cdot & \cdots & \cdots & \triangleright & |b \rangle \ ~ & ~ & | & ~ & | \ |0 \rangle & \triangleright & \oplus & \cdots & \oplus & \cdots & \fbox{NOT} & \triangleright & | f(x) \rangle \end{matrix} & \text{where} & |0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} & |1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
First we establish that this circuit yields the results shown in the table, and then demonstrate that it also carries out a parallel calculation in one step using all four input values of x. The truth tables for the various quantum gates are provided in the Appendix.
$\begin{matrix} |000 \rangle & \xrightarrow{CnNOT} & |0 \rangle |00 \rangle & \xrightarrow{I \otimes CNOT} & |00 \rangle |0 \rangle & \xrightarrow{I \otimes I \otimes NOT} & |00 \rangle |1 \rangle & f(0) = 1 \ |010 \rangle & ~ & |0 \rangle |10 \rangle & ~ & |01 \rangle |1 \rangle & ~ & |01 \rangle |0 \rangle & f(1) = 0 \ |100 \rangle & ~ & |1 \rangle |01 \rangle & ~ & |10 \rangle |1 \rangle & ~ & |10 \rangle |0 \rangle & f(2) = 0 \ |110 \rangle & ~ & |1 \rangle |11 \rangle & ~ & |11 \rangle |0 \rangle & ~ & |11 \rangle |1 \rangle & f(3) = 1 \end{matrix} \nonumber$
A superposition of all four input states behaves as follows under the operation of the quantum circuit.
$\begin{matrix} \frac{1}{2} \begin{pmatrix} |00 \rangle |0 \rangle \ + \ |01 \rangle |0 \rangle \ + \ |10 \rangle |0 \rangle \ + \ |11 \rangle |0 \rangle \end{pmatrix} \xrightarrow{CnNOT} \frac{1}{2} \begin{pmatrix} |0 \rangle |00 \rangle \ + \ |0 \rangle |10 \rangle \ + \ |1 \rangle |01 \rangle \ + \ |1 \rangle |11 \rangle \end{pmatrix} \xrightarrow{I \otimes CNOT} \frac{1}{2} \begin{pmatrix} |00 \rangle |0 \rangle \ + \ |01 \rangle |1 \rangle \ + \ |10 \rangle |1 \rangle \ + \ |11 \rangle |0 \rangle \end{pmatrix} \xrightarrow{I \otimes I \otimes NOT} \frac{1}{2} \begin{pmatrix} |00 \rangle |1 \rangle \ + \ |01 \rangle |0 \rangle \ + \ |10 \rangle |0 \rangle \ + \ |11 \rangle |1 \rangle \end{pmatrix} \xrightarrow{Decimal} \frac{1}{2} \begin{pmatrix} |0 \rangle |1 \rangle \ + \ |1 \rangle |0 \rangle \ + \ |2 \rangle |0 \rangle \ + \ |3 \rangle |1 \rangle \end{pmatrix} \end{matrix} \nonumber$
Haroche and Raimond describe this process as follows: "By superposing the inputs of a computation, one operates the machine 'in parallel', making it compute simultaneously all the values of a function and keeping its state, before any final bit detection is performed, suspended in a coherent superposition of all the possible outcomes."
It appears that we get four calculations in a single operation of the circuit. However, there's a catch, and that is that any inquiry into the results of the calculation can yield only one of the results. On measurement the quantum superposition collapses, as it always does, to one value and the others are irretrievably lost. As Haroche and Raimond write, "It is, however, one thing to compute potentially at once all the values of f and quite another to be able to exploit this quantum parallelism and extract from it more information than from a mundane classical computation. The final stage of information acquisition must always be a measurement."
In summary, this tutorial has demonstrated how a quantum circuit can function as an algorithm for the evaluation of a mathematical function, and how the same algorithm is capable of parallel evaluations of that function. The exploitation of quantum parallelism for practical outcomes such as factorization requires more elaborate quantum circuits than the one presented here.
Appendix
$\begin{matrix} \text{NOT} & \text{CNOT} & \text{CnNOT} \ \begin{pmatrix} 0 & ' & 1 \ 1 & ' & 0 \end{pmatrix} & \begin{pmatrix} \text{Decimal} & \text{Binary} & ' & \text{Binary} & \text{Decimal} \ 0 & 00 & ' & 00 & 0 \ 1 & 01 & ' & 01 & 1 \ 2 & 10 & ' & 11 & 3 \ 3 & 11 & ' & 10 & 2 \end{pmatrix} & \begin{pmatrix} \text{Decimal} & \text{Binary} & ' & \text{Binary} & \text{Decimal} \ 0 & 000 & ' & 000 & 0 \ 1 & 001 & ' & 001 & 1 \ 2 & 010 & ' & 010 & 2 \ 3 & 011 & ' & 011 & 3 \ 4 & 100 & ' & 101 & 5 \ 5 & 101 & ' & 100 & 4 \ 6 & 110 & ' & 111 & 7 \ 7 & 111 & ' & 110 & 6 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.71%3A_A_Simple_Quantum_Circuit_for_Parallel_Computation.txt
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The following circuit produces the table of results to its right. The top wires carry the value of x and the circuit places f(x) on the bottom wire. As is shown in the previous tutorial this circuit can also operate in parallel accepting as input all x-values and returning on the bottom wire a superposition of all values of f(x).
$\begin{matrix} \begin{pmatrix} \text{x} & 0 & 1 & 2 & 3 \ \text{f(x)} & 1 & 0 & 0 & 1 \end{pmatrix} & \text{where} & \begin{matrix} |0 \rangle = |0 \rangle |0 \rangle \ |1 \rangle = |0 \rangle |1 \rangle \ |2 \rangle = |1 \rangle |0 \rangle \ |3 \rangle = |1 \rangle |1 \rangle \end{matrix} \end{matrix} \nonumber$
The function belongs to the balanced category because it produces 0 and 1 with equal frequency. A modification of this circuit (Deutsch-Jozsa algoritm, p. 298 in The Quest for the Quantum Computer, by Julian Brown) answers the question of whether the function is constant or balanced. Naturally we already know the answer, so this is a simple demonstration that the circuit works.
The input is |0>|0>|1> followed by a Hadamard gate on each wire, as shown in the circuit on page 3. As is well known the Hadamard operation creates the following superposition states.
$\begin{matrix} \text{H} \begin{pmatrix} 1 \ 0 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & \text{H} \begin{pmatrix} 0 \ 1 \end{pmatrix} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} \end{matrix} \nonumber$
Therefore the Hadamard operation transforms the input state to the following three-qubit state which is fed to the quantum circuit.
$\frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -1 \end{pmatrix} = \frac{1}{ 2 \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} \nonumber$
The following matrices are required to execute the circuit.
$\begin{matrix} \text{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & \text{NOT} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{H} = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & \text{CnNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
After the portion of the quantum circuit shown above, Hadamard gates are added to the top two wires, as shown in the circuit on page 3. The matrix representing the circuit is assembled using tensor matrix multiplication and then allowed to operate on the wave function. The full circuit is shown below.
$\text{QuantumCircuit} = \text{kronecker(H, kronecker(H, I)) kronecker(I, kronecker(I, NOT)) kronecker(I, CNOT) CnNOT} \nonumber$
$\begin{matrix} \text{QuantumCircuit} = ~ \frac{1}{2} \begin{pmatrix} 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \ 0 & 1 & -1 & 0 & 1 & 0 & 0 & -1 \ 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \ 0 & 1 & 1 & 0 & -1 & 0 & 0 & -1 \ 1 & 0 & 0 & 1 & 0 & -1 & -1 & 0 \ 0 & 1 & -1 & 0 & -1 & 0 & 0 & 1 \ 1 & 0 & 0 & -1 & 0 & -1 & 1 & 0 \end{pmatrix} & \text{QuantumCircuit} \frac{1}{2 \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.707 \ 0.707 \end{pmatrix} & \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix} \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Next the qubits on the top two wires are measured. If both are |0> the function is constant, but if at least one is |1> the function is balanced. The measurement on the top wires is implemented with projection operators |0><1|, and confirms that the function is not constant but belongs to the balanced category.
$\begin{matrix} \text{The first qubit is not |0} \rangle. & \text{kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T \text{kronecker(I, I)} \right] \text{QuantumCircuit} \frac{1}{2 \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \ \text{The second qubit is not |0} \rangle. & \text{kronecker} \left[ \text{I, kronecker} \left[ \begin{pmatrix} 1 \ 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}^T,~ \text{I} \right] \right] \text{QuantumCircuit} \frac{1}{2 \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \ \text{The first qubit is |1} \rangle. & \text{kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T \text{kronecker(I, I)} \right] \text{QuantumCircuit} \frac{1}{2 \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.707 \ 0.707 \end{pmatrix} \ \text{The second qubit is |1} \rangle. & \text{kronecker} \left[ \text{I, kronecker} \left[ \begin{pmatrix} 0 \ 1 \end{pmatrix} \begin{pmatrix} 0 \ 1 \end{pmatrix}^T, ~ \text{I} \right] \right] \text{QuantumCircuit} \frac{1}{2 \sqrt{2}} \begin{pmatrix} 1 \ -1 \ 1 \ -1 \ 1 \ -1 \ 1 \ -1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ -0.707 \ 0.707 \end{pmatrix} \end{matrix} \nonumber$
The following illustarted an algebraic analysis of the Deutsch-Jozsa algorithm.
$\begin{matrix} \text{Initial} & ~ & 1 & ~ & 2 & ~ & 3 & ~ & 4 & ~ & 5 & ~ & \text{Final} \ |0 \rangle & \triangleright & \fbox{H} & \cdots & \cdot & \cdots & \cdots & \cdots & \cdots & \cdots & \fbox{H} & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & ~ & | \ |0 \rangle & \triangleright & \fbox{H} & \cdots & | & \cdots & \cdot & \cdots & \cdots & \cdots & \fbox{H} & \triangleright & \text{Measure, 0 or 1} \ ~ & ~ & ~ & ~ & | \ |1 \rangle & \triangleright & \fbox{H} & \cdots & \oplus & \cdots & \oplus & \cdots & \fbox{NOT} & \cdots & \cdots \end{matrix} \nonumber$
$\begin{matrix} \text{H} |0 \rangle \rightarrow \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] & \text{H} |1 \rangle \rightarrow \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] \end{matrix} \nonumber$
$\begin{matrix} \text{NOT} & \text{CNOT} & \text{CnNOT} \ \begin{pmatrix} 0 & ' & 1 \ 1 & ' & 0 \end{pmatrix} & \begin{pmatrix} \text{Decimal} & \text{Binary} & ' & \text{Binary} & \text{Decimal} \ 0 & 00 & ' & 00 & 0 \ 1 & 01 & ' & 01 & 1 \ 2 & 10 & ' & 11 & 3 \ 3 & 11 & ' & 10 & 2 \end{pmatrix} & \begin{pmatrix} \text{Decimal} & \text{Binary} & ' & \text{Binary} & \text{Decimal} \ 0 & 000 & ' & 000 & 0 \ 1 & 001 & ' & 001 & 1 \ 2 & 010 & ' & 010 & 2 \ 3 & 011 & ' & 011 & 3 \ 4 & 100 & ' & 101 & 5 \ 5 & 101 & ' & 100 & 4 \ 6 & 110 & ' & 111 & 7 \ 7 & 111 & ' & 110 & 6 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} |000 \rangle \ \text{H} \otimes \text{H} \otimes \text{H} \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle + |1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] = \frac{1}{ 2 \sqrt{2}} \left[ |000 \rangle - |001 \rangle + |010 \rangle - |011 \rangle + |100 \rangle - |101 \rangle + |110 \rangle - |111 \rangle \right] \ \text{CnNOT} \ \frac{1}{2 \sqrt{2}} \left[ |000 \rangle - |001 \rangle + |010 \rangle - |011 \rangle + |101 \rangle - |100 \rangle + |111 \rangle - |110 \rangle \right] \ \text{I} \otimes \text{CNOT} \ \frac{1}{2 \sqrt{2}} \left[ |000 \rangle - |001 \rangle + |011 \rangle - |010 \rangle + |101 \rangle - |100 \rangle + |110 \rangle - |111 \rangle \right] \ \text{I} \otimes \text{I} \otimes \text{NOT} \ \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |0 \rangle - |1 \rangle \right] \frac{1}{ \sqrt{2}} \left[ |1 \rangle - |0 \rangle \right] \ \text{H} \otimes \text{H} \otimes \text{H} \otimes \text{I} \ |1 \rangle |1 \rangle \frac{1}{ \sqrt{2}} \left( |1 \rangle - |0 \rangle \right) \end{matrix} \nonumber$
Since the top wires contain |1 > the function is balanced. This algorithm illustrates the roles of superposition, entanglement and interference in quantum computation. Regarding the latter, it is destructive interference in the last step that eliminates unwanted outcomes yielding the final result on the last line. One pass through the quantum circuit answers the question (is the function balanced or constrant) that would take four calculations on a classical computer.
The interference that occurs in the last step is illustrated by letting |a> = |0> and |b> = |1> and carrying out Hadamard transforms on the first two qubits.
$\frac{1}{4 \sqrt{2}} \begin{bmatrix} (a_1 +b_1) (a_2 +b_2)b_3 - (a_1 + b_1)(a_2 +b_2)a_3 \cdots \ +(a_1 + b_1)(a_2 - b_2) b_3 - (a_1 + b_1)(a_2 - b_2) a_3 \cdots \ +(a_1 -b_1)(a_2 +b_2)b_3 - (a_1 - b_1)(a_2 + b_2) a_3 \cdots \ + (a_1 - b_1)(a_2 - b_2) b_3 - (a_1 - b_1)(a_2 -b_2)a_3 \end{bmatrix} ~ \text{simplify} ~ \rightarrow - \frac{ \sqrt{2} a_1 a_2 (a_3 - b_3)}{2} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.72%3A_An_Illustration_of_the_Deutsch-Jozsa_Algorithm.txt
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