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This is an outline of the Mathcad implementation of a two‐electron SCF calculation published in JCE by Snow and Bills. [Snow, R. L.; Bills, J. L. J. Chem. Educ. 1975, 52, 506.]
Under the orbital approximation, Ψ(1,2) = Φ(1)Φ(2), the two‐electron Schrödinger equation for helium can be decoupled into two, one‐electron Hartree differential equations.The individual helium atom electrons are assumed to occupy an orbital which is a linear combination of two Slater 1s orbitals,
$\Phi = C_1 f_1 + C_2 f_2 \nonumber$
$\begin{matrix} f_1 = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha r) & \text{and} & f_2 = \sqrt{ \frac{ \beta^3}{ \pi}} \text{exp} ( - \beta r) \end{matrix} \nonumber$
The effective Hamiltonian for the ith electron is,
$H_i = - \frac{1}{2r^2} \frac{d}{dr} \left( r^2 \frac{d}{dr} \blacksquare \right) - \frac{2}{r_i} + \int_0^{ \infty} \left( C_{j1} f_1 + C_{j2} f_2 \right) \frac{1}{r_{ij} } \left( C_{j1} f_1 + C_{j2} f_2 \right) d \tau \nonumber$
where the Cjs are the coefficients of the jth electron.
Assuming that the ith electron is also in an orbital of the form given in equation (1), the variational method yields the following expression for the energy of the ith electron.
$\varepsilon_i = \frac{ \int_0^{ \infty} \left( C_{j1} f_1 + C_{j2} f_2 \right) H_i \left( C_{j1} f_1 + C_{j2} f_2 \right) d \tau}{ \int_0^{ \infty} \left( C_{j1} f_1 + C_{j2} f_2 \right)^2 d \tau} = \frac{Ci_1^2 + H_{11} 2 Ci_i Ci_2 H_{12} + Ci_2^2 H_{22}}{Ci_1^2 + 2Ci_1 Ci_2 S_{12} + Ci_2^2} \nonumber$
The optimum values for the orbital scale factors are given below. The user can change these values to demonstrate that they do indeed yield a minimum energy for the trial wavefunction chosen.
$\begin{matrix} \alpha = 1.45 & \beta = 2.90 \end{matrix} \nonumber$
Evaluation of integrals given values for the scale factors, α and β, of the Slater type orbitals follows (See Snow and Bills for details):
$\begin{matrix} \tau = \frac{ \alpha - \beta}{ \alpha + \beta} & \tau 1 = \frac{ \alpha - \beta}{3 \alpha + \beta} & \tau 2 = \frac{ \alpha - \beta}{ \alpha + 3 \beta} \end{matrix} \nonumber$
Kinetic energy integrals:
$\begin{matrix} T_{11} = \frac{ \alpha^2}{2} & T_{22} = \frac{ \beta^2}{2} & T_{12} = \frac{1}{8} \left( \alpha + \beta \right)^2 \left( 1 - \tau^2 \right)^{2.5} \end{matrix} \nonumber$
Electron‐nucleus potential energy integrals:
$\begin{matrix} V_{11} = -2 \alpha & V_{22} = -2 \beta & V_{12} = - ( \alpha + \beta ) \left( 1 - \tau \right)^{1.5} \end{matrix} \nonumber$
Electron‐electron repulsion integrals:
$\begin{matrix} V_{1111} = \frac{5}{8} \alpha & V_{2222} = \frac{5}{8} \alpha & V_{1212} = \frac{5}{16} \left( 1 - \tau^2 \right)^3 ( \alpha + \beta ) \end{matrix} \nonumber$
$\begin{matrix} V_{1122} = \frac{1}{16} \left( 1 - \tau^2 \right) \left( 5 - \tau^2 \right) ( \alpha + \beta ) & V_{1222} = \frac{1}{32} \left( 1 - \tau^2 \right)^{1.5} \left( 1 - tau2^2 \right) \left( 5 - \tau 2^2 \right) ( \alpha + 3 \beta ) \ V_{1112} = \frac{1}{32} \left( 1 - \tau^2 \right)^{1.5} \left( 1 - \tau 1^2 \right) \left( 5 - \tau 1^2 \right) ( 3 \alpha + \beta ) & \text{Overlap integral: } S_{12} = \left( 1 - \tau^2\ \right)^{1.5} \end{matrix} \nonumber$
Having evaluated the integrals, the next step is the calculation of the matrix elements that appear in equation (4).
$\begin{matrix} H_{11} = T_{11} + V_{11} + Cj_1^2 V_{1111} + 2Cj_1 Cj_2 V_{1112} + Cj_2^2 V_{1122} \ H_{12} = T_{12} + V_{12} + Cj_1^2 V_{1112} + 2Cj_1 Cj_2 V_{1212} + Cj_2^2 V_{1222} \ H_{22} = T_{22} + V_{22} + Cj_1^2 V_{1122} + 2Cj_1 Cj_2 V_{1222} + Cj_2^2 V_{2222} \end{matrix} \nonumber$
Given initial values for the coefficients of the jth electron (these are declared below in a more convient location using Mathcadʹs global equal sign), minimization of the orbital energy simultaneously with respect to the coefficients of the ith electron yields the orbital energy of the ith electron and its coefficients. These output coefficients become the input coefficients of the next iteration when the orbital energy of the j th electron is calculated. The procedure is repeated until self‐consistency is achieved. This occurs when the output coefficients are equal to the input coefficients, or to put it another way, when the coefficients of the two electrons are identical.
In the numeric mode Mathcad requires seed values for all variable which appear in the expression to be evaluated in a Given/Find solve block. The seed values for the coefficients shown below are arbitrarily set at 0.5.
$\begin{matrix} Ci_1 = .5 & Ci_2 = .5 \end{matrix} \nonumber$
The variational integral for the electron orbital energy:
$\varepsilon \left( Ci_1,~ Ci_2 \right) = \frac{Ci_1^2 H_{11} + 2Ci_1 H_{12} + Ci_2^2 H_{22}}{Ci_1^2 + 2Ci_1 Ci_2 S_{12} + Ci_2^2} \nonumber$
Minimization of the variational integral simultaneously with respect to the coefficients plus the normalization condition
$\begin{matrix} \text{Given} & \frac{d}{dCi_1} \varepsilon \left( Ci_1,~Ci_2 \right) = 0 & \frac{d}{dCi_2} \varepsilon \left( Ci_1,~Ci_2 \right) = 0 & Ci_1^2 + 2 Ci_1 Ci_2 S_{12} + Ci_2^2 = 1 \end{matrix} \nonumber$
yields the output coefficients and the orbital energy:
$\begin{matrix} \begin{pmatrix} Ci_1 \ Ci_2 \end{pmatrix} = \text{Find} \left( Ci_1,~Ci_2 \right) & \begin{pmatrix} Ci_1 \ Ci_2 \end{pmatrix} = \begin{pmatrix} 0.809249 \ 0.219060 \end{pmatrix} \end{matrix} \nonumber$
In the SCF method the output coefficients become the input coefficients in the next iteration.
Display the orbital energy:
$\varepsilon \left( Ci_1,~Ci_2 \right) = -0.984326 \nonumber$
The energy of the atom at this point in the calculation is the orbital energy of the ith electron plus the kinetic and nuclear potential energy of the jth electron.
$\begin{matrix} E_{atom} = \varepsilon \left( Ci_1,~Ci_2 \right) + Cj_1^2 \left( T_{11} + V_{11} \right) + 2C_1 Cj_2 \left( T_{12} + V_{12} \right) + Cj_2^2 \left( T_{22} + V_{22} \right) & E_{atom} = -2.833076 \end{matrix} \nonumber$
Input coefficients:
$\begin{pmatrix} Cj_1 \ Cj_2 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
Paste new input coefficients after each iteration.
Summary of the SCF results for the following initial input coefficients:
$\begin{pmatrix} Cj_1 \ Cj_2 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$\begin{pmatrix} \text{Iteration} & Ci_1 & Ci_2 & \varepsilon \left( Ci_1,~Ci_2 \right) & E_{atom} \ 1 & 0.809249 & 0.21906 & -0.984326 & -2.833076 \ 2 & 0.847034 & 0.176951 & -0.90556 & -2.860616 \ 3 & 0.839638 & 0.185241 & -0.920653 & -2.86163 \ 4 & 0.841091 & 0.183615 & -0.917676 & -2.86167 \ 5 & 0.840806 & 0.183934 & -0.918258 & -2.861672 \ 6 & 0.840862 & 0.183871 & -0.918144 & -2.861672 \ 7 & 0.840851 & 0.183884 & -0.918167 & -2.861672 \ 8 & 0.840852 & 0.183881 & -0.918161 & -2.861673 \ 9 & 0.840852 & 0.183882 & -0.918164 & -2.861672 \ 10 & 0.840852 & 0.183882 & -0.918164 & -2.861672 \end{pmatrix} \nonumber$
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Coordinate Space Operators
Position operator: $x \cdot \Box$ Momentum operator: $\mathrm{p}=\frac{1}{\mathrm{i}} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \Box$ Integral: $\int_{0}^{\infty} \Box d x$
Kinetic energy operator: $\mathrm{KE}=-\frac{1}{2} \cdot \frac{\mathrm{d}^{2}}{\mathrm{d} \mathrm{x}^{2}} \Box$ Potential energy operator: $\mathrm{PE}=\frac{-1}{\mathrm{x}} \Box$
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \cdot \frac{\mathrm{d}^{2}}{\mathrm{d} \mathrm{x}^{2}} \Box-\frac{1}{\mathrm{x}} \cdot \Box \nonumber$
The ground state wave function in coordinate space is:
$\Psi(x) :=2 \cdot x \cdot \exp (-x) \nonumber$
Display the coordinate‐space distribution function:
The ground state energy is ‐0.5 Eh.
$\frac{-1}{2} \cdot \frac{\mathrm{d}^{2}}{\mathrm{d} \mathrm{x}^{2}} \Psi(\mathrm{x})-\frac{1}{\mathrm{x}} \cdot \Psi(\mathrm{x})=\mathrm{E} \cdot \Psi(\mathrm{x}) \text { solve, } \mathrm{E} \rightarrow \frac{-1}{2} \nonumber$
The coordinate wave function is normalized:
$\int_{0}^{\infty} \Psi(x)^{2} d x=1 \nonumber$
The expectation value for position:
$\int_{0}^{\infty} x \cdot \Psi(x)^{2} d x=1.5 \nonumber$
The expectation value for momentum:
$\int_{0}^{\infty} \Psi(x) \cdot \frac{1}{i} \cdot \frac{d}{d x} \Psi(x) d x=0 \nonumber$
The expectation value for kinetic energy:
$\int_{0}^{\infty} \Psi(x) \cdot \frac{-1}{2} \cdot \frac{d^{2}}{d x^{2}} \Psi(x) d x=0.5 \nonumber$
The expectation value for potential energy:
$\int_{0}^{\infty} \frac{-1}{x} \cdot \Psi(x)^{2} d x=-1 \nonumber$
The momentum wave function is generated by the following Fourier transform of the coordinate space wave function.
$\Phi(p) :=\frac{1}{\sqrt{2 \cdot \pi}} \int_{0}^{\infty} \exp (-i \cdot p \cdot x) \cdot \Psi(x) d x \rightarrow \frac{2^{\frac{1}{2}}}{\pi^{\frac{1}{2}} \cdot(i \cdot p+1)^{2}} \nonumber$
Momentum Space Operators
Momentum space integral: $\int_{-\infty}^{\infty} \Box \mathrm{dp}$ Momentum operator: $p \cdot \Box$
Kinetic energy operator: $\frac{\mathrm{p}^{2}}{2}$ Position operator: $i \cdot \frac{\mathrm{d}}{\mathrm{dp}} \Box$
The same calculations made with the momentum space wave function:
The momentum wave function is normalized:
$\int_{-\infty}^{\infty}(|\Phi(p)|)^{2} d p=1 \nonumber$
The expectation value for position:
$\int_{-\infty}^{\infty} \overline{\Phi(p)} \cdot i \cdot \frac{d}{d p} \Phi(p) d p=1.5 \nonumber$
The expectation value for momentum:
$\int_{-\infty}^{\infty} p \cdot(|\Phi(p)|)^{2} d p=0 \nonumber$
The expectation value for kinetic energy:
$\int_{-\infty}^{\infty} \frac{\mathrm{p}^{2}}{2} \cdot(|\Phi(\mathrm{p})|)^{2} \mathrm{dp}=0.5 \nonumber$
The Wigner function for the hydrogen atom ground state is generated using the momentum wave function.
$\mathrm{W}(\mathrm{x}, \mathrm{p}) :=\frac{1}{2 \cdot \pi} \cdot \int_{-\infty}^{\infty} \overline{\Phi\left(\mathrm{p}+\frac{\mathrm{s}}{2}\right)} \cdot \exp (-\mathrm{i} \cdot \mathrm{s} \cdot \mathrm{x}) \cdot \Phi\left(\mathrm{p}-\frac{\mathrm{s}}{2}\right) \mathrm{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\mathrm{N} :=60 \qquad \mathrm{i} :=0 . . \mathrm{N} \qquad \mathrm{x}_{\mathrm{i}} :=\frac{6 \cdot \mathrm{i}}{\mathrm{N}} \ \mathrm{j} :=0 . . \mathrm{N} \qquad \mathrm{p}_{\mathrm{j}} :=-5+\frac{10 \cdot \mathrm{j}}{\mathrm{N}} \qquad \text{Wigner}_{\text {i ,j }} :=\mathrm{W}\left(\mathrm{x}_{i}, \mathrm{p}_{\mathrm{j}}\right) \nonumber$
One of the interesting features of doing quantum mechanics with the Wigner distribution is that the position and momentum operators retain their classical forms; they are both multiplicative operators. By comparison in the coordinate representation position is multiplicative and momentum is differential. In the momentum representation itʹs the reverse. This is illustrated below with the following calculations.
Phase space integral: $\int_{-\infty}^{\infty} \int_{0}^{\infty} \Box d x d p$ Position operator: $x \cdot \Box$ Momentum operator: $p \cdot \Box$
Kinetic energy operator: $\mathrm{KE}=\frac{\mathrm{p}^{2}}{2} \cdot \Box$ Potential energy operator: $\mathrm{PE}=\frac{-1}{\mathrm{x}} \cdot \Box$
Phase space calculations using the Wigner distribution:
The Wigner distribution is normalized:
$\int_{-\infty}^{\infty} \int_{0}^{\infty} W(x, p) d x d p=1 \nonumber$
The expectation value for position:
$\int_{-\infty}^{\infty} \int_{0}^{\infty} x \cdot W(x, p) d x d p=1.5 \nonumber$
The expectation value for momentum:
$\int_{-\infty}^{\infty} \int_{0}^{\infty} p \cdot W(x, p) d x d p=0 \nonumber$
The expectation value for kinetic energy:
$\int_{-\infty}^{\infty} \int_{0}^{\infty} \frac{p^{2}}{2} \cdot W(x, p) dx dp=0.5 \nonumber$
The expectation value for potential energy:
$\int_{-\infty}^{\infty} \int_{0}^{\infty} \frac{-1}{x} \cdot W(x, p) d x d p=-1 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/02%3A_Atomic_Structure/2.54%3A_Quantum_Calculations_on_the_Hydrogen_Atom_in_Coordinate_Momentum_and_Phase_Space.txt
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This tutorial presents three pictures of the 1s state of the one‐dimensional hydrogen atom using its position, momentum and phase‐space representations.
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$
The ground state eigenstate is:
$\begin{matrix} \Psi (x) = 2 x \text{exp} (-x) & \int_0^{ \infty} \Psi (x)^2 dx = 1 \end{matrix} \nonumber$
The ground state energy is -0.5 Eh.
$\frac{-1}{2} \frac{d^2}{x^2} \Psi (x) - \frac{1}{x} \Psi (x) = E \Psi (x) \text{ solve, E} \rightarrow \frac{-1}{2} \nonumber$
The momentum wave function is generated by the following Fourier transform of the coordinate space wave function.
$\Phi (p) = \frac{1}{ \sqrt{2 \pi}} \int_0^{ \infty} \text{exp(-i p x)} \Psi (x) dx \rightarrow \frac{2^{ \frac{1}{2}}}{ \text{(i p + 1)} \pi^{ \frac{1}{2}}} \nonumber$
The Wigner function for the hydrogen atom ground state is generated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \overline{ \Phi \left( \text{p} + \frac{ \text{s}}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\begin{matrix} N = 60 & i = 0 .. N & x_i = \frac{6i}{N} & j = 0 .. N & p_j = -5 + \frac{10j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~ p_j \right) \end{matrix} \nonumber$
2.56: The Wigner Distribution for the 2s State of the 1D Hydrogen Atom
This tutorial presents three pictures of the 2s state of the one‐dimensional hydrogen atom using its position, momentum and phase‐space representations.
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$
The 2s wave function is:
$\begin{matrix} \Psi (x) = \frac{1}{ \sqrt{8}} \text{x} (2 - \text{x}) \text{exp} \left( - \frac{x}{2} \right) & \int_0^{ \infty} \Psi (x)^2 dx = 1 \end{matrix} \nonumber$
The 2s state energy is -0.125 Eh.
$\frac{ \frac{-1}{2} \frac{d^2}{dx^2} \Psi (x) - \frac{1}{x} \Psi (x)}{ \Psi(x)} \text{simplify} \rightarrow \frac{-1}{8} \nonumber$
The momentum wave function is generated by the following Fourier transform of the coordinate space wave function.
$\Phi (p) = \frac{1}{ \sqrt{2 \pi}} \int_0^{ \infty} \text{exp(-i p x)} \Psi ( \text{x}) \text{dx} \rightarrow \frac{2}{ \pi^{ \frac{1}{2}}} \frac{2 \text{i p} - 1}{ \pi^{ \frac{1}{2}} \text{(2 i p + 1)^3}} \nonumber$
The Wigner function (phase‐space representation) for the 2s state is generated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int_{ - \infty}^{ \infty} \overline{ \Phi \left( \text{p} + \frac{ \text{s}}{2} \right) } \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\begin{matrix} N = 100 & i = i .. N & x_i = \frac{15i}{N} & j = 0 .. N & p_j = -3 \frac{6j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~ p_j \right) \end{matrix} \nonumber$
If we rotate this figure to look below the plane, we see that for the 2s state of the 1D hydrogen atom the Wigner distribution takes on negative values.
2.57: The Wigner Distribution for the 2p State of the 1D Hydrogen Atom
This tutorial presents three pictures of the 2p state of the one‐dimensional hydrogen atom using its position, momentum and phase‐space representations.
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare + \frac{ \text{L(L+1)}}{2x^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$
The 2p wave function is:
$\begin{matrix} \Psi (x) = \frac{1}{ \sqrt{24}} x^2 \text{exp} \left( - \frac{x}{2} \right) & \int_0^{ \infty} \Psi (x)^2 dx = 1 \end{matrix} \nonumber$
The 2p state energy is -0.125 Eh.
$\frac{-1}{2} \frac{d^2}{dx^2} \Psi (x) + \frac{1}{x^2} \Psi (x) - \frac{1}{x} \Psi (x) = \text{E} \Psi (x) \text{solve, E} \rightarrow \frac{-1}{8} = -0.125 \nonumber$
The momentum wave function is generated by the following Fourier transform of the coordinate space wave function.
The Wigner function (phase‐space representation) for the 2p state is generated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int_{ - \infty}^{ \infty} \overline{ \left( \text{p} + \frac{ \text{s}}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\begin{matrix} N = 100 & i = 0 .. N & x_i = \frac{15i}{N} & j = 0 .. N & p_j = -3 + \frac{6j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~p_j \right) \end{matrix} \nonumber$
2.58: The Wigner Distribution for the 3s State of the 1D Hydrogen Atom
This tutorial presents three pictures of the 3s state of the one‐dimensional hydrogen atom using its position, momentum and phase‐space representations.
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$
The position 3s wave function is:
$\begin{matrix} \Psi (x) = \frac{2}{243} \sqrt{3} \text{x} \left( 27-18x + 2x^2 \right) \text{exp} \left( - \frac{x}{3} \right) & \int_0^{ \infty} \Psi (x)^2 dx = 1 \end{matrix} \nonumber$
The 3s energy is -0.065 Eh.
$\Phi (p) = \frac{1}{ \sqrt{2 \pi}} \int_0^{ \infty} \text{exp(-i p x)} \Psi (x) \text{dx} \rightarrow \left( -2^{ \frac{1}{2}} \right) 3^{ \frac{1}{2}} \frac{9p^2 + 6ip - 1}{(3ip + 1)^4 \pi^{ \frac{1}{2}}} \nonumber$
The Wigner function (phase‐space representation) for the hydrogen atom 3s state is generated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \overline{ \Phi \left( \text{p} + \frac{}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\begin{matrix} N = 100 & i = 0 .. N & x_i = \frac{30i}{N} & j = 0 .. N & p_j = -2 + \frac{4j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~ p_j \right) \end{matrix} \nonumber$
Just as for the 2s state, the Wigner distribution for the 3s state takes on negative values.
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This tutorial presents three pictures of the 3p state of the one‐dimensional hydrogen atom using its position, momentum and phase‐space representations.
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare + \frac{ \text{L(L+1)}}{2x^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$
The 3p wave function is:
$\begin{matrix} \Psi (x) = \frac{8}{27 \sqrt{6}} \left( 1 - \frac{x}{6} \right) x^2 \text{exp} \left( \frac{-x}{2} \right) & \int_0^{ \infty} \Psi (x)^2 dx = 1 \end{matrix} \nonumber$
The 3p state energy is ‐0.0556 Eh.
$\frac{-1}{2} \frac{d^2}{dx^2} \Psi (x) + \frac{1}{x^2} \Psi (x) - \frac{1}{x} \Psi (x) = \text{E} \Psi (x) \text{ solve, E} \rightarrow \frac{-1}{18} = -0.0556 \nonumber$
The momentum wave function is generated by the following Fourier transform of the coordinate space wave function.
$\Phi (p) = \frac{1}{ \sqrt{2 \pi}} \int_0^{ \infty} \text{exp(-i p x)} \Psi (x) dx \rightarrow \frac{2}{3} 2^{ \frac{1}{2}} 6^{ \frac{1}{2}} \frac{(-1) + 6 \text{i p}}{(3 \text{i p} +1)^4 \pi^{ \frac{1}{2}}} \nonumber$
The Wigner function (phase‐space representation) for the 3p state is generated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \overline{ \Phi \left( \text{p} + \frac{ \text{s}}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\begin{matrix} N = 150 & i = 0 .. N & x_i = \frac{35i}{N} & j = 0 .. N & p_j = -2 + \frac{4j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~p_j \right) \end{matrix} \nonumber$
2.60: The Wigner Distribution for the 4s State of the 1D Hydrogen Atom
This tutorial presents three pictures of the 4s state of the one‐dimensional hydrogen atom using its position, momentum and phase‐space representations.
The energy operator for the one‐dimensional hydrogen atom in atomic units is:
$\frac{-1}{2} \frac{d^2}{dx^2} \blacksquare - \frac{1}{x} \blacksquare \nonumber$
The position 4s wave function is:
$\begin{matrix} \Psi (x) = \frac{x}{4} \left( 1 - \frac{3}{4} x + \frac{1}{8} x^2 - \frac{1}{192} x^3 \right) \text{exp} \left( \frac{-x}{4} \right) & \int_0^{ \infty} \Psi (x)^2 \text{dx} = 1 \end{matrix} \nonumber$
The 4s energy is -0.03125 Eh.
$\frac{-1}{2} \frac{d^2}{dx^2} \Psi (x) - \frac{1}{x} \Psi (x) = \text{E} \Psi (x) \text{solve, E} \rightarrow \frac{-1}{32} = -0.03125 \nonumber$
The momentum wave function is generated by the following Fourier transform of the coordinate wave function.
$\Phi (p) = \frac{1}{ \sqrt{2 \pi}} \int_0^{ \infty} \text{exp(-i p x)} \Psi (x) dx \rightarrow (-2) 2^{ \frac{1}{2}} \frac{64 \text{i p}^3 - 48 \text{p}^2 - 12 \text{i p} + 1}{(4 \text{i p} + 1)^5 \pi^{ \frac{1}{2}}} \nonumber$
The Wigner function (phase‐space representation) for the hydrogen atom 4s state is generated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int_{- \infty}^{ \infty} \overline{ \Phi - \left( \text{p} + \frac{ \text{s}}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
The Wigner distribution is displayed graphically.
$\begin{matrix} N = 100 & i = 0 .. N & x_i = \frac{50i}{N} & j = 0 .. N & p_j = -2 + \frac{4j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~ p_j \right) \end{matrix} \nonumber$
Just as for the 2s and 3s states, the Wigner distribution for the 4s state takes on negative values.
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The One‐dimensional Hydrogen Atom with a Delta Function Potential Energy Interaction Between the Proton and Electron
The energy Hamiltonian and a normalized wave function for the hydrogen atom with delta function interaction between the electron and proton is given below.
$\begin{matrix} \text{H} = - \frac{1}{2} \frac{d^2}{dx^2} \blacksquare - \Delta (x) \blacksquare & \Psi (x) = \text{exp} \left( - |x| \right) & \int_{- \infty}^{ \infty} \Psi (x)^2 dx \rightarrow 1 \end{matrix} \nonumber$
The wave function is not well‐behaved in coordinate space making the evaluation of kinetic energy tricky. However, the evaluation of potential energy is straight forward. So the strategy is to evaluate potential energy in coordinate space and kinetic energy in momentum space. The latter requires a Fourier transform of the coordinate space wave function.
Calculation of Potential Energy in Coordinate Space
$V = \int_{- \infty}^{ \infty} \Psi (x) - \Delta (x) \Psi (x) \text{dx} \rightarrow - 1 \nonumber$
Fourier Transform of the Coordinate Wave Function into Momentum Space
$\Psi (p) = \int_{- \infty}^{ \infty} \frac{ \text{exp(-i p x)}}{ \sqrt{2 \pi}} \Psi (x) \text{dx simplify} \rightarrow \frac{ \sqrt{2}}{ \sqrt{ \pi} \left( p^2 + 1 \right)} \nonumber$
Calculation of Kinetic Energy in Momentum Space
$\text{T} = \int_{- \infty}^{ \infty} \Phi (p) \frac{p^2}{2} \Phi (p) \text{dp simplify} \rightarrow \frac{1}{2} \nonumber$
Calculation of Total Energy
$\text{E = T + V} \rightarrow \text{E} = - \frac{1}{2} \nonumber$
The Wigner Function
The Wigner phase‐space distribution function is calculated using the momentum wave function.
$\text{W(x, p)} = \frac{1}{2 \pi} \int{- \infty}^{ \infty} \overline{ \Phi \left( \text{p} + \frac{ \text{s}}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber$
$\begin{matrix} N = 50 & i = 0 .. N & x_i = -4 + \frac{8i}{N} & j = 0 .. N & p_j = -7 + \frac{14j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~ p_j \right) \end{matrix} \nonumber$
Next we look at two variational calculations on the same system. The first involves a gaussian trial wave function, and the second a trigonmetric trial wave function.
Gaussian Trial Wave Function
$\begin{matrix} \Psi_1 (x,~ \beta ) = \left( \frac{2 \beta}{ \pi} \right)^{ \frac{1}{4}} \text{exp}\left( - \beta x^2 \right) & \int_{- \infty}^{ \infty} \Psi_1 (x,~ \beta)^2 \text{dx assume, } \beta >0 \rightarrow 1 \end{matrix} \nonumber$
Evaluation of Variational Energy Integral
$\text{E} ( \beta ) = \int_{- \infty}^{ \infty} \Psi_1 (x,~ \beta ) - \frac{1}{2} \frac{d^2}{dx^2} \Psi (x,~ \beta ) \text{dx... assume, } \beta > 0 \rightarrow \frac{ \beta}{2} - \frac{ \sqrt{2} \sqrt{3}}{ \sqrt{ \pi}} + \int_{- \infty}^{ \infty} - \Delta (x) \Psi_1 (x,~ \beta)^2 dx \nonumber$
Energy Minimization With Respect to Variational Parameter
$\begin{matrix} \frac{d}{d \beta} \text{E} ( \beta = 0 \text{ solve, } \beta \rightarrow \frac{2}{ \pi} & \begin{array}{c|c} \text{E} ( \beta ) & _{ \text{simplify}}^{ \text{substitute, } \beta = \frac{2}{ \pi}} \rightarrow - \frac{1}{ \pi} = -0.318 \end{array} \end{matrix} \nonumber$
Error
$\left| \frac{-.318+ .5}{-.5} \right| = 36.4 \% \nonumber$
Trigonometric Trial Wave Function
$\begin{matrix} \Psi_2 ( \text{x}, ~ \beta ) = \sqrt{ \frac{ \beta}{2}} \text{sech} ( \beta \text{x} ) & \end{matrix} \nonumber$
Evaluation of Variational Energy Integral
$\begin{matrix} \text{E} ( \beta ) = \begin{array}{c|c} \int_{ - \infty}^{ \infty} \Psi_2 ( \text{x},~ \beta ) - \frac{1}{2} \frac{d^2}{dx^2} \Psi_2 ( \text{x, } \beta ) dx ... & _{ \text{simplify}}^{ \text{assume, } \beta > 0} \rightarrow \frac{ \beta ( \beta -3}{6} \ + \int_{ - \infty}^{ \infty} - \Delta ( \text{x}) \Psi_2 ( \text{x, } \beta )^2 dx \end{array} \end{matrix} \nonumber$
Energy Minimization With Respect to Variational Parameter
$\begin{matrix} \frac{d}{d \beta} \text{E} ( \beta ) = 0 \text{ solve, } \beta \rightarrow \frac{3}{2} & \begin{array}{c|c} \text{E} ( \beta ) & _{ \text{simplify}}^{ \text{substitute, } \beta = \frac{3}{2}} \rightarrow - \frac{3}{8} = -0.375 \end{array} \end{matrix} \nonumber$
Error
$\left| \frac{-.375 + .5}{-.5} \right| = 25 \% \nonumber$
Graphical Comparison of Exact Wave Function with Trial Wave Function
This graphical comparison is consistent with the variational results presented above. The trigonometric trial function more closely resembles the exact wave function.
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The following expression for the atomic structure factor in the coordinate and momentum representations can be found in a paper by Fearnside and Matthew [AJP 65(8), 795-796 (1997)].
$f ( \mu ) = \int \text{exp} (i \mu ~ r ) | \Psi (r)|^2 dr = \int \Psi (p) \Psi ^* (P + \mu ) dp \nonumber$
The purpose of this tutorial is to establish the validity of this equation using Dirac notation and actual calculations using the hydrogen atom ground state eigenfunction in the position and momentum representations. See the Appendix for a graphical representation of what follows.
$\begin{matrix} \int \text{exp} ( i \mu ~ r) | \Psi (r) |^2 dr = \int \langle \Psi | r \rangle \langle r | \mu \rangle \langle r | \Psi \rangle dr = \int \int \langle \Psi | r \rangle \langle r | \mu \rangle \langle r | p \rangle \langle p | \Psi \rangle dr dp \int \int \langle \Psi | r \rangle \langle r | \mu \rangle \langle r | p \rangle \langle p | \Psi \rangle drdp = \int \int \langle p | \Psi \rangle \langle \Psi | r \rangle \langle r | p + \mu \rangle drdp = \int \int \langle p | \Psi \rangle \langle \Psi | p + \mu \rangle dp \end{matrix} \nonumber$
The following expressions have been used in mathematical manipulations alone.
$\begin{matrix} \int | p \rangle \langle p | dp = 1 & \int | r \rangle \langle r | dr = 1 & \langle r | \mu \rangle \langle r | p \rangle = \langle r | p + \mu \rangle \end{matrix} \nonumber$
Detail on the latter relationship is as follows.
$\langle r | mu \rangle \langle r | p \rangle = \text{exp} (i \mu ~ r ) \text{exp} (ip ~ r) = \text{exp} (i (p + \mu ) r) = \langle r | p + \mu \rangle \nonumber$
Now for the actual calculations (in atomic units) using the hydrogen atom ground state.
Ground state in the coordinate representation:
$\Psi (r) = \frac{1}{ \sqrt{ \pi}} \text{exp} (-r) \nonumber$
$\begin{array}{c|c} f ( \mu ) = \int_0^{ \infty} \int_0^{ \pi} \int_0^{2 \pi} \Psi (r)^2 \text{exp} (i \cos r \cos ( \theta )) r^2 \sin ( \theta ) d \phi d \theta dr & _{ \text{simplify}}^{ \text{complex}} \rightarrow \frac{16}{16 + 8 \mu^2 + \mu^4} \end{array} \nonumber$
Ground state in the momentum representation:
$\Psi (p) = \frac{2 \sqrt{2}}{ \pi} \frac{1}{ \left( 1 + p^2 \right)^2} \nonumber$
$\begin{array} f( \mu ) = \frac{8}{ \pi^2} \int_0^{ \infty} \int_0^{ \pi} \int_0^{2 \pi} \frac{2 \sqrt{2}}{ \pi} \frac{1}{ \left( 1 + p^2 \right)^2 \left[ 1 + \left( p^2 + \mu^2 + 2 p \mu \cos ( \theta ) \right) \right]^2} p^2 \sin ( \theta ) d \phi d \theta dp & _{ \text{simplify}}^{ \text{complex}} \rightarrow \frac{16}{16 + 8 \mu^2 + \mu^4} \end{array} \nonumber$
where
$\Psi (p + \mu) = \frac{2 \sqrt{2}}{ \pi} \frac{1}{ \left[ 1 + \left( p^2 + \mu^2 + 2 p \mu \cos ( \theta ) \right) \right]^2} \nonumber$
Appendix
$\begin{matrix} \int \Psi (p) \Psi^* (p + \mu ) dp & = & \int \Psi ^* (r) \text{exp} (i \mu r) \Psi (r) dr \ \updownarrow ( Dirac) & & \updownarrow ( Dirac) \ \int \langle p | \Psi \rangle \langle \Psi | p + \mu \rangle dp & & \int \langle \Psi | r \rangle \langle r | \mu \rangle \langle r | \Psi \rangle dr \ \updownarrow \left( \int | r \rangle \langle r | dr = 1 \right) & & \updownarrow \left( \int | p \rangle \langle p | dp = 1 \right) \ \int \int \langle \Psi | r \rangle \langle r | p + \mu \rangle \langle p | \Psi \rangle dpdr & \overset{ \langle r | p + \mu \rangle = \langle r | \mu \rangle \langle r | p \rangle}{\longleftrightarrow} & \int \int \langle \Psi | r \rangle \langle r | \mu \rangle \langle r | p \rangle \langle p | \Psi \rangle dpdr \end{matrix} \nonumber$
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The behavior of electrons in molecules and atoms is described by quantum mechanics; classical (Newtonian) mechanics cannot be used because the de Broglie wavelengths (λ=h/mv) of the electrons are comparable with molecular (and atomic) dimensions.
The relevant quantum-mechanical ideas are as follows:
• Electrons are characterized by their entire distributions (called wave functions or orbitals) rather than by instantaneous positions and velocities: an electron may be considered always to be (with appropriate probability) at all points of its distribution (which does not vary with time).
• The kinetic energy of an electron decreases as the volume occupied by the bulk of its distribution increases, so delocalization lowers its kinetic energy.
$KE = \frac{p^2}{2m} = \frac{h^2}{2m \lambda^2} \approx \frac{A}{D^2} \approx \frac{A}{V^{ \frac{2}{3}}} \nonumber$
• The potential energy of interaction between an electron and other charges is as calculated by classical physics, using the appropriate distribution (wave function) for the electron: an electron distribution is therefore attracted by nuclei and its potential energy decreases as the average electron-nuclear distance decreases.
$PE \approx - \frac{B}{D} \approx - \frac{B}{V^{ \frac{2}{3}}} \nonumber$
• A minimum-energy electron distribution represents the best compromise between concentration near the nuclei (to reduce potential energy) and delocalization (to reduce kinetic energy).
$E \approx KE + PE \approx \frac{A}{V^{ \frac{2}{3}}} - \frac{B}{V^{ \frac{2}{3}}} \nonumber$
A bond will form between two atoms when the electron distribution of the combined atoms (molecular orbital) yields a significantly lower energy than the separate-atom distributions (atomic orbitals). An example is a covalent bond, in which two electrons, one originally on each atom, change their distributions so that each extends over both atoms.
*Taken from "Molecules" in The Encyclopedia of Physics by Frank E. Harris (with some additions and modifications by Frank Rioux)
3.02: The Covalent Bond in the Hydrogen Molecule
Introduction
The covalent chemical bond is a difficult concept that is frequently oversimplified as a purely electrostatic phenomenon in textbooks at all levels of the undergraduate chemistry curriculum. Therefore, the purpose of this paper is to provide an elementary quantum-mechanical analysis of the covalent bond appropriate for an undergraduate course in physical or advanced inorganic chemistry. It is important to emphasize that there is no acceptable classical electrostatic explanation for the covalent bond, just as there is no classical explanation for atomic stability or atomic structure. Quantum-mechanical principles are required to understand atomic and molecular stability and structure.
Background
Forty years ago, Ruedenberg and his collaborators undertook a detailed study of the covalent bond in H2+ [1-3]. This thorough and incisive theoretical analysis revealed that chemical-bond formation was not simply an electrostatic phenomenon as commonly thought, but that electron kinetic energy also played an essential role. Ruedenberg's contribution to our current understanding of the physical nature of the chemical bond has been discussed in a number of publications in the pedagogical literature in chemistry and physics [4-9]. In addition, there are excellent critiques and summaries elsewhere that are accessible to the interested nonspecialist [10-12]. It should be noted that Slater also recognized the importance of electron kinetic energy in chemical bond formation in a benchmark paper published in the inaugural volume of the Journal of Chemical Physics [13]. He returned to the subject subsequently [14], but never pursued it at the depth that Ruedenberg and his colleagues did.
Ruedenberg chose to study H2+ because, as the simplest molecule, he could easily extract detailed information about all the contributions to the total energy from its one electron wave function. In the present study the simplest electron-pair bond, H2, will be examined. The analysis is carried out at a much more elementary level, but the same message emerges-electron kinetic energy plays a crucial role in chemical bond formation. Theoretical analysis shows that H-H bond formation, 2H → H2, is an exothermic process that obeys the virial theorem: ∆E = ∆V/2 = -∆T [15]. Noncritical use of the virial theorem, therefore, may lead one to believe that stable bond formation is solely an electrostatic potential-energy effect, and that consideration of kinetic energy is neither relevant nor necessary. However, H-H bond formation can be thought of as a very simple chemical reaction, and we know that it is never justified to assume that the balanced chemical equation is also the mechanism for the reaction. For example, even a simple first-order isomerization reaction (R → P) requires the formation of an activated form of the reactant (R → R* → P ).
Similarly, to study the covalent bond it is instructive to postulate a "mechanism" for bond formation, a sequence of hypothetical steps that are equivalent to the overall process 2H → H2. Unlike a traditional chemical mechanism, it cannot be tested empirically and, therefore, its value or validity rests on the clarity and cogency of its basic premises. There are actually several plausible mechanisms, but our attention will be restricted to one that might appear especially cogent to undergraduate audiences. For example, when we describe the bonding in methane to students, we generally invoke a mechanism that uses the concepts of atomic promotion [2s22p2 → 2s12p3], hybridization [2s12p3 → (sp3 hybrids)4], and bond formation through the overlap of atomic orbitals. The H2 bond-formation mechanism described in this paper will consist of two steps: atomic promotion and overlap of atomic orbitals (2H → 2H* → H2). This simple mechanism has previously been used to analyze the bonding in H2+ [12]. It should be noted that the individual steps of the mechanism do not satisfy the virial theorem, but, of course, collectively they do. To carry out a quantum-mechanical analysis of bond formation in H2, it is necessary to decide at what level of theory to work. In this analysis scaled hydrogenic 1s orbitals will be used for the atomic orbitals. At the molecular level, both molecular orbital (MO) and valence bond (VB) wave functions will be considered. Labeling the nuclei a and b, and the electrons 1 and 2, the atomic orbitals are,
$\begin{matrix} 1s_{a} = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} (- \alpha r_a ) & 1s_{b} = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} (- \alpha r_b ) \end{matrix} \nonumber$
Table 1. Variational Results for the MO and VB Wave Functions
MO VB
α
1.197 1.166
T/Eb 1.1282 1.1389
V/Eb -2.2564 -2.2778
E/Eb -1.1282 -1.1389
% Error 3.95 3.04
Re0 1.38 1.41
% Error 1.43 0.71
Table 2. Bond Formation Mechanism Results for the Molecular Orbital Wave Function
2 H(α=1) 2H*(α=1.197)a H2(α=1.197)
T/Eh 1.00 0.4328 1.4328 -0.3046 1.1282
V/Eh -2.00 -0.3940 -2.3940 0.1376 -2.2564
E/Eh -1.00 0.0388 -0.9612 -0.1670 -1.1282
aEq 4 is used to calculate the entries in this column.
Table 3. Bond Formation Mechanism Results for the Valence Bond Wave Function
2 H(α=1) 2H*(α=1.166)a H2(α=1.166)
T/Eh 1.00 0.3596 1.3596 -0.2207 1.1389
V/Eh -2.00 -0.3320 -2.3320 0.0542 -2.2778
E/Eh -1.00 0.0276 -0.9724 -0.1665 -1.1389
aEq 4 is used to calculate the entries in this column.
Using this basis set, the MO and VB wave functions are (neglecting spin),
$\Psi_{MO} = N_{MO} \left[ 1s_a (1) + 1s_b (1) \right] \left[ 1s_a (2) + 1s_b (2) \right] \nonumber$
$\Psi_{VB} = N_{VB} \left[ 1s_a (1) 1s_b (2) + 1s_a(2) 1s_b(1)\right] \nonumber$
In these equations the scale factor, α is a variational parameter which controls how rapidly the atomic wave function decays to zero, and NMO and NVB are the appropriate normalization constants.
The Hydrogen Atom
Using a scaled hydrogenic wave function (see eq 1) in a variational calculation for the hydrogen atom yields the following expression for the energy in atomic units (me = e = ħ = 1),
$E_H = \frac{ \alpha^2}{2} - \alpha \nonumber$
where the first term is the electron kinetic energy, and the second term is the electron-nucleus potential energy. Minimization of EH with respect to α yields,
$\alpha =1 \rightarrow E_H = <T> + <V> = \frac{1}{2} -1 = - \frac{1}{2} \nonumber$
where and represent the expectation values for kinetic and potential energy, respectively.
The Hydrogen Molecule. The results for the variational calculations on the hydrogen molecule using the molecularorbital and valence-bond wave functions, and the Born-Oppenheimer energy operator,
$\hat{H} = - \frac{1}{2} \nabla_1^2 - \frac{1}{2} \nabla_2^2 - \frac{1}{r_{a1}} - \frac{1}{r_{a2}} - \frac{1}{r_{b1}} - \frac{1}{r_{b2}} + \frac{1}{r_{12}} + \frac{1}{R_{ab}} \nonumber$
are presented in Table 1. The table provides the optimum value of α, the total energy, and the equilibrium bond length for both wave functions as reported in the literature [16]. The kinetic and potential contributions to the total energy are obtained using the virial theorem. The experimental values [17] for the ground-state energy and the equilibrium bond distance are, respectively, -1.1746 Eh (1 Eh = 27.2114 eV = 2.6255 MJ mol-1) and 1.40 ao (1 ao = 52.92 pm).
The Mechanism
For both wave functions, the bond formation mechanism is the same. In the first step, atomic promotion, the hydrogen atom orbitals prepare for bonding by contracting from α = 1 to the optimum α value of the final molecular wave function. This step is atomic and endothermic, increasing the kinetic energy more than it decreases the potential energy. The potential energy decreases because the electrons move closer on average to their respective nuclei. The kinetic energy increases because of the greater confinement of the electrons in the contracted orbitalsókinetic energy is inversely proportional to the square of the average distance of the electron from the nucleus.
The second step consists of the formation of a molecular wave function by overlap of the promoted atomic orbitals. The constructive interference that accompanies orbital overlap brings about charge delocalization and charge redistribution. Charge delocalization distributes the electron density over the molecule as a whole (each electron now belongs to both nuclei) and brings about a significant decrease in kinetic energy. Charge redistribution transfers some electron density from the neighborhood of the nuclei into the inter-nuclear region, which involves an increase in electron potential energy. The second step is exothermic because kinetic energy decreases more than potential energy increases-charge delocalization funds the redistribution of charge from the area around the nuclei into the bond region. The results for both molecular wave functions are summarized in Tables 2 and 3. In summary, this simple two-step mechanism clearly shows that covalent bond formation in H2 is driven by a decrease in kinetic energy brought about by the charge delocalization that accompanies the overlap of atomic orbitals. This is also the basic conclusion of Ruedenbergís more-detailed and sophisticated analysis for H2+.
Acknowledgment. I wish to acknowledge a number of very helpful recommendations made by an anonymous reviewer.
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Abstract
Slater's method of analyzing the covalent bond using the virial theorem is presented for the hydrogen molecule. The overall conclusion reached is the same as that reached with Ruedenberg's well-known abinitio quantum mechanical study - electron kinetic energy plays an important role in chemical bond formation.
Introduction
In the second edition of his classic monograph, Molecular Quantum Mechanics, Peter Atkins begins the chapter on molecular structure with the following sentences (1),
Now we come to the heart of chemistry. If we can understand what holds atoms together as molecules we may also start to understand why, under certain conditions, old arrangements change in favor of new ones. We shall understand structure, and through structure, the mechanism of change.
Few will argue with Atkins' eloquent assertion that the chemical bond is at the heart of chemistry, however finding a satisfactory discussion of chemistry's central concept is difficult. For example, a survey of widely adopted general chemistry texts showed the following errors in the description of the covalent bond to be prevalent. Surprisingly the origin of some of these errors can be traced to two authoritative monographs on chemical bonding of enormous influence (2).
• The covalent bond is presented as a purely electrostatic phenomenon. Electron kinetic energy is never mentioned, even though the total energy of a molecule is a sum of kinetic and potential energy contributions, and atomic and molecular stability cannot be understood solely in terms of potential energy.
• Closely related to this is that what is actually an energy curve is called a potential energy curve. What is shown in introductory texts is the total molecular energy as a function of internuclear separation under the Born-Oppenheimer approximation. In other words, nuclear kinetic energy is frozen, but electron kinetic energy contributes to the total molecular energy.
• It is claimed that the electron density in the inter-nuclear (bond) region has a lower potential energy because it is attracted to two nuclei. Actually using simple electrostatic arguments it is easy to show that the electron-nuclear potential energy is higher in the internuclear region than it is closer to the nuclei. On the basis of potential energy alone the electrons would prefer to be in the nucleus.
• An energy minimum, or molecular ground state, is achieved because of increases in nuclear-nuclear and electron-electron repulsions as the internuclear separation decreases. As will be shown later, the immediate cause of the molecular ground state is a sharp increase in electron kinetic energy.
• The amount of electron density transferred to the bonding region is greatly overstated, sometimes implying that a pair of electrons is shared between two nuclei rather than bytwo nuclei.
Unfortunately, to find accurate analyses of the physical origin of the covalent bond one must go to the research or pedagogical literature; chemistry textbooks will not, in general, provide enlightenment. In the 1960s and 70s Ruedenberg and his collaborators carried out a detailed quantum mechanical study of the covalent bond in H2+ (4 - 6). The most important conclusion of this thorough and insightful study was that electron kinetic energy plays a crucial role in the formation of a chemical bond. Ruedenberg's contributions to the understanding of the chemical bond have been summarized in the pedagogical literature (7 - 12) and in review articles (13 - 15). There are also at least two encyclopedia entries which give accurate and clear interpretations of covalent bond formation (16, 17).
It is surprising that none of these efforts to make Ruedenberg's work accessible to the non-specialist have had any noticeable impact on the way chemical bonding is presented by the authors of chemistry textbooks currently used in the undergraduate curriculum. While physical chemistry texts avoid the errors cited above, they generally do not attempt to provide an "explanation" of the chemical bond. For example, after outlining the mathematical techniques required to solve Schr�dinger's equation for H2+ and H2, physical chemistry texts do not interpret the calculations other than to say something to the effect that the stability of the chemical bond is a quantum mechanical effect that has no classical analog or explanation. Atkins' physical chemistry text is a notable exception; in a footnote he alerts the reader to the subtleties of the chemical bond and briefly outlines the interplay between kinetic and potential energy during the formation of the covalent bond in H2+(3).
It is not widely appreciated that John C. Slater used the virial theorem to come to a similar conclusion about the importance of electron kinetic energy in chemical bond formation in the early days of quantum mechanics. In a paper published in the inaugural volume of the Journal of Chemical Physics Slater pioneered the use of the virial theorem in interpreting the chemical bond. With regard to the virial theorem he said (18),
"... this theorem gives a means of finding kinetic and potential energy separately for all configurations of the nuclei, as soon as the total energy is known, from experiment or theory." (emphasis added)
The purpose of this paper is to outline Slater's method of analysis for the hydrogen molecule, the simplest example of the traditional two-electron chemical bond.
Background Theory
The Hamiltonian energy operator (in atomic units) for the hydrogen molecule consists of the following ten contributions,
$\hat{H} = - \frac{1}{2M} \nabla_a^2 - \frac{1}{2M} \nabla_b^2 - \frac{1}{2} \nabla_1^2 - \frac{1}{2} \nabla_2^2 - \frac{1}{r_{a1}} - \frac{1}{r_{a2}} - \frac{1}{r_{b1}} - \frac{1}{r_{b2}} + \frac{1}{r_{12}} + \frac{1}{R} \nonumber$
where a and b label the nuclei, 1 and 2 label the electrons, and R = Rab, the internuclear separation. Under the Born-Oppenheimer approximation the nuclear kinetic energy operators drop out, and the molecular electronic energy as a function of internuclear separation, E(R), can be obtained by solving Schr�dinger's equation variationally for the resulting eight-term electronic Hamiltonian, Hel, which in this analysis includes nuclear-nuclear repulsion.
$E(R) = \frac{ \langle \Psi (R;~1,2) | \hat{H}_{el} | \Psi (R;~1,2) \rangle}{ \langle \Psi (R;~1,2) | \Psi (R;~1,2) \rangle} \nonumber$
Now one can return to equation (1) with the calculated E(R) and treat it as the molecular "potential" field in which the nuclei move. The nuclear kinetic energy operators are re-written in terms of the nuclear center-of-mass coordinate and the internal coordinate R. After discarding the term involving the motion of the center of mass, the Hamiltonian energy operator becomes
$\hat{H} = - \frac{1}{2 \mu} \nabla^2 + E(R) \nonumber$
Following Slater, a more empirical approach is taken here and the ab initio E(R) is replaced by an analytical surrogate such as the Morse function (19) which has a similar R-dependence (20).
$E(R) = D_e \left[ 1 - \text{exp} (- \beta (R-R_e )) \right]^2 - D_e \nonumber$
In other words the Morse function represents the total energy of a diatomic molecule assuming the Born-Oppenheimer approximation.
Substitution of equation (4) into (3) yields an energy Hamiltonian for which Schrödinger's equation has an exact solution with the following quantized nuclear vibrational states (2),
$G(v) = \left( v + \frac{1}{2} \right) \overline{ \nu}_e - \left( v + \frac{1}{2} \right)^2 \overline{ \nu}_e x_e \nonumber$
where v = 0, 1, 2, ..., and the Morse parameters, De and b, are defined as,
$\begin{matrix} D_e = \frac{hc \overline{ \nu}_e (1- x_e^2)}{4x_e} & \beta = \sqrt{ \frac{ \mu}{D_e}} \pi \overline{ \nu}_e c\end{matrix} \nonumber$
Fitting equation (5) to spectroscopic data for H2 yields the following values for the Morse parameters: De= 7.92x10-19 joule; b = 0.0191 pm-1; and Re = 74.1 pm (21).
The Virial Theorem
At this point the virial theorem is used to obtain a total energy profile for covalent bond formation in H2. The first step in this approach is to acknowledge that the energy of a molecule is a sum of kinetic and potential energy contributions.
$E(R) = T(R) + V(R) \nonumber$
The virial theorem for a diatomic molecule as a function of internuclear separation, R, is (14)
$2T(R) + V(R) + R \frac{dE(R)}{dR} = 0 \nonumber$
Equations (7) and (8) can now be used to obtain expressions for the kinetic and potential energy in terms of the total energy and its first derivative.
$T(R) = -E(R) - R \frac{dE(R)}{dR} \nonumber$
$V(R) = 2E(R) + R \frac{dE(R)}{dR} \nonumber$
Quoting Slater again (18)
These important equations determine the mean kinetic and potential energies as a function of r, one might almost say, experimentally, directly from the curves of E as a function of r which can be found from band spectra. The theory is so simple and direct that one can accept the results without question, remembering only the limitation of accuracy mentioned above. (Here Slater is referring to the neglect of zero point vibration in his analysis.)
Thus, when equation (4), parameterized as indicated above, is used in equations (9) and (10) the energy profile for covalent bond formation in H2 shown in Figure 1 is obtained.
Analysis
This energy profile shows that as the inter-nuclear separation decreases, the potential energy rises, falls, and then rises again. The kinetic energy first decreases and then increases at about the same value of R that the potential energy begins to decrease.
The quantum mechanical interpretation of the energy profile is as follows (4, 18). As the molecular orbital is formed at large R constructive interference between the two overlapping atomic orbitals on the hydrogen atoms draws electron density away from the nuclear centers into the inter-nuclear region. The potential energy rises as electron density is removed from the region around the nuclei, but the total energy decreases because of a larger decrease in kinetic energy due to charge delocalization - the electrons now belong to the molecule and not the individual atoms. Thus, a decrease in kinetic energy funds the initial build-up of charge between the nuclei that is traditionally associated with chemical bond formation.
Following this initial phase, at an inter-nuclear separation of about 150 pm, the potential energy begins to decrease and the kinetic energy increases, both sharply eventually, while the total energy continues to decrease gradually. This is an atomic affect, not a molecular one, as Ruedenberg clearly showed. The initial transfer of charge away from the nuclei and into the bond region allows the atomic orbitals to contract causing a large decrease in potential energy because the electron density is moved, on average, closer to the nuclei. The kinetic energy increases because the atomic orbitals are smaller and kinetic energy is inversely proportional to the square of the average orbital radius. This is an atomic affect because the orbital contraction actually causes some electron density to be withdrawn from the bonding region and returned to the nuclei.
An energy minimum is reached while the potential energy is still in a significant decline, indicating that kinetic energy, which is increasing rapidly, is the immediate cause of a stable bond and the molecular ground state in H2. The final increase in potential energy which is mainly due to nuclear-nuclear repulsion doesn't begin until the inter-nuclear separation is less than 50 pm, while the equilibrium bond length is 74 pm. Thus the common explanation that an energy minimum is reached because of nuclear-nuclear and electron-electron repulsion does not have merit. As Ruedenberg (4) has noted "there are no ground states in classical mechanics or electrostatics."
Conclusion
In conclusion a review of some basic principles and some additional observations are offered. Under the Born-Oppenheimer approximation the quantum mechanical energy operator for H2 consists of the final eight terms in equation (1), which fall into four types of energy contributions: electron kinetic energy, electron-electron, nuclear-nuclear, and electron-nuclear potential energy. Electron-nuclear potential energy is the only negative contribution to the total energy and is essential in understanding the stability of the chemical bond. H2 is a stable molecule because it has a lower total energy than its constituent atoms, not simply because there is a build-up of charge in the inter-nuclear region. As shown above, by itself this charge build-up actually increases the energy rather than decreasing it as is popularly, but incorrectly, believed.
We all recognize that electron kinetic energy is important at the computational level through its presence in the energy operator given in equation (1). In other words ground states are calculated by minimizing the total energy, the sum of the kinetic and potential energy contributions. However, many ignore kienticenergy when it comes to interpretation because they believe it is irrelevant. The origin of this error is a common misapplication of the virial theorem.
For example, it is common to argue from the virial theorem in the form,
$\Delta E = \frac{ \Delta V}{2} = - \Delta T \nonumber$
that bond stability is due solely to potential energy, because it decreases with bond formation and kinetic energy actually increases by twice as much. However, this form of the virial theorem is valid only for R = R�� and R = Re, the initial and final states. It tells us nothing about what is occuring when the bond is actually being formed - for that you need equations (7 - 10). As has been shown above, what is actually occuring during bond formation is a rather subtle role reversal between kinetic and potential energy. It is ironic that Slater correctly used the virial theorem to show that electron kinetic energy is essential to understanding the chemical bond in the 1930s, while today many use it incorrectly to show that electron kinetic energy is irrelevant.
It has been known since the early years of the 20th century that no dynamic or static array of charged particles is stable on the basis of classical electrostatic principles alone. Therefore, the quantum mechanical picture of a wave-particle duality for the electron and the peculiar quantum mechanical nature of kinetic energy are essential in understanding atomic and molecular stability and structure. The importance of kinetic energy runs counter to conventional opinion regarding the covalent bond in two seemingly paradoxical ways. First, a decrease in kinetic energy due to incipient molecular orbital formation funds the transfer of charge density into the internuclear region, lowering the total energy. Second, a large increase in kinetic energy associated with the subsequent atomic orbital contraction prevents the collapse of the molecule and causes an energy minimum and a stable molecular ground state.
In a previous publication (10b) the formation of the covalent bond in H2 was analyzed in terms of a two-step mechanism involving: a) orbital contraction ("hybridization") followed by b) orbital overlap (charge delocalization and redistribution). It was shown that step a) is an atomic and endothermic process, while b) is a molecular and exothermic effect. The present study is consistent with this prior analysis, but reverses the order of the contributing effects; atomic orbital overlap precedes atomic orbital contraction. Both mechanisms clearly reveal the important role that electron kinetic energy plays in chemical bond formation.
In summary I concur with Kutzelnigg's observation (22), "The chemical bond is a highly complex phenomenon which eludes all attempts at simple description." This means we must be careful how we teach the chemical bond to our students. It is not acceptable to present incorrect models of bonding to undergraduates because they are easier to understand and therefore easier to teach.
Literature Cited:
1. Atkins, P. W. Molecular Quantum Mechanics, Oxford University Press, Oxford, UK, 1983, p. 250.
2. Pauling, L. The Nature of the Chemical Bond, 3rd ed., Cornell University Press, Ithaca, 1960, pp. 19-21. Coulson, C. A. Valence, 2nd ed., Oxford University Press, London, 1961, pp. 85-86.
3. Atkins, P. W. Physical Chemistry, 6th ed., W. H. Freeman and Co., New York, 1998, p. 397.
4. Ruedenberg, K. Rev. Mod. Phys. 1962, 34, 326-352.
5. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 54, 1495-1511. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 55, 5804-5818.
6. Ruedenberg, K. In Localization and Delocalization in Quantum Chemistry; Chalvet, O. et al., Eds.; Reidel: Dordrecht, The Netherlands, 1975; Vol. I, pp 223-245.
7. Baird, N. C. J. Chem. Educ. 1986, 63, 660-664.
8. Harcourt, R. D. Am. J. Phys. 1988, 56, 660-661.
9. Nordholm, S. J. Chem. Educ. 1988, 65, 581-584.
10. Bacskay, G. G.; Reimers, J. R.; Nordholm, S. J. Chem. Educ. 1997, 74, 1494-1502.
11. Rioux, F. a) Chem. Educator 1997, 2(6), 1-14; b) Chem. Educator 2001, 6(5), 288-290.
12. Harcourt, R. D.; Solomon, H.; Beckworth, J.; Chislett, L. Am. J. Phys. 1982, 50, 557-559.
13. Kutzelnigg, W. Angew. Chem. Int. Ed. Eng. 1973, 12, 546-562.
14. Melrose, M. P.; Chauhan, M.; Kahn, F. Theor. Chim. Acta 1994, 88, 311-324.
15. Gordon, M. S.; Jensen, J. H. Theor. Chem. Acc. 2000, 103, 248-251.
16. Gordon, M. S.; Jensen, J. H. Encyclopedia of Computational Chemistry, Schleyer, P. v. R., Ed.; John Wiley & Sons: New York, 1998, pp 3198-3214.
17. Harris, F. E. In Encyclopedia of Physics, 2nd Edition; Lerner, R. G.; Trig, G. L., Eds.; VCH Publishers, Inc.: New York, 1991, pp 762-764.
18. Slater, J. C. J. Chem. Phys. 1933, 1, 687-691.
19. Morse, P. M. Phys. Rev. 1929, 34, 57.
20. For a virial theorem analysis using the best ab initio results for H2 see, Winn, J. S. J. Chem. Phys.1981, 74, 608-611.
21. Huber, K. P.; Herzberg Molecular Spectra and Molecular Structure, vol IV, Constants of Diatomic Molecules; Van Nostrand Reinhold: New York, 1979.
22. Kutzelnigg, W. Angew. Chem. Int. Ed. Eng. 1984, 23, 292.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.03%3A_The_Covalent_Bond_Clarified_Through_the_Use_of_the_Virial_Theorem.txt
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Abstract: The chemical bond is a concept of unique importance in chemistry that students first study seriously in a college or university general chemistry course. At first glance, the descriptions of covalent bond formation found in general chemistry texts seem plausible and comprehensible; however, when they are examined more carefully, it is found that they violate both classical and quantum mechanical principles.
Introduction
In the second edition of his classic text, Molecular Quantum Mechanics, Peter Atkins begins the chapter on molecular structure with the following sentences [1]:
Now we come to the heart of chemistry. If we can understand what holds atoms together as molecules we may also start to understand why, under certain conditions, old arrangements change in favor of new ones. We shall understand structure, and through structure, the mechanism of change.
Few will argue with Atkins' eloquent assertion that the chemical bond is at the heart of chemistry, but where do we find an accurate explanation of chemistry's central concept?-certainly not in chemistry textbooks written for undergraduate audiences. For example, a survey of widely adopted general chemistry texts showed the following errors in the description of the covalent bond to be prevalent.
• The covalent bond is presented as a purely electrostatic phenomenon. Electron kinetic energy is never mentioned, even though the total energy of a molecule is a sum of kinetic and potential energy contributions, and atomic and molecular stability cannot be understood solely in terms of potential energy.
• Closely related to this is what is actually an energy curve is called a potential energy curve. What is shown in introductory texts is the total molecular energy as a function of internuclear separation under the Born-Oppenheimer approximation. In other words, nuclear kinetic energy is frozen, but electron kinetic energy contributes to the total molecular energy.
• It is claimed that the electron density in the internuclear (bond) region has a lower potential energy because it is attracted to two nuclei. Actually using simple electrostatic arguments (see the appendix) it is easy to show that the electron-nuclear potential energy is higher in the internuclear region than it is closer to either of the nuclei. On the basis of potential energy alone the electrons would prefer to be in the nucleus.
• It is claimed that an energy minimum, or molecular ground state, is achieved because of increases in nuclear- nuclear and electron-electron repulsions as the internuclear separation decreases. As will be shown later, the immediate cause of the molecular ground state is a sharp increase in electron kinetic energy.
• The amount of electron density transferred to the bonding region is greatly overstated, sometimes implying that a pair of electrons is shared in the space between two nuclei rather than by two nuclei.
By comparison, these errors are not often found in physical chemistry textbooks. Most physical chemistry texts set up the Schrödinger equation for the H2+ and H2 molecules and outline the solutions, but avoid interpreting the calculations other than saying something to the effect that the chemical bond is a quantum mechanical phenomenon that has no classical analog or explanation. A notable exception is the text by Atkins and de Paula, which gives a correct analysis of the bond in H2+ in a footnote and the text by Raff, which summarizes the best ab initio results for H2 [2]. Thus, in order to find an accurate analysis of covalent bond formation, it is necessary to consult advanced quantum chemistry texts, the research literature, or the pedagogical literature.
In the 1960s and 70s Ruedenberg and his collaborators carried out a detailed quantum mechanical study of the covalent bond in H2 + [3-5]. The most important conclusion of this thorough and insightful study was that electron kinetic energy plays a crucial role in chemical bond formation. Ruedenberg's contributions to the understanding of the chemical bond have been summarized in the pedagogical literature [6-11] and in review articles [12-14]. It is unfortunate that none of these efforts to make Ruedenberg's work accessible to the nonspecialist have had any noticeable impact on the way chemical bonding is presented by authors of chemistry textbooks at the introductory or intermediate level.
It is not widely appreciated that John Slater came to similar conclusions about covalent bond formation thirty years before Ruedenberg using a more empirical approach based directly on the virial theorem [15-17]; therefore, the purpose of this paper is to outline Slater's method for the hydrogen molecule, the simplest example of the traditional two-electron chemical bond. It should be noted, however, that Slaterís approach is appropriate for any diatomic molecule for which Morse parameters are available.
Background Theory
In the early days of the quantum revolution Slater used the virial theorem to analyze the chemical bond and was the first to notice the importance that electron kinetic energy played incovalent bond formation. With regard to the virial theorem he said [15]:
...this theorem gives a means of finding kinetic and potential energy separately for all configurations of the nuclei, as soon as the total energy is known, from experiment or theory.
Under the Born-Oppenheimer approximation, the molecular energy as a function of internuclear separation, E(R), is represented well by a Morse function [18] whose parameters for H2 are based on spectroscopic data: De = 7.93 aJ; β = 0.0190 pm-1; Re = 74.1 pm [19].
$E(R) = D_e \left[ 1 - \text{exp} \left( - \beta (R- R_e ) \right) \right]^2 - D_e \nonumber$
The virial theorem for diatomic molecules [15] can be used with eq 1 to obtain expressions for the average values of kinetic and potential energy as a functions of the internuclear separation, R.
$T(R) = - E(R) - R \frac{dE(R)}{dR} \nonumber$
$V(R) = 2E(R) + R \frac{dE(R)}{dR} \nonumber$
Quoting Slater again [15]:
These important equations determine the mean kinetic and potential energies as functions of R, one might almost say, experimentally, directly from the curves of E as a function of R which can be found from band spectra. The theory is so simple and direct that one can accept the results without question...
When eq 1, parameterized as indicated earlier, is used in eqs 2 and 3, the energy profile for covalent bond formation in H2 shown in the Figure 1 is obtained.
Analysis
As Slater pointed out, the molecular energy curve can be obtained from experiment (spectroscopy) or theory (ab initio quantum mechanics). The energy profile obtained is similar in both approaches. For example, Winn [20] offered the following analysis of the ab initio results of Kolos and Wolniewicz [21] for H2.
As the atoms approach, the potential energy rises (electrons are moving away from nuclei) and the kinetic energy falls (as delocalization begins). In the vicinity of R/Re = 2, this trend reverses. The kinetic energy increases as the electronic wave function is localized further, raising the momentum, but the potential energy falls, as charge is now brought nearer both nuclei. Only at R/Re < 0.5 does nuclear repulsion cause the potential energy to increase and contribute to the total repulsion energy.
This analysis will serve as the basis for a simple quantum mechanical model for covalent bond formation consistent with the spectroscopic energy profile shown in Figure 1 that uses qualitative concepts accessible to introductory students. It consists of two steps: (1) molecular orbital formation through the overlap of atomic orbitals, followed by (2) atomic orbital contraction. It is assumed step (1) ends as kinetic energy reaches a minimum (151 pm), and step (2) ends when the total energy reaches a minimum (74.1 pm). This approach yields the following quantitative analysis of bond formation in H2.
Step 1 Step 2 Overall
ΔT/aJ
-0.495 1.288 0.793
ΔV/aJ
0.159 -1.745 -1.586
ΔE/aJ
-0.336 -0.457 -0.793
R/pm 151 74.1 74.1
1. The constructive interference that accompanies molecular orbital formation brings about charge delocalization and charge redistribution. This step is molecular in character and is driven by a decrease in kinetic energy as the table above shows.
Charge delocalization occurs because each electron now belongs to both nuclei and occupies a larger volume than in the atomic state. This effect is accompanied by a significant decrease in electron kinetic energy [22]. Charge redistribution occurs because atomic overlap transfers some electron density away from the nuclear centers into the internuclear (overlap) region, which involves an increase in potential energy. Overall this step is exothermic because kinetic energy decreases more than potential energy increases. Thus, charge delocalization funds the redistribution of charge into the internuclear region that is normally associated with bond formation.
2. The reduction of electron density near the nuclei that occurs in step 1 allows the atomic orbitals to contract returning some electron density to the nuclear centers from the internuclear region. This step is atomic in character and is driven by a large decrease in potential energy.
The decrease in potential energy occurs because the electrons are brought closer to the nuclei, and the increase in kinetic energy occurs because orbital contraction has reduced the volume occupied by the electrons, thereby reducing the level of delocalization. It has been shown that at the equilibrium bond distance the net transfer of electron density into the internuclear region is only 16% [5].
According to the energy profile, an energy minimum is reached while the potential energy is still in a significant decline, showing that kinetic energy, which is increasing rapidly, is the immediate cause of a stable bond and molecular ground state in H2. The final increase in potential energy, which is mainly due to nuclear-nuclear repulsion, doesn't begin until the internuclear separation is less than 50 pm and the equilibrium bond length is 74 pm. Thus, the common explanation that an energy minimum is reached because of nuclear-nuclear and electron-electron repulsion does not have merit. As Ruedenberg [5] has succinctly noted "there are no ground states in classical mechanics or electrostatics".
Conclusion
In summary, an "empirical" analysis of the covalent bond in H2 has been presented based on the virial theorem and a Morse molecular energy function parameterized using spectroscopic data. This analysis shows that the role of electron kinetic energy in covalent bond formation runs counter to conventional wisdom in two seemingly paradoxical ways. First, a decrease in kinetic energy due to incipient molecular orbital formation funds the transfer of charge density into the internuclear region, lowering the total energy. Second, a large increase in kinetic energy accompanying the subsequent atomic orbital contraction prevents the collapse of the molecule, and causes an energy minimum and a stable molecular ground state.
Clearly, using rigorous quantum mechanics in introductory chemistry courses in order to avoid error in describing the chemical bond is not a practical solution; however, teaching simple, easily digestible, but incorrect models for the covalent bond is pedagogically unacceptable. It is therefore necessary to offer general chemistry students an explanation of the nature of the covalent bond that is both correct and accessible. To achieve this end a rudimentary two-step quantum mechanical mechanism for covalent bond formation in H2 has been proposed.
Appendix
When asked what motivated the creation of his famous model of the atom Bohr replied "the stability of matter, a pure miracle when considered from the standpoint of classical physics." The following simple calculation will demonstrate what Bohr meant by this statement. This calculation will be carried out in atomic units where the charge on the electron is ñ1, the charge on the nucleus +1, and distances are measured in bohr, ao.
Two nuclei (Z = 1) are placed at x = 0.0 and 2.0, respectively. An electron is located exactly between them at x = 1.0, where we instinctively, but incorrectly, think it would want to be on the basis of electrostatic considerations. The potential energy consists of three interactions (nuclear-nuclear repulsion and two electron-nuclear attractions) and is calculated as follows:
$V = \frac{(+1)(+1)}{2} + \frac{(-1)(+1)}{2} + \frac{(-1)(+1)}{1} = -1.50 \nonumber$
Now move the electron 0.5 ao closer to one of the nuclei.
$V = \frac{(+1)(-1)}{2} + \frac{(-1)(+1)}{0.5} + \frac{(-1)(+1)}{1.5} = -2.17 \nonumber$
And so it goes, on the basis of electrostatic considerations, until the electron is inside one nucleus or the other. Although the electron was treated as a point charge in this calculation, a rigorous quantum mechanical calculation tells the same story-moving charge to the internuclear region increases electrostatic potential energy.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.04%3A_Brief_Version_of_the_Covalent_Bond_Clarified_Through_the_Use_of_the_Virial_Theorem.txt
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$\begin{matrix} \text{Total energy:} & E(R) = T(R) + V(R) \ \text{Virial theorem:} & 2 T(R) + V(R) = -R \frac{d}{dR} E(R) \ \text{Morse parameters:} & D_e = 0.761 aJ & \beta = 0.0193 pm^{-1} & R_e = 74.1 pm \end{matrix} \nonumber$
These parameters are taken from Fig. 5.6 on page 165 of McQuarrie and Simon's Physical Chemistry: A Molecular Approach.
$\begin{matrix} \text{Morse Function:} & E(R) = D_e \left[ 1 - \text{exp} \left[ - \beta \left( R - R_e \right) \right] \right]^2 - D_e \end{matrix} \nonumber$
Using equation (1) to eliminate V(R) in equation (2) yields an equation for kinetic energy as a function of the internuclear separation.
$\begin{matrix} \text{Kinetic energy:} & T(R) = -E(R) - R \frac{d}{dR} E(R) \end{matrix} \nonumber$
Using equation (1) to eliminate T(R) in equation (2) yields an equation for potential energy as a function of the internuclear separation.
$\begin{matrix} \text{Potential energy:} & T(R) = 2E(R) + R \frac{d}{dR} E(R) \end{matrix} \nonumber$
John C. Slater [J.Chem. Phys. 1933, 1, 687-691] used the virial theorem to analyze the chemical bond and was among the first to note the importance of kinetic energy in covalent bond formation. He wrote,
...this theorem gives a means of finding kinetic and potential energy separately for all configurations of the nuclei, as soon as the total energy is known, from experiment or theory.
These important equations (4 and 5) determine the mean kinetic and potential energies as functions of R, one might almost say experimentally, directly from the curves of E as a function of R which can be found from band spectra. The theory is so simple and direct that one can accept the results without question....
John Winn [J. Chem. Phys. 1981, 74, 608-611] offered the following interpretation of the energy profile shown below.
As the atoms approach, the potential energy rises (electrons are moving away from nuclei) and the kinetic energy falls (as delocalization begins). In the vicinity of R/Re = 2, this trend reverses. The kinetic energy increases as the wave function is localized ..., but the potential energy falls as the charge is now brought nearer both nuclei. Only at R/Re < 0.5 does nuclear repulsion cause the potential energy to increase and contribute to the total repulsion energy.
$\begin{matrix} \text{Plot range:} & R = 30 pm,~35 pm .. 400 pm \end{matrix} \nonumber$
The following table provides quantitative support for Winn's analysis.
$\begin{pmatrix} o & \text{Step 1} & \text{Step 2} & \text{Overall} \ \frac{ \Delta T}{aJ} & -0.472 & 1.233 & 0.761 \ \frac{ \Delta V}{aJ} & 0.166 & -1.688 & -1.522 \ \frac{ \Delta E}{aJ} & -0.306 & -0.455 & -0.761 \ \frac{R}{pm} & 151 & 74.1 & 74.1 \ \text{Type} & \text{Molecular} & \text{Atomic} & \text{Molecular + Atomic} \end{pmatrix} \nonumber$
This type of analysis is valid for any diatomic molecule for which Morse paramters are available.
$\begin{matrix} \text{Conversion factors:} & pm = 10^{-12} m & aJ = 10^{-18} joule \end{matrix} \nonumber$
3.06: The Covalent Bond According to Slater and Ruedenberg
John Slater pioneered the use of the virial theorem in interpreting the chemical bond in a benchmark paper published in the inaugural volume of the Journal of Chemical Physics (1). This early study indicated that electron kinetic energy played an important role in bond formation. Thirty years later Klaus Ruedenberg and his collaborators published a series of papers (2,3,4) detailing the crucial role that kinetic energy plays in chemical bonding, thereby completing the project that Slater started. This Mathcad worksheet recapitulates Slater's use of the virial theorem in studying chemical bond formation and summarizes Ruedenberg final analysis.
Equation [1] gives the virial theorem for a diatomic molecule as a function of internuclear separation, while equation [2] is valid at the energy minimum.
$2T(R) +V(R) = =R \frac{d}{dR} E(R) \nonumber$
$E \left( R_e \right) = \frac{V \left( R_e \right)}{2} = -T \left( R_e \right) \nonumber$
At non-equilibrium values for the internuclear separation the virial equation can be used, with E = T + V, to obtain equations for the kinetic and potential energy as a function of the inter-nuclear separation.
$T(R) = -E(R) - R \frac{d}{dR} E(R) \nonumber$
$V(R) = 2 E(R) + R \frac{d}{dR} E(R) \nonumber$
Thus, if E(R) is known one can calculate T(R) and V(R) and provide a detailed energy profile for the formation of a chemical bond. E(R) can be provided from spectroscopic data or from ab initio quantum mechanics. In this examination of the chemical bond we employ the empirical approach and use spectroscopic data for the hydrogen molecule to obtain the parameters (highlighted below) for a model of the chemical bond based on the Morse function (5). However, it should be noted that quantum mechanics tells the same story and yields an energy profile just like that shown in the following figure.
$\begin{matrix} R = 30, 30.2 .. 400 & D_e = 0.761 & \beta = 0.0193 & R_e = 74.1 & E(R) = \left[ D_e \left[ 1 - \text{exp} \left[ - \beta \left( R - R_e \right) \right] \right] ^2 - D_e \right] \end{matrix} \nonumber$
The dissociation energy, De, is given in atta (10-18) joules, the internuclear separation in picometers, and the constant β in inverse picometers.
This energy profile shows that as the internuclear separation decreases, the potential energy rises, falls, and then rises again. The kinetic energy first decreases and then increases at about the same internuclear distance that the potential energy begins to decrease.
As the molecular orbital is formed at large R constructive interference between the two overlaping atomic orbitals draws electron density away from the nuclear centers into the internuclear region. The potential energy rises as electron density is drawn away from the nuclei, but the total energy decreases because of a larger decrease in kinetic energy due to charge delocalization. Thus a decrease in kinetic energy funds the initial build up of charge between the nuclei that we normally associate with chemical bond formation.
Following this initial phase, at an internuclear separation of about 180 pm the potential energy begins to decrease and the kinetic energy increases, both sharply (eventually), while the total energy continues to decrease gradually. This is an atomic effect, not a molecular one as Ruedenberg so clearly showed. The initial transfer of charge away from the nuclei and into the bond region allows the atomic orbitals to contract significantly (α increases) causing a large decrease in potential energy because the electron density has moved, on average, closer to the nuclei. The kinetic energy increases because the orbitals are smaller and kinetic energy increases inversely with the square of the orbital radius.
An energy minimum is reached while the potential energy is still in a significant decline (6), indicating that kinetic energy is the immediate cause of a stable bond and the molecular ground state in H2. The final increase in potential energy which is due mainly to nuclear-nuclear repulsion, and not electron-electron repulsion, doesn't begin until the internuclear separation is less than 50 pm, while the equilibrium bond length is 74 pm . Thus the common explanation that an energy minimum is reached because of nuclear-nuclear repulsion does not have merit.
The H2 ground state, E = -0.761 aJ, is reached at an internuclear separation of 74 pm (1.384 a0). In light of the previous arguments it is instructive to partition the total H2 electron density into atomic and molecular contributions. Each electron is in a molecular orbital which is a linear combination of hydrogenic 1s orbitals, as is shown below.
$\begin{matrix} \Psi_{mo} = \frac{ \Psi_{1sa} + \Psi_{1sb}}{ \sqrt{2 + 2S_{ab}}} & \text{where} & \Psi_{1sa} = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} \left( - \alpha r_a \right) & \Psi_{1sb} = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} \left( - \alpha r_b \right) \end{matrix} \nonumber$
The total electron density is therefore 2ΨMO2. The atomic, or non-bonding, electron density is given by the following equation, which represents the electron density associated with two non-interacting atomic orbitals.
$\rho_n = 2 \left( \left| \frac{ \Psi_{1sa} + i \Psi_{1sb}}{ \sqrt{2}} \right| \right) = \Psi_{1sa}^2 + \Psi_{1sb}^2 \nonumber$
Clearly, the bonding electron density must be the difference between the total electron density and the non-bonding, or atomic, electron density.
$\rho_b = \Psi_{MO}^2 - \rho_n = \rho_t - \rho_n \nonumber$
These three terms are plotted along the bond axis in the figure below. Alpha is the optimum orbital scale factor, Sab is the overlap integral, and R is the equilibrium internuclear distance in atomic units.
$\begin{matrix} \alpha = 1.197 & S_{ab} = 0.681 & R = 1.384 & y = 0 & z = 0 & x = -3, -2.99 .. 5 \end{matrix} \nonumber$
$\rho_t ( \alpha,~x,~y,~z) = \frac{ \frac{ \alpha^2}{ \pi} \left[ \text{exp} \left( - \alpha \sqrt{x^2 + y^2+z^2} \right) + \text{exp} \left[ - \alpha \sqrt{(x-R)^2 +y^2+z^2} \right] \right]^2}{1 + S_{ab}} \nonumber$
$\rho_n ( \alpha,~x,~y,~z) = \frac{ \alpha^2}{ \pi} \left[ \text{exp} \left( - \alpha \sqrt{x^2 + y^2+z^2} \right) + \text{exp} \left[ - \alpha \sqrt{(x-R)^2 +y^2+z^2} \right] \right]^2 \nonumber$
$\rho_b ( \alpha,~x,~y,~z) = \rho_t ( \alpha,~x,~y,~z) - \rho_n ( \alpha,~x,~y,~z) \nonumber$
The bonding electron density illustrates that constructive interference accompanying atomic orbital overlap transfers charge from the nuclei into the internuclear region, while the non-bonding density clearly shows the subsequent atomic orbital contraction which draws some electron density back toward the nuclei.
Literature cited:
1. Slater, J. C. J. Chem. Phys. 1933, 1, 687-691.
2. Ruedenberg, K. Rev. Mod. Phys. 1962, 34, 326-352.
3. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 54, 1495-1511; 1971, 55, 5804-5818.
4. Feinberg, M. J.; Ruedenberg, K.; Mehler, E. L. Adv. Quantum Chem. 1970, 5, 27-98.
5. McQuarrie, D. A.; Simon, J. D. Physical Chemistry: A Molecular Approach, University Science Books, Sausalito, CA, 1997, p. 165.
6. Slater, J. C. Quantum Theory of Matter, Krieger Publishing, Huntington, N.Y., 1977, pp. 405-408.
7. Rioux, F. The Chemical Educator, 1997, 2, No. 6.
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John C. Slater [J. Chem. Phys. 1933, 1, 687-691] used the virial theorem to analyze the chemical bond and was the first to recognize the importance of kinetic energy in covalent bond formation. He wrote,
...this theorem gives a means of finding kinetic and potential energy separately for all configurations of the nuclei, as soon as the total energy is known, from experiment or theory.
The kinetic energy, potential energy and total energy of the hydrogen molecule are calculated as a function of R using the virial theorem and a Morse function for the total energy based on experimental parameters.
Total energy:
$E(R) = T(R) + V(R) \nonumber$
Virial theorem:
$2 T(R) + V(R) = - R \frac{d}{dR} E(R) \nonumber$
Morse parameters:
$\begin{matrix} D_e = 0.761 aJ & \beta = 0.0193 pm^{-1} & R_e = 74.1 pm \end{matrix} \nonumber$
These parameters are taken from Fig. 5.6 on page 165 of McQuarrie and Simon's Physical Chemistry: A Molecular Approach.
Morse function:
$E(R) = D_e \left[ 1 - \text{exp} \left[ - \beta \left( R- R_e \right) \right] \right]^2 - D_e \nonumber$
Using equation (1) to eliminate V(R) in equation (2) yields an equation for kinetic energy as a function of the internuclear separation. Using equation (1) to eliminate T(R) in equation (2) yields an equation for potential energy as a function of the internuclear separation.
$T(R) = -E(R) - R \frac{d}{dR} E(R) \nonumber$
$V(R) = 2E(R) + R \frac{d}{dR} E(R) \nonumber$
These important equations determine the mean kinetic and potential energies as functions of R, one might almost say experimentally, directly from the curves of E as a function of R which can be found from band spectra. The theory is so simple and direct that one can accept the results without question....
The covalent bond in H2 can be thought of as arising from the overlap of 1s atomic orbitals forming a molecular orbital. As the atoms approach, the potential energy rises because constructive interference of the overlapping atomic orbitals draws electron density away from the nuclear centers into the bond region. Kinetic energy decreases due to delocalization of the electrons over the entire molecular orbital. At an inter-nuclear separation of 151 pm this trend reverses as the atomic orbitals shrink bringing electron density closer to both nuclei. It is the sharp increase in kinetic energy at 74.1 pm due to the shrinking of the molecular orbital that causes the energy minimum and the formation of a stable bond; potential energy is still in a steep decline and doesn't begin to increase (mainly due to nuclear repulsion) until 40 pm. This analysis is based in part on a paper by John Winn [J. Chem. Phys. 1981, 74, 608-611].
The following table provides quantitative support for Winn's analysis. Step 1 occurs from large inter-nuclear separation to 151 pm and is molecular in character. Step 2 is the change from 151 pm to 74.1 pm and is basically atomic as electron density is drawn closer to the nuclei and out of the inter-nuclear bond region.
$\begin{pmatrix} ' & \text{Step 1} & \text{Step 2} & \text{Overall} \ \frac{ \Delta T}{aJ} & -0.472 & 1.233 & 0.761 \ \frac{ \Delta V}{aJ} & 0.166 & -1.688 & -1.522 \ \frac{ \Delta E}{aJ} & -0.306 & -0.455 & -0.761 \ \frac{R}{pm} & 151 & 74.1 & 74.1 \ \text{Type} & \text{Molecular} & \text{Atomic} & \text{Molecular + Atomic} \end{pmatrix} \nonumber$
Conversion factors:
$\begin{matrix} pm = 10^{-12} m & aJ = 10^{-18} joule \end{matrix} \nonumber$
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Atomic and molecular stability and spectroscopy, and the nature of the chemical bond cannot be explained using classical physics: quantum mechanical principles are required. Bohr was a pioneer in an effort to apply an early version of quantum theory to these important issues with his models of the hydrogen atom and hydrogen molecule. Of course, Bohr's approach became "old" quantum mechanics and was abandoned in the 1920s with the creation of a "new" quantum mechanics by Heisenberg and Schrödinger and their collaborators. This tutorial deals exclusively with the chemical bond and summarizes Slater's contribution to our current understanding of the energetics of its formation using the virial theorem. Since the acceptance of the "new" quantum mechanics many others have contributed to the interpretation of chemical bond formation, but like Bohr, Slater was an insightful pioneer.
In a seminal paper, J. Chem. Phys. 1933, 1, 687-691, John C. Slater used the virial theorem to analyze chemical bond formation and was the first (in my opinion) to recognize the importance of kinetic energy in covalent bond formation. Regarding the universally valid virial theorem he wrote,
...this theorem gives a means of finding kinetic and potential energy separately for all configurations of the nuclei, as soon as the total energy is known, from experiment or theory.
The purpose of this tutorial is to demonstrate the validity of this assertion. The experimental method employs a Morse function for the energy of the hydrogen molecule ion parametrized using spectroscopic data. The theoretical approach is based on an ab initio LCAO-MO calculation of the energy based on a molecular orbital consisting a superposition of scaled hydrogen atomic orbitals.
Experimental Method
The kinetic energy, potential energy and total energy of the hydrogen molecule ion are calculated as a function of R using the virial theorem and a Morse function for the total energy based on experimental parameters.
Total energy:
$E(R) = T(R) + V(R) \nonumber$
Virial theorem:
$2T(R) + V(R) = -R \frac{d}{dR} E(R) \nonumber$
Morse parameters:
$\begin{matrix} D_e = 0.103 E_h & R_e = 2.003 a_0 & \beta = 0.708 a_0^{-1} \end{matrix} \nonumber$
These parameters are from Levine, I. N. Quantum Chemistry, 6th ed., p. 400.
Morse function:
$E(R) = D_e \left[ 1 - \text{exp} \left[ - \beta \left( R - R_e \right) \right] \right]^2 - D_e \nonumber$
Using equation (1) to eliminate V(R) in equation (2) yields an equation for kinetic energy as a function of the internuclear separation. Using equation (1) to eliminate T(R) in equation (2) yields an equation for potential energy as a function of the internuclear separation.
$T(R) - =E(R) - R \frac{d}{dR} E(R) \nonumber$
$V(R) = 2E(R) + R \frac{d}{dR} E(R) \nonumber$
These important equations determine the mean kinetic and potential energies as functions of R, one might almost say experimentally, directly from the curves of E as a function of R which can be found from band spectra. The theory is so simple and direct that one can accept the results without question....
While the Morse calculation has an empirical flavor to it, its results are interpreted in terms of rudimentary quantum theory concepts. This energy profile shows that as the protons approach, the potential energy rises because molecular orbital formation (constructive interference) draws electron density away from the nuclei into the internuclear region. The kinetic energy decreases because molecular orbital formation brings about electron delocalization.
At about 4.1 a0 this trend reverses as atomic orbital contraction begins. The kinetic energy rises because the volume occupied by the electron decreases and the potential energy decreases for the same reason: a smaller electronic volume brings the electron closer to the nuclei on average. Only at R ~ 1a0 does nuclear repulsion cause the total potential energy to increase and contribute to the repulsion energy of the molecule. In other words, the kinetic energy increase is the immediate cause of the energy minimum or ground state.
Theoretical Method
The following LCAO-MO calculation uses a superposition of scaled 1s atomic orbitals and yields the following result for the energy of the hydrogen molecule ion as a function of the internuclear distance and the orbital scale factor.
$\begin{matrix} 1s_a = \sqrt{ \frac{ \alpha^3}{ \pi} \text{exp} ( - \alpha r_a )} & s_b = \sqrt{ \frac{ \alpha^3}{ \pi} \text{exp} ( - \alpha r_b )} & S_{ab} = \int 1s_a 1s_b d \tau & \Psi_{mo} = \frac{1s_a + 1s_b}{ \sqrt{2 + 2S_{ab}}} \end{matrix} \nonumber$
$E ( \alpha,~R) = \frac{ - \alpha^2}{2} + \frac{ \left[ \alpha^2 - \alpha - \frac{1}{R} +\frac{1 + \alpha R}{R} \text{exp} (-2 \alpha R) + \alpha ( \alpha - 2) (1 + \alpha R) \text{exp} ( - \alpha R) \right]}{ \left[ 1 + \text{exp} \left( - \alpha,~R \right) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right) \right]} + \frac{1}{R} \nonumber$
$\begin{matrix} \alpha = 1 & \text{Energy = -2} & \text{Given} & \text{Energy} = E( \alpha,~ R) & \frac{d}{d \alpha} E( \alpha,~R) = 0 & \text{Energy(R) = Find}( \alpha,~\text{Energy}) \end{matrix} \nonumber$
As noted above, using $E = T+V$ with the virial theorem $2T + V = -R \frac{d}{dR}E$ leads to the following expressions for T and V.
$\begin{matrix} T(R) = - \text{Energy}(R)_1 -R \frac{d}{dR} \text{Energy}(R)_1 & V(R) = 2 \text{Energy}(R)_1 + R \frac{d}{dR} \text{Energy} (R)_1 & R = .2, .25 .. 10 \end{matrix} \nonumber$
It is clear that both methods lead to very similar energy profiles. The only significant difference is that the Morse calculation leads to a lower energy minimum, -0.1026 Eh versus -0.0865 Eh for the molecular orbital calculation. Therefore, the interpretation provided for the Morse energy profile is valid for the LCAO-MO profile.
$\begin{matrix} \text{Conversion factors:} & a_o = 5.29177(10)^{-11} m & E_h = 4.359748(10)^{-18} \text{ joule} \end{matrix} \nonumber$
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A simple charge cloud model for H2 is shown below (1,2). Two protons, separated by distance D, are symmetrically embedded in a uniform‐density, two‐electron spherical charge cloud of radius R.
In atomic units (h/2π = me = e = 4πε0 = 1) the total electronic energy based on this model is,
$E = \frac{9}{4R^2} - \frac{24}{5R} + \frac{D^2}{2R^3} + \frac{1}{D} \nonumber$
The various energy components are identified below (3).
$\begin{matrix} \text{Electron kinetic energy:} & T = \frac{9}{4R^2} \ \text{Electron-nucleus potential energy:} & V_{ne} = \frac{-6}{R} + \frac{D^2}{2R^3} \ \text{Electron-electron potential energy:} & V_{ee} = \frac{6}{5R} \ \text{Nucleus-nucleus potential energy:} & V_{nn} = \frac{1}{D} \end{matrix} \nonumber$
The total electronic energy is minimized with respect to R for a series of internuclear separations, D, to generate an energy curve that can be used with the virial theorem to generate kinetic and potential energy contributions.
$\begin{matrix} \text{Seed values for the minimization algorithm:} & R = 1 & E = -2 \end{matrix} \nonumber$
$\begin{matrix} \text{Given} & E = \frac{9}{4R^2} - \frac{24}{5R} + \frac{D^2}{2R^3} + \frac{1}{D} & \frac{d}{dR} \left( \frac{9}{4R^2} - \frac{24}{5R} + \frac{D^2}{2R^3} + \frac{1}{D} \right) = 0 & \text{E(D) = Find(R, E)} \end{matrix} \nonumber$
Substitution of $E(D) = T(D) + V(D)$ into the virial relation $2T(D) + V(D) = -D \frac{d}{dD} E(D)$ yields expressions for the kinetic and potential energies given below.
$\begin{matrix} D = .1, .2 .. 6 & \text{Kinetic energy:} & T(D) = -E(D)_1 - D \frac{d}{dD} E(D)_1 \ ~ & \text{Potential energy:} & V(D) = 2E(D)_1 + D \frac{d}{dD} E(D)_1 \end{matrix} \nonumber$
This graphic display of the total energy and its kinetic and potential energy components clearly shows that molecular stability depends on electron kinetic energy. The immediate cause of the total energy minimum is a rise in electron kinetic energy. In other words potential energy is still decreasing at D = 1.364, and doesnʹt start to rise sharply, due to electron and nuclear repulsion, until D = 0.75. So, although it is commonly thought and taught, electron and nuclear repulsion are not the immediate cause of atomic and molecular stability.
Minimization of the total energy simultaneously with respect to D and R locates the energy ground state and the equilibrium values of D and R precisely.
$\begin{matrix} \text{Seed values for the energy minimization:} & D = 1 & R = 3 \end{matrix} \nonumber$
$\begin{matrix} E(R,~D) = \frac{9}{4R^2} - \frac{24}{5R} + \frac{D^2}{2R^3} + \frac{1}{D} & \begin{pmatrix} R \ D \end{pmatrix} = \text{Minimize(E, R, D)} & \begin{pmatrix} R \ D \end{pmatrix} = \begin{pmatrix} 1.364 \ 1.364 \end{pmatrix} & \text{E(R, D) = -1.21} \end{matrix} \nonumber$
Show that the virial theorem is obeyed:
$\left| \frac{V(R)}{T(D)} \right| = 2 \nonumber$
The calculated internuclear separation, 1.364 a0, is in good agreement with the experimental bond length of 1.40 a0. However, the calculated ground state energy ‐1.21 Eh is lower than the experimental value ‐1.17 Eh. Thus in this calculation the virial theorem is satisfied, but the variational principle is violated.
It is the authorʹs opinion that the charge cloud approach is a primitive version of density functional theory (DFT), because as can be seen above, all energy components are calculated in terms of the electron density, the signature of the DFT method.
Literature cited:
1. G. F. Neumark, Microfilm Abstr. 11, 834 (1951).
2. L. M. Kleiss, Diss. Abstr. 14, 1562 (1954).
3. F. Rioux, Am. J. Phys. 44, 56 (1976)
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Introduction
Ruedenbergʹs innovative analysis of the covalent bond (1‐2) has received considerable attention in the pedagogical literature (3‐13) as well as a number of excellent reviews in the primary literature that are accessible to non‐specialists (14‐16). The purpose of this study is analyze three mechanisms for covalent bond formation in H2+ that are suitable for an undergraduate course in quantum chemistry and reveal Ruedenbergʹs central message ‐ a full understanding of the nature of the chemical bond requires consideration of the role of electron kinetic energy.
All three mechanisms postulate a single intermediate state for the reaction for H2+ bond formation.
$H + H^+ \rightarrow H_2^+ \nonumber$
Scaled hydrogenic wave functions will be used to calculate the initial atomic state, the final molecular state and the intermediate states. Mathematical details can be found in the Appendix.
The Hydrogen Atom
Using the following 1s orbital
$1s( \alpha,~R) = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( -\alpha,~R) \nonumber$
in a variational calculation on the hydrogen atom yields the following result for the energy in atomic units.
$E_H = \frac{ \alpha^2}{2} - \alpha \nonumber$
The first term is electron kinetic energy and the second term is electron‐nucleus potential energy. Minimization of EH with respect to α yields α = 1, and the following result for the ground state energy.
$E_H = T+V = 0.50 - 1.0 = -0.50 \nonumber$
The Hydrogen Ion Molecule
The H2+ molecular orbital is written as a linear superposition of scaled hydrogenic orbitals centered on the two hydrogen nuclei, a and b.
$\Psi_{bmo} = \frac{1s_a ( \alpha,~r_a) + 1s_b ( \alpha,~r_b)}{ \sqrt{2 + 2S_{ab} ( \alpha,~R_{ab})}} \nonumber$
Sab is the overlap integral,
$S_{ab} ( \alpha,~R_{ab} = \text{exp} \left( - \alpha R_{ab} \right) \left[ 1 + \alpha R_{ab} + \frac{ \alpha^2 R^2}{3} \right] \nonumber$
A variational calculation using Ψbmo yields the results in the following table.
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ o & H_{atom} + H_{ion} & o & \Psi_{mo} = \frac{a + b}{ \sqrt{2 + 2S}} \ o & \alpha = 1 & \Delta & \frac{ \alpha = 1.238}{R = 2.003} \ T & 0.5 & 0.0865 & 0.5865 \ V & -1.0 & -0.1730 & -1.1730 \ V_{ne} & -1.0 & -0.6723 & -1.623 \ V_{nn} & 0 & 0.4993 & 0.4993 \ E & -0.50 & -0.0865 & -0.5865 \end{pmatrix} \nonumber$
Table 1 shows that the atomic and molecular states individually satisfy the virial theorem (E = V/2 = ‐T), as does the bond formation process (ΔE = ΔV/2 = -ΔT). At first glance Table 1 and the virial theorem suggest that chemical bonding is governed solely by electrostatics. In bond formation, in the transition from atoms to a molecule, kinetic energy increases, potential energy decreases, and total energy decreases. From this perspective, clearly potential energy must be the key factor in the formation of a stable molecule, because it has the same sign as the change in total energy, and a decrease in energy is the signature of stability. That this sort of reasoning leads to an incorrect interpretation of the covalent bond will be illustrated by the mechanistic analyses that follow.
Mechanism I
The first mechanism is one that might appear familiar to general chemistry students. For example, when we describe the bonding in methane to undergraduates we generally invoke a mechanism that uses the concepts of atomic promotion [2s22p2 ‐‐> 2s12p3], hybridization [2s12p3 ‐‐> (sp3 hybrids)4], and bond formation through the overlap of atomic orbitals. The H2+ bond formation mechanism described here consist of just two steps: atomic promotion and covalent bond formation (H + H+ ‐‐> H* + H+ ‐‐> H2+). This simple mechanism has previously been discussed in the review literature (16) and has been invoked in a study of the covalent bond in H2 (10).
$H ( \alpha = 1) + H^+ \rightarrow ( \alpha = 1.238) + H^+ \rightarrow H_2^+ ( \alpha = 1.238,~R = 2.003) \nonumber$
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Intermediate}}{ \text{Excited Atomic State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ o & H_{atom} + H_{ion} & o & H_{atom} + H_{ion} & o & \Psi_{mo} = \frac{a + b}{ \sqrt{2 + 2S}} \ o & \alpha = 1 & \Delta & \alpha = 1.238 & \Delta & \frac{ \alpha = 1.238}{R = 2.003} \ T & 0.50 & 0.2663 & 0.7663 & -0.1798 & 0.5865 \ V & -1.0 & -0.2380 & -1.2380 & 0.0650 & -1.1730 \ V_{ne} & -1.0 & -0.2380 & -1.2380 & -0.4343 & -1.6723 \ V_{nn} & 0 & 0 & 0 & 0.4993 & 0.4993 \ E & -0.50 & 0.0283 & -0.4717 & -0.1148 & -0.5865 \end{pmatrix} \nonumber$
In the first step the hydrogen atom orbitals prepare for bonding by contracting from α = 1 to α = 1.238, the optimum value of the final molecular wave function. This step is atomic and endoergic (+0.0283 Eh), increasing the kinetic energy more (+0.2663 Eh) than it decreases the potential energy (‐0.2380 Eh). The potential energy decreases because the electron is drawn closer to the nucleus. The kinetic energy increases because of the increased confinement of the electron in the contracted atomic orbital ‐ kinetic energy is inversely proportional to the square of the average distance of the electron from the nucleus, or inversely proportional to volume of the electron distribution raised to the 2/3 power, V‐2/3.
The second step consists of the formation of a molecular wave function by the superposition (linear combination) of the promoted atomic orbitals. The electron density of this molecular orbital is,
$\left( \left| \Psi_{bmo} \right| \right)^2 = \left( \left| \frac{a+b}{ \sqrt{2 + 2S}} \right| \right)^2 = \frac{a^2 + 2ab + b^2}{2 + 2s} \nonumber$
Molecular orbital formation distributes the electron density over the entire molecule and this charge delocalization brings about a significant decrease in kinetic energy (‐0.1798 Eh). Potential energy increases slightly (0.0650 Eh) because nuclear repulsion (0.4993 Eh) more than offsets the increase in electron‐nuclear attraction (‐0.4343 Eh) due to the fact that the electron now interacts with both nuclei. This step is exoergic (‐0.1148 Eh) because kinetic energy decreases more than potential energy increases. In other words, in this mechanism charge delocalization funds (drives) the formation of the chemical bond.
Mechanism II
Mechanism II postulates an intermediate non‐bonding molecular state that has the equilibrium values of α and R. The non‐bonding character this state is revealed by examining its electron density.
$\begin{matrix} \left( \left| \Psi_{nmo} \right| \right)^2 = \left( \left| \frac{a+ib}{ \sqrt{2}} \right| \right)^2 = \frac{a^2}{2} + \frac{b^2}{2} \end{matrix} \nonumber$
The electron density has no interference term and simply places half the electron density in each atomic orbital. The energy of the non‐bonding state as a function of α and R and is provided in the Appendix. This intermediate state subsequently relaxes to the final molecular bonding state. The calculations for this mechanism are summarized in Table III.
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Intermediate}}{ \text{Non Bonding State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ o & H_{atom} + H_{ion} & o & \Psi_{nmo} = \frac{a+ib}{ \sqrt{2}} & o & \Psi_{mo} = \frac{a + b}{ \sqrt{2 + 2S}} \ o & \alpha = 1 & \Delta & \frac{ \alpha = 1.238}{R = 2.003} & \Delta & \frac{ \alpha = 1.238}{R = 2.003} \ T & 0.50 & 0.2663 & 0.7663 & -0.1798 & 0.5865 \ V & -1.0 & -0.2258 & -1.2258 & 0.0528 & -1.1730 \ V_{ne} & -1.0 & -0.7251 & -1.7251 & 0.0528 & -1.6723 \ V_{nn} & 0 & 0.4993 & 0.4993 & 0 & 0.4993 \ E & -0.50 & 0.0407 & -0.4595 & -0.1270 & -0.5865 \end{pmatrix} \nonumber$
Formation of the non‐bonding molecular state is endoergic (+0.0407 Eh) because orbital contraction increases kinetic energy (+0.2664 Eh) more than it decreases potential energy (‐0.2258 Eh). The kinetic energy increase for this intermediate state is identical to that in Mechanism I. The potential energy decrease is less than that for Mechanism I because nuclear‐nuclear repulsion (+0.4993 Eh) in the intermediate state is greater than ʺelectron‐other‐nucleusʺ attraction [‐0.4871 Eh = ‐0.7251 Eh ‐ (‐0.2380 Eh)]
The subsequent relaxation to the bonding molecular state is exoergic (‐0.1270 Eh) because charge delocalization decreases kinetic energy (‐0.1799 Eh) more than charge redistribution to the bond region increases electron‐nucleus potential energy (+0.0528 Eh). As in Mechanism I we have an endoergic atomic step followed by a exoergic molecular step.
This mechanism clarifies an important issue. It is commonly thought, and frequently taught, that from a potential energy point of view the internuclear region is the most favorable place for electron density because it is midway between to nuclei. However, this mechanism clearly shows that the redistribution of charge into the internuclear region due to the interference term in the bonding electron density occurs with an increase in potential energy. That this is not some arcane quantum mechanical phenomenon is revealed by a simple classical electrostatic calculation [11]. From a purely electrostatic perspective the prefered location for an electron is in the nucleus.
If a stable chemical bond is due to the build‐up of charge density in the internuclear region, then this mechanism shows that kinetic energy plays a crucial role in bringing about this effect.
Mechanism III
The third mechanism postulates an initial molecular state, rather than a promoted atomic state or a non‐bonding molecular state for the intermediate. The initial molecular state has the equilibrium bond length, but the atomic value for the orbital scale factor, α. The calculations based on this mechanism are shown below.
$H ( \alpha = 1) + H^+ \rightarrow H_2^+ ( \alpha = 1,~R = 2.003) \rightarrow H_2^+ ( \alpha = 1.238,~R=2.003) \nonumber$
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Intermediate}}{ \text{Bonding State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ o & H_{atom} + H_{ion} & o & \Psi_{imo} = \frac{a+b}{ \sqrt{2+2S}} & o & \Psi_{mo} = \frac{a + b}{ \sqrt{2 + 2S}} \ o & \alpha = 1 & \Delta & \frac{ \alpha = 1}{R = 2.003} & \Delta & \frac{ \alpha = 1.238}{R = 2.003} \ T & 0.50 & -0.1138 & 0.3862 & 0.2003 & 0.5865 \ V & -1.0 & 0.0599 & -0.9401 & -0.2329 & -1.1730 \ V_{ne} & -1.0 & -0.4394 & -1.4394 & -0.2329 & -1.6723 \ V_{nn} & 0 & 0.4993 & 0.4993 & 0 & 0.4993 \ E & -0.5 & -0.0539 & -0.5539 & -0.0326 & -0.5865 \end{pmatrix} \nonumber$
The intermediate molecular state is calculated at the equilibrium bond length (R = 2.003 ao) for a molecular orbital with α = 1.0. Charge delocalization over the two nuclear centers on the formation of the molecular orbital brings about a large decrease in kinetic energy (‐0.1138 Eh). Potential energy (+0.0599 Eh) increases because nuclear repulsion (0.4993 Eh) is larger, as in Mechanism II, than electron‐other‐nucleus attraction (‐0.4394 Eh). This step is exoergic and we see again that it is kinetic energy drives covalent bond formation.
In the second step of this mechanism the atomic orbitals making up the molecular orbital contract (α increases from 1 to 1.238) to achieve the final equilibrium molecular state. This orbital contraction decreases electron‐nucleus potential energy (‐0.2329 Eh) more than it increases kinetic energy (+0.2003 Eh), so this step is also exoergic. However, as Ruedenberg pointed out, this step is essentially atomic in character. Orbital contraction draws electron density out of the bond region back toward the nuclei. In other words it returns some of the charge density transferred to the bond region in the first step back to the nuclear centers.
Summary
The three mechanisms examined postulate three different intermediate states: (1) promoted atomic; (2) non‐bonding molecular; (3) intermediate molecular. However, they all tell the same story; kinetic energy plays a crucial role in chemical bond formation, and therefore viable models for the covalent bond require a consideration of both kinetic and potential energy. The virial theorem notwithstanding (as Ruedenberg has said), the formation of a chemical bond is not simply an electrostatic phenomena.
Appendix: Computational Details
$\colorbox{yellow}{Atomic energy contributions}$
$\begin{matrix} E_H ( \alpha) = \frac{ \alpha^2}{2} - \alpha & T_H ( \alpha) = \frac{ \alpha^2}{2} & V_H ( \alpha) = - \alpha \end{matrix} \nonumber$
$\colorbox{yellow}{Bonding molecular energy contributions:}$
$\begin{matrix} S ( \alpha,~R) = \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right) & T_{bmo} ( \alpha,~R) = \frac{ \alpha^2}{2} \frac{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R - \frac{ \alpha^2 R^2}{3} \right)}{1 + S( \alpha,~R)} \end{matrix} \nonumber$
$\begin{matrix} V_{bmo} ( \alpha,~R) = \frac{- \alpha - \frac{1}{R} + \frac{1}{R} (1 + \alpha R) \text{exp} (-2 \alpha R) - 2 \alpha (1+ \alpha R_ \text{exp} ( - \alpha R)}{1 + S ( \alpha,~R)} + \frac{1}{R} \ E_{bmo} ( \alpha,~R) = T_{bmo} ( \alpha,~R) + V_{bmo} ( \alpha,~R) \end{matrix} \nonumber$
$\colorbox{yellow}{Non-bonding molecular energy contributions:}$
$\begin{matrix} T_{nbmo} ( \alpha,~R) = \frac{ \alpha^2}{2} & V_{nbmo} ( \alpha,~R) = - \alpha + \left( \alpha + \frac{1}{R} \right) \text{exp} (-2 \alpha R) & E_{nbmo} ( \alpha,~R) = T_{nbmo} ( \alpha,~R) + V_{nbmo} ( \alpha,~R) \end{matrix} \nonumber$
$\colorbox{yellow}{Initial Atomic State}$
$\begin{matrix} T_H (1) = 0.5000 & V_H (1) = -1.0000 & E_H (1) = -0.5000 \end{matrix} \nonumber$
$\colorbox{yellow}{Final Molecular State:}$
$\begin{matrix} T_{bmo} (1.238,~2.003) = 0.5865 & V_{bmo} (1.238,~2.003) = -1.1730 \ E_{bmo} (1.238,~2.003) = -0.5865 \end{matrix} \nonumber$
$\colorbox{yellow}{Mechanism I - Intermediate - Excited Atomic State}$
$\begin{matrix} T_H (1.238) = 0.7663 & V_H (1.228) = -1.2380 & E_H (1.238) = -0.4717 \end{matrix} \nonumber$
$\colorbox{yellow}{Mechanism II - Intermediate = Nonbonding Molecular State}$
$\begin{matrix} E_{nbmo} (1.238,~2.0033) = -0.4595 & T_{nbmo} (1.238,~2.0033) = 0.7663 & V_{nbmo} (1.238,~2.0033) = -1.2258 \end{matrix} \nonumber$
$\colorbox{yellow}{Mechanism III - Intermediate Bonding Molecular State}$
$\begin{matrix} T_{bmo} (1,~2.003) = 0.3862 & V_{bmo} (1,~2.003) = -0.9401 & E_{bmo} (1,~2.003) = -0.5539 \end{matrix} \nonumber$
Literature cited:
1. Ruedenberg, K. Rev. Mod. Phys. 1962, 34, 326‐352. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 54, 1495‐1511. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 55, 5804‐5818.
2. Ruedenberg, K. In Localization and Delocalization in Quantum Chemistry; Chalvet, O. et al., Eds.; Reidel: Dordrecht, The Netherlands, 1975; Vol. I, pp 223‐245.
3. Harcourt, R. D.; Solomon, H.; Beckworth. J. Am. J. Phys. 1982, 50, 557‐559.
4. Baird, N. C. J. Chem. Educ. 1986, 63, 660‐664.
5. Harcourt, R. D. Am. J. Phys. 1988, 56, 660‐661.
6. Nordholm, S. J. Chem. Educ. 1988, 65, 581‐584.
7. Bacskay, G. G.; Reimers, J. R.; Nordholm, S. J. Chem. Educ. 1997, 74, 1494‐1502.
8. Weinhold, F. J. Chem. Educ. 1999, 76(8), 1141‐1145.
9. Rioux, F. Chem. Educator 1997, 2(6), 1‐14.
10. Rioux, F. Chem. Educator 2001, 6(6), 288‐290.
11. Rioux, F. Chem. Educator 2003, 8(1), 10‐12.
12. Ashkenazai, G; Kosloff, R. Chem. Educator 2006, 11(2), 1‐10.
13. Nordholm, S.; Back, A.; Bacskay, G. B. J. Chem. Educ. 2007, 84(7), 1201‐1203.
14. Kutzelnigg, W. Angew. Chem. Int. Ed. Eng. 1973, 12, 546‐562.
15. Melrose, M. P.; Chauhan, M.; Kahn, F. Theor. Chim. Acta 1994, 88, 311‐324.
16. Gordon, M. S.; Jensen, J. H. Theor. Chem. Acc. 2000, 103, 248‐251.
Added references: Recently Klaus Ruedenberg and former student (now colleague) Mike Schmidt published two exhaustive studies on the quantum mechanical principles of the covalent bond.
• ʺWhy Does Electron Sharing Lead to Covalent Bonding? A Variational Analysisʺ J. Comput. Chem. 2007, 28, 391‐410.
• ʺPhysical Understanding through Variational Reasoning: Electron Sharing and Covalent Bondingʺ J. Phys. Chem. A, 2009. 113(10), 1954‐1968.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.10%3A_Three_Mechanisms_for_Bond_Formation_in_the_Hydrogen_Molecule_Ion.txt
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The centrality of the covalent bond to chemistry is captured in a short and eloquent statement by Peter Atkins [1].
Now we come to the heart of chemistry. If we can understand what holds atoms together as molecules we may also start to understand why, under certain conditions, old arrangements change in favor of new ones. We shall understand structure, and through structure, the mechanism of change.
The barrier to understanding is that, in spite of what our textbooks teach, ʺThe chemical bond is a highly complex phenomenon which eludes all attempts at simple description [2].ʺ Charles Coulson, one of the pioneering theorists in the application of quantum mechanics to the study of chemical bonding wrote the following [3].
Sometimes it seems to me that a bond between two atoms has become so real, so tangible, so friendly that I can almost see it. And then I awake with a little shock: for a chemical bond is not a real thing: it does not exist: no one has ever seen it, no one ever can. It is a figment of our own imagination.... Here is a strange situation. The tangible, the real, the solid, is explained by the intangible, the unreal, the purely mental.
Many chemists and physicists have grappled with the intellectual challenge of understanding the chemical bond. Among those who have made significant contributions are Hellmann [4], Pauling [5], Slater [6, 7], Coulson [8] and Ruedenberg [9-12]. Among these, Ruedenbergʹs work is the most comprehensive and cogent. It has received considerable attention in the pedagogical literature [13‐23] as well as a number of excellent reviews in the primary literature that are accessible to non specialists [24-26].
Unfortunately none of these efforts to make Ruedenbergʹs analysis of the covalent bond accessible to educators has had any noticeable affect on the way the chemical bond is presented in introductory and intermediate level undergraduate textbooks [27]. Therefore, the purpose of this paper is to try again by providing a simple two-step mechanistic approach to covalent bond formation in H2+ that reveals Ruedenbergʹs central message - a full understanding of the nature of the chemical bond requires consideration of the role of electron kinetic energy, or more appropriately confinement energy [28].
The mechanism postulates a single intermediate molecular state for the H2+ bond formation reaction.
$H + H^+ \rightarrow H_2^+ \nonumber$
Scaled hydrogenic wave functions will be used to calculate the initial atomic state, and the intermediate and final molecular states. Computational details for the intermediate and final molecular states are available in the appendix.
The Initial State
Schrödingerʹs equation for the hydrogen atom, of course, is exactly soluble and the result in atomic units (h/2π = me = e = 4πε0 = 1; Eh = 4.3597x10-18 J) is given below. By convention the energy of the hydrogen ion is zero.
$E_H = T + V = 0.50E_h - 1.0E_h = -0.50E_h \nonumber$
The Hydrogen Molecule Ion
The H2+ molecular orbital is written as a linear superposition of scaled hydrogenic orbitals centered on the two hydrogen nuclei, where S is the overlap integral. (a and b label the atomic orbitals, A and B the nuclear centers which are separated by a distance R.
$\Psi_{mo} = \frac{a+b}{ \sqrt{2 + 2S}} \nonumber$
where
$\begin{matrix} a = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha r_A ) & b = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha r_B ) & S = \int ab d \tau \end{matrix} \nonumber$
This trial wave function and the appropriate energy operator lead to the following variational energy integral, expressed in Dirac notation.
$E = \left( a + b \left| T - \frac{1}{r_A} - \frac{1}{r_B} + \frac{1}{R} \right| a + b \right) [2 (1+S)]^{-1} \nonumber$
Expansion, after symmetry considerations, yields the following expression,
$E = \frac{(a |T| a) + (a|T| b) + \left( a \left| \frac{-1}{r_A} \right| a \right) + \left( b \left| \frac{-1}{r_A} \right| b \right) +2 \left( a \left| \frac{-1}{r_A} \right| b \right)}{1+S} + \frac{1}{R} \nonumber$
which is written below in short-hand notation.
$E = \frac{T_{aa} + T_{ab} + V_{Aaa} + V_{Abb} + 2V_{Vab}}{1+S} + \frac{1}{R} \nonumber$
Minimization of the energy of H2+ yields the optimum values for α and R, its ground state energy and energy components. Table 1 summarizes the results for the formation of the molecule ion from a hydrogen atom and hydrogen ion.
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ o & H_{atom} + H_{ion} & o & \Psi_{mo} = \frac{a+b}{ \sqrt{2 + 2S}} \ o & \alpha = 1 & \Delta & \frac{ \alpha = 1.238}{R = 2.003} \ T & 0.5 & 0.0865 & 0.5865 \ V & -1.0 & -0.1730 & -1.1730 \ V_{ne} & -1.0 & -0.6723 & -1.6723 \ V_{ne} (Aaa) & -1.0 & 0.1539 & -0.8461 \ V_{ne} (Abb) & 0 & -0.3329 & -0.3329 \ V_{ne} (Aab) & 0 & -0.4933 & -0.4933 \ V_{nn} & 0 & 0.4993 & 0.4993 \ E & -0.5 & -0.0865 & -0.5865 \end{pmatrix} \nonumber$
Vne(Aaa) is the interaction of the electron density centered on nucleus A with nucleus A. Vne(Abb) is the interaction of the electron density centered on nucleus B with nucleus A. Vne(Aab) is the interaction of the overlap electron density with nucleus A. These energy contributions include the companion terms involving nucleus B as justified by molecular symmetry.
Table 1 shows that the atomic and molecular states individually satisfy the virial theorem (E = V/2 = -T), and therefore so does the bond formation process (ΔE = ΔV/2 = -ΔT). At first glance Table 1 and the virial theorem suggest that chemical bonding is governed solely by electrostatics. In bond formation, in the transition from atoms to a molecule, kinetic energy increases, potential energy decreases, and total energy decreases. From this perspective it appears that potential energy must be the key factor in the formation of a stable molecule, because it has the same sign as the change in total energy, and a decrease in energy is the signature of stability. That this view is an over-simplification is shown by the energy profile (in atomic units) provided by an ab initio calculation of covalent bond formation in the hydrogen molecule ion using the trial molecular orbital given above.
This energy profile shows that consideration of kinetic energy is essential in understanding bond formation and molecular stability. First, the initial drop in total energy as the nuclei approach each other is due to a decrease in kinetic energy, because the potential energy initially increases. Second, the energy minimum (ground state) is achieved while potential energy is in a steep decline. It is a sharp increase in kinetic energy that causes the energy minimum and provides the ʺrepulsiveʺ effect necessary to counter the (still) attractive potential energy term. As the profile shows nuclear repulsion doesnʹt become dominant until well after the energy minimum has been reached. Clearly a valid model for the covalent bond requires consideration of both kinetic and potential energy.
These ab initio results are supported by a more empirical approach based on the virial theorem, a method of analysis first demonstrated by Slater in 1933 [5]. A serviceable representation of molecular energy for diatomic molecules (excluding nuclear kinetic energy, i.e. vibrational degrees of freedom) is provided by the Morse function.
$E(R) = D_e \left[ 1 - \text{exp} \left[ - \beta \left( R - R_e \right) \right] \right]^2 - D_e \nonumber$
The Morse parameters (De, Re and β) can be obtained from analysis of spectroscopic data. Therefore, using the Morse function with the general expression for the virial theorem (valid for both the classical and quantum mechanical domains)
$2T(R) + V(R) = - R \frac{d}{dR} E(R) \nonumber$
and the expression for total energy
$E(R) = T(R) + V(R) \nonumber$
leads to the following equations for kinetic and potential energy as a function of internuclear separation.
$\begin{matrix} T(R) = -E(R) - R \frac{d}{dR} E(R) & V(R) = 2E(R) + R \frac{d}{dR} E(R) \end{matrix} \nonumber$
Using spectroscopic parameters from the literature [29] for H2+ allows the generation of the following energy profile in atomic units.
The basic agreement between Figure 1 (theoretical, ab initio) and Figure 2 (experimental, spectroscopic) supports the view that both kinetic and potential energy considerations are required for a viable model of the covalent bond; models that consider only electrostatic potential energy effects are invalid and misleading.
In Figures 1 and 2 the fact that the potential energy increases, then decreases, and finally increases again as R decreases suggests that it can be partitioned into three terms. The initial decrease in kinetic energy is followed by an increase indicating that it consists of two opposing contributions. Thus, Ruedenberg analyzed the H2+ bond formation in terms of five contributions to the binding energy. In an effort to make Ruedenbergʹs ideas accessible to undergraduates, a simple two-step mechanistic approach will be used to interpret covalent bond formation.
As shown below, the mechanism postulates an intermediate molecular state at the equilibrium bond length, but with the atomic value for the orbital scale factor, α. Computational details on the intermediate state are provided in the appendix.
$H ( \alpha = 1) + H^+ \rightarrow H_2^+ ( \alpha = 1,~R = 2.003) \rightarrow H_2^+ ( \alpha = 1.238,~R = 2.003) \nonumber$
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Intermediate}}{ \text{Molecular State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ o & H_{atom} + H_{ion} & o & \Psi_{imo} = \frac{a+b}{ \sqrt{2+2S}} & o & \Psi_{mo} = \frac{a+b}{ \sqrt{2+2S}} \ o & \alpha = 1 & \Delta & \frac{ \alpha = 1}{R=2.003} & \Delta & \frac{ \alpha=1.238}{R=2.003} \ T & 0.5 & -0.1138 & 0.3862 & 0.2003 & 0.5865 \ V & -1.0 & 0.0599 & -0.9401 & -0.2329 & -1.1730 \ V_{ne} & -1.0 & -0.4394 & -1.4394 & -0.2329 & -1.6723 \ V_{ne}(Aaa) & -1.0 & 0.3693 7 -0.6307 & -0.2154 & -0.8461 \ V_{ne}(Abb) & 0 & -0.2976 & -0.2976 & -0.0353 & -0.3329 \ V_{ne}(Aab) 0 & -0.511 & -0.511 & 0.0178 & -0.4933 \ V_{nn} 0 & 0.4993 & 0.4993 & 0 & 0.4993 \ E & -0.5 & -0.0539 & -0.5539 & -0.0326 & -0.5865 \end{pmatrix} \nonumber$
The first step is exoergic and is driven by a decrease in electron kinetic energy. Charge delocalization over the two nuclear centers on the formation of the molecular orbital brings about a large decrease in kinetic energy (-0.1138 Eh) because of the larger molecular volume now available to the electron [28]. Potential energy increases (+0.0599 Eh) because nuclear repulsion (+0.4993 Eh) is larger than the decrease electron-nucleus potential energy (-0.4394 Eh), which consists of the three contributions shown in Table 2. Constructive interference between atomic orbitals during molecular orbital formation draws some electron density into the internuclear region where potential energy is higher than in the region around the nucleus. Thus, the significant increase in Vne(Aaa) (+0.3693 Eh) is the main reason that the decrease in the attractive Vne term is less than the increase in the repulsive Vnn term.
The second step is also exoergic, but is driven by a decrease in potential energy. In this step the atomic orbitals making up the molecular orbital contract (α increases from 1 to 1.238) to achieve the final equilibrium molecular state. Orbital contraction draws electron density from the bond region back toward the nuclei - it returns some of the charge density transferred to the bond region in the first step back to the nuclear centers. This is evidenced by the significant decrease in Vne(Aaa) (-0.2154 Eh) and the change in the overlap integral from 0.5856 for α = 1 to 0.4632 for α = 1.238 (see the appendix). The changes in Vne(Abb) and Vne(Aab) are relatively minor in this step. Potential energy decreases (-0.2329 Eh) because the electron is on average closer to the nuclei. Kinetic energy increases (+0.2003 Eh) because orbital contraction decreases the volume occupied by the electron [28].
This mechanism is consistent with the computational results summarized in Figures 1 and 2, and therefore, reiterates that a consideration of both kinetic and potential energy effects are required to understand the nature of the covalent bond. Electron-nucleus potential energy is the only negative (attractive) energy term and is the ʺglueʺ that holds the molecule together. Paradoxically kinetic energy plays two contradictory roles in covalent bond formation. Its initial decrease due to charge delocalization funds the build up of charge in the internuclear region and is responsible for the early drop in total energy, while its subsequent sharp increase due to orbital contraction insures a ground state and a stable molecular entity.
To summarize further we can say that the first step is molecular in character and is driven by a decrease in electron kinetic energy. The second step, orbital contraction, is atomic in character (drawing electron density back to the nuclei) and driven by a decrease in electron-nucleus potential energy.
Literature cited:
1. Atkins, P. W. Molecular Quantum Mechanics, 2nd ed.; Oxford University Press, Oxford, UK, 1983, p.250.
2. Kutzelnigg, W. Angew. Chem. internat. Edit. 1973, 12, 546-562.
3. Coulson, C. A. J. Chem. Soc. 1955, 2069.
4. Hellmann, H. Z. Phys. 1933, 35, 180.
5. Pauling, L. The Nature of the Chemical Bond, 3rd ed.; Cornell University Press, Ithaca, NY, 1960.
6. Slater, J. C. J. Chem. Phys. 1933, 1, 687-691.
7. Slater, J. C. Quantum Theory of Matter, Krieger Publishing: Untington, NY, 1977; pp 405-408.
8. Coulson, C. A. Valence, 2nd ed.; Oxford University Press, Oxford, UK, 1961.
9. Ruedenberg, K. Rev. Mod. Phys. 1962, 34, 326-352.
10. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 54, 1495-1511. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971, 55, 5804-5818.
11. Ruedenberg, K. In Localization and Delocalization in Quantum Chemistry; Chalvet, O. et al., Eds.; Reidel: Dordrecht, The Netherlands, 1975; Vol. I, pp 223-245.
12. Ruedenberg, K.; Schmidt, M. W. J. Comput. Chem. 2007, 28, 391-410. Ruedenberg, K.; Schmidt, M. W. J. Phys. Chem. A 2009, 113, 1954-1968.
13. Harcourt, R. D.; Solomon, H.; Beckworth. J. Am. J. Phys. 1982, 50, 557-559.
14. Baird, N. C. J. Chem. Educ. 1986, 63, 660-664.
15. Harcourt, R. D. Am. J. Phys. 1988, 56, 660-661.
16. Nordholm, S. J. Chem. Educ. 1988, 65, 581-584.
17. Bacskay, G. G.; Reimers, J. R.; Nordholm, S. J. Chem. Educ. 1997, 74, 1494-1502.
18. Weinhold, F. J. Chem. Educ. 1999, 76(8), 1141-1145.
19. Rioux, F. Chem. Educator 1997, 2(6), 1-14.
20. Rioux, F. Chem. Educator 2001, 6(5), 288-290.
21. Rioux, F. Chem. Educator 2003, 8(1), 10-12.
22. Ashkenazi, G; Kosloff, R. Chem. Educator 2006, 11(2), 67-76.
23. Nordholm, S.; Back, A.; Bacskay, G. B. J. Chem. Educ. 2007, 84, 1201-1203.
24. Kutzelnigg, W. Angew. Chem. Int. Ed. Eng. 1973, 12, 546-562.
25. Melrose, M. P.; Chauhan, M.; Kahn, F. Theor. Chim. Acta 1994, 88, 311-324.
26. Gordon, M. S.; Jensen, J. H. Theor. Chem. Acc. 2000, 103, 248-251.
27. An exception is the physical chemistry text by Jeff Davis which provides a concise and coherent quantum mechanically correct interpretation of the covalent bond. Unfortunately it is long out of print. Davis, J. C., Advanced Physical Chemistry: Molecules, Structure, and Spectra; The Ronald Press, New York, 1965; pp 427-428.
28. Electrons confined by attractive interaction with nuclei in atoms and molecules occupy, due to their wave nature, stationary states. They are not executing classical trajectories and consequently the term kinetic energy is inappropriate. Substitution of the de Broglie relation (p = h/λ) into the classical expression for kinetic energy yields the electronʹs confinement energy, h2/2mλ2. Because the wavelength of a confined electron is proportional to the confinement dimension, it is also proportional to cube root of the confinement volume, V1/3. Therefore the electronʹs confinement energy is proportional to V-2/3.
29. Levine, I. N. Quantum Chemistry, 6th ed., Pearson Prentice Hall, Upper Saddle River, NJ, 2009; p 400.
Appendix
$\colorbox{yellow}{Overlap integral:}$
$S( \alpha,~R) = \text{exp}( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right) \nonumber$
$\colorbox{yellow}{Kinetic energy integrals:}$
$\begin{matrix} Taa ( \alpha,~R) = \frac{ \frac{1}{2} \alpha^2}{1 + S( \alpha,~R)} & Tab( \alpha,~R) = \frac{ \frac{ \alpha^2}{2} \text{exp} ( - \alpha R) \left( 1 + \alpha R - \frac{ \alpha^2 R^2}{3} \right)}{1 + S ( \alpha,~R)} \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Potential energy integrals:} & VAaa ( \alpha,~R) = \frac{ - \alpha}{1 + S( \alpha,~R)} \end{matrix}$
$\begin{matrix} VAbb( \alpha,~R) = \frac{ - \alpha \left[ \frac{1}{ \alpha R} - \text{exp} (-2 \alpha R) \left( 1 + \frac{1}{ \alpha R} \right) \right]}{1 + S( \alpha,~R)} & VAab ( \alpha,~R) = \frac{- \alpha \text{exp} ( - \alpha R) (1 + \alpha R)}{1 + S( \alpha,~R)} \end{matrix} \nonumber$
$\colorbox{yellow}{Intermediate Molecular State}$
$\begin{matrix} \colorbox{yellow}{Values of variational parameters:} & \alpha = 1 & R = 2.003 & \colorbox{yellow}{Overlap integral:} & S( \alpha,~R) = 0.5856 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Total energy:} & Taa ( \alpha,~R) + Tab ( \alpha,~R) + VAaa ( \alpha,~R) + VAbb ( \alpha,~R) + 2 VAab ( \alpha,~R) + \frac{1}{R} = -0.5539 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Kinetic energy:} & Taa ( \alpha,~R) + Tab ( \alpha,~R) = 0.3862 & Taa ( \alpha,~R) = 0.3153 & Tab ( \alpha,~R) = 0.0709 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Electron-nucleus potential energy:} & VAaa ( \alpha,~R) + VAbb ( \alpha,~R) + 2 VAab ( \alpha,~R) = -1.4394 \end{matrix} \nonumber$
$\begin{matrix} VAaa ( \alpha,~R) = =0.6307 & VAbb ( \alpha,~R) = -0.2976 & 2VAab ( \alpha,~R) = -0.5111 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Nuclear potential energy:} & \frac{1}{R} = 0.4993 \ \colorbox{yellow}{Total potential energy:} & VAaa ( \alpha,~R) + VAbb ( \alpha,~R) + 2VAab ( \alpha,~R) + \frac{1}{R} = -0.9401 \end{matrix} \nonumber$
$\colorbox{yellow}{Final Molecular State}$
$\begin{matrix} \colorbox{yellow}{Values of variational parameters:} & \alpha = 1.238 & R = 2.003 & \colorbox{yellow}{Overlap integral:} & S ( \alpha,~R) = 0.4632 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Total Energy:} & Taa ( \alpha,~R) + Tab ( \alpha,~R) + VAaa ( \alpha,~R) + VAbb ( \alpha,~R) + 2VAab ( \alpha,~R) + \frac{1}{R} = -0.5865 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Kinetic energy:} & Taa ( \alpha,~R) + Tab ( \alpha,~R) = 0.5865 & Taa ( \alpha,~R) = 0.5237 & Tab ( \alpha,~R) = 0.0627 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Electron-nucleus potential energy:} & Vaa ( \alpha,~R) + VAbb ( \alpha,~R) + 2 VAab ( \alpha,~R) = -1.6722 \end{matrix} \nonumber$
$\begin{matrix} VAaa ( \alpha,~R) = -0.8461 & VAbb ( \alpha,~R) = -0.3329 & 2VAab ( \alpha,~R) = -0.4933 \end{matrix} \nonumber$
$\begin{matrix} \colorbox{yellow}{Nuclear potential energy:} & \frac{1}{R} = 0.4993 \ \colorbox{yellow}{Total potential energy:} & VAaa ( \alpha,~R) + VAbb ( \alpha,~R) + 2VAab ( \alpha,~R) + \frac{1}{R} = -1.1730 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.11%3A_A_Mechanistic_Approach_to_Bond_Formation_in_the_Hydrogen_Molecule_Ion.txt
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Appendix
This study of the nature of the H2+ covalent bond forms a molecular orbital as a linear combination of scaled hydrogenic orbitals, LCAO-MO. The quantum mechanical integrals necessary to carry out the required calculations are provided immediately below. Consult the first five sections of Chapter 23 of our textbook for an overview of the covalent bond in the hydrogen molecule ion. The reaction for its formation is,
$H + H^+ \rightarrow H_2^+ \nonumber$
For the hydrogen atom: <E> = -0.5; <T> = 0.5; <V> = -1.0.
The Virial Theorem is expressed in a variety of ways:
$\begin{matrix} \left| \frac{ \Delta V}{ \Delta T} \right| = 2 & \left| \frac{ \Delta E}{ \Delta T} \right| = 1 & \frac{ \Delta E}{ \Delta V} = \frac{1}{2} \end{matrix} \nonumber$
The evaluated integrals required for this exercise are provided below:
Overlap integral:
$S( \alpha,~R) = \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right) \nonumber$
Kinetic energy integrals:
$\begin{matrix} Taa ( \alpha,~R) = \frac{ \alpha^2}{2} & Tab ( \alpha,~R) = \frac{ \alpha^2}{2} \text{exp} ( - \alpha,~R) \left( 1 + \alpha R - \frac{ \alpha^2 R^2}{3} \right) \end{matrix} \nonumber$
Potential energy integrals:
$Vaa ( \alpha,~R) = - \left[ \alpha - \text{exp} ( -2 \alpha R) \left( \frac{1}{R} + \alpha \right) \right] \nonumber$
Total energy integral:
$E ( \alpha,~R) = \frac{Haa ( \alpha,~R) + Hab ( \alpha,~R)}{1+S( \alpha,~R)} \nonumber$
Bonding molecular orbital:
$\Psi_{bmo} ( \alpha,~x,~y,~z,~R) = \frac{ \sqrt{ \frac{ \alpha^3}{ \pi}} \left[ \text{exp} \left[ - \alpha \sqrt{ \left( x - \frac{R}{2} \right)^2 + y^2 + z^2} \right] + \text{exp} \left[ - \alpha \sqrt{ \left( x + \frac{R}{2} \right)^2 + y^2 + z^2} \right] \right]}{ \sqrt{2 + 2S ( \alpha,~R)}} \nonumber$
Demonstrate wavefunction is normalized:
$\int_{ - \infty}^{ \infty} \int_{ - \infty}^{ \infty} \int_{ - \infty}^{ \infty} \Psi_{bmo} (1,~x,~y,~z,~2)^2 \text{dx dy dz} = 1.0000 \nonumber$
Anti-bonding molecular orbital:
$\Psi_{amo} ( \alpha,~x,~y,~z,~R) = \frac{ \sqrt{ \frac{ \alpha^3}{ \pi}} \left[ \text{exp} \left[ - \alpha \sqrt{ \left( x - \frac{R}{2} \right)^2 + y^2 + z^2} \right] - \text{exp} \left[ - \alpha \sqrt{ \left( x + \frac{R}{2} \right)^2 + y^2 + z^2} \right] \right]}{ \sqrt{2 - 2S ( \alpha,~R)}} \nonumber$
Demonstrate wavefunction is normalized:
$\int_{ - \infty}^{ \infty} \int_{ - \infty}^{ \infty} \int_{ - \infty}^{ \infty} \Psi_{amo} (1,~x,~y,~z,~2)^2 \text{dx dy dz} = 1.0000 \nonumber$
Non-bonding molecular orbital:
$\Psi_{nmo} ( \alpha,~x,~y,~z,~R) = \frac{ \sqrt{ \frac{ \alpha^3}{ \pi}} \left[ \text{exp} \left[ - \alpha \sqrt{ \left( x - \frac{R}{2} \right)^2 + y^2 + z^2} \right] + \text{i exp} \left[ - \alpha \sqrt{ \left( x + \frac{R}{2} \right)^2 + y^2 + z^2} \right] \right]}{ \sqrt{2}} \nonumber$
Demonstrate wavefunction is normalized:
$\int_{ - \infty}^{ \infty} \int_{ - \infty}^{ \infty} \int_{ - \infty}^{ \infty} (| \Psi_{nmo} (1,~x,~y,~z,~2)^2 |)^2 \text{dx dy dz} = 1.0000 \nonumber$
The Covalent Bond in H2+
Variational Calculation of the Molecular Ground State and Bond Energy
The molecular orbital for the hydrogen molecule ion is formed as a linear combination of scaled hydrogenic 1s orbitals centered on the nuclei, a and b.
$\Psi = \frac{a+b}{ \sqrt{2 + 2S}} \nonumber$
where
$\begin{matrix} a = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha r_a ) & b = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha r_b ) & S = \int a ~b ~d \tau \end{matrix} \nonumber$
The energy operator:
$H = \frac{-1}{2} \left[ \frac{d}{dR} \left( r^2 \frac{d}{dR} \blacksquare \right) \right] - \frac{1}{r_a} - \frac{1}{r_b} + \frac{1}{R} \nonumber$
The energy integral to be minimized by the variation method:
$E = \frac{ \int (a+b) H(a+b) d \tau}{2+2S} = \frac{Haa + Hab}{1+S} \nonumber$
where
$\begin{matrix} Haa = (a|T| a) + \left( a \left| \frac{-1}{r_a} \right| a \right) + \left( a \left| \frac{-1}{r_b} \right| a \right) + \left( a \left| \frac{1}{R} \right| a \right) \ Hab = (a|T| b) + \left( a \left| \frac{-1}{r_a} \right| b \right) + \left( a \left| \frac{-1}{r_b} \right| b \right) + \left( a \left| \frac{1}{R} \right| b \right) \end{matrix} \nonumber$
and Haa = Taa + Vaa and Hab = Tab + Vab and Hba = Hab. These integrals and the overlap integral S can be found in the appendix above.
1. Plot E(α,R) vs R, for α = 1 for values of R ranging from 0.2 to 10 in increments of 0.05. You can make space by pressing . You can eliminate space by pressing . Use the markers for the x- and y-axis to find the optimum internuclear separation, R, and the ground-state energy.
Notice how shallow the energy minimum initially is. You can exagerate the minimum by plotting from -0.6 to -0.4 on the y-axis.
2. Minimize the energy with respect to R with α = 1. This will require a Given/Find solve block and seed values for α and R. Report the optimum R value (equilibrium bond distance) and the ground-state energy. Compare these calculated values to the literature values of 2.00 ao for R and -0.6029 Eh for the E.
$\begin{matrix} \alpha = 1 & R = 2 & \text{Given } \frac{d}{dR} E ( \alpha,~R) = 0 & \text{R = Find(R)} & R = 2.4928 & E ( \alpha,~R) = -0.5648\end{matrix} \nonumber$
$\begin{matrix} \frac{R-2.00}{2.00} = 24.6415 \% & & \left| \frac{E ( \alpha,~R) + .6029)}{ -.6029} \right| = 6.3143 \% & \frac{E ( \alpha,~R)}{-.6029} = 93.6875 \% \end{matrix} \nonumber$
Demonstrate that the calculated hydrogen molecule ion energy does not satisfy the Virial Theorem.
$\begin{matrix} T = \frac{Taa ( \alpha,~R) + Tab | \alpha,~R)}{1+S ( \alpha,~R)} & T = 0.3827 & V = \frac{Vaa ( \alpha,~R) + Vab ( \alpha,~R)}{1 + S( \alpha,~R)} & V = -0.9475 \ \left| \frac{V}{T} \right| = 2.4759 & \left| \frac{E ( \alpha,~R)}{T} \right| = 1.4759 & \left| \frac{E( \alpha,~R)}{V} \right| = 0.5961 \end{matrix} \nonumber$
3. Write the equation for breaking the H2+ bond. Then use the theoretical value of the ground state energy to calculate the bond energy of the hydrogen molecule ion and compare it to the experimental value, 0.1029 Eh. You may recall that the hydrogen atom ground state energy is -0.5 hartree.
$\begin{matrix} \text{Hydrogen molecule ion =} & \text{Hydrogen atom +} & \text{Hydrogen ion} \ -0.5648 & -0.500 & 0 & \text{Bond energy} = 0.0648 \end{matrix} \nonumber$
4. The first three exercises refer to a variational calculation based on the hydrogen 1s orbital (α = 1). Now treat both α and R as variational by minimizing the energy simultaneously with respect to α and R.
$\begin{matrix} \text{Given} & \frac{d}{dR} E ( \alpha,~R) = 0 & \begin{pmatrix} \alpha \ R \end{pmatrix} = \text{Find}( \alpha,~R) & \begin{pmatrix} \alpha \ R \end{pmatrix} = \begin{pmatrix} 1.2380 \ 2.0033 \end{pmatrix} & E ( \alpha,~R) = -0.5865 \end{matrix} \nonumber$
Demonstrate that the calculated hydrogen molecule ion energy does satisfy the Virial Theorem.
$\begin{matrix} T = \frac{Taa ( \alpha,~R) + Tab | \alpha,~R)}{1+S ( \alpha,~R)} & T = 0.5865 & V = \frac{Vaa ( \alpha,~R) + Vab ( \alpha,~R)}{1 + S( \alpha,~R)} & V = -1.1730 \ \left| \frac{V}{T} \right| = 2.0000 & \left| \frac{E ( \alpha,~R)}{T} \right| = 1.0000 & \left| \frac{E( \alpha,~R)}{V} \right| = 0.5000 \end{matrix} \nonumber$
5. Compare E and R with the literature values. Recalculate the H2 + bond energy and compare it with the experimental value. Comment on the agreement between theory and experiment as compared with that found in part 3.
$\begin{matrix} \text{Hydrogen Molecule Ion =} & \text{Hydrogen Atom +} \text{Hydrogen Ion} \ -0.5865 & -0.5000 & 0 & \text{Bond Energy} = 0.0865 \end{matrix} \nonumber$
$\begin{matrix} \frac{.1029 - .0865}{.1029} = 15.9378 \% & \frac{.0865}{.1029} = 84/0622 \% & \left| \frac{E ( \alpha,~R) + .6029}{-.6029} \right| = 2.7191 \% \end{matrix} \nonumber$
6. The vibrational force constant is the second derivative of the energy with respect to R evaluated at the energy minimum. Calculate the force constant and compare it to the experimental value of .103 in atomic units.
$\begin{matrix} k = \frac{d^2}{dR^2} E ( \alpha,~R) & k = 0.1409 & \frac{k}{.103} = 136.8240 \% \end{matrix} \nonumber$
7. The appendix contains the bonding and antibonding orbitals for H2+. On one graph plot Ψ for the bonding and antibonding orbitals along the bond axis (i.e. set y and z to zero). On another graph plot Ψ2 for the bonding and antibonding orbitals. Compare these orbitals.
These figures show that the electron density is greatest at the nuclear centers, and that the bonding molecular orbital has significant electron density in the internuclear region. Just how much charge is transferred to the bond region will be calculated later. The antibonding molecular orbital has a node exactly between the two nuclei and thus removes charge from the internuclear region. This, of course, is why it called an anti-bonding molecular orbital.
A Mechanism for Covalent Bond Formation in H2+
The following two-step mechanism is proposed for covalent bond formation:
• The hydrogen atom and proton form an intermediate, nonbonding state at the equilibrium bond length (R= 2.0033) with the optimum value of the decay constant (α = 1.238).
• The non-bonding state, Nnmo(a + ib), morphs into the final molecular state, Nbmo (a + b).
$\begin{pmatrix} \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Intermediate}}{ \text{Non Bonding State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} \ H_{atom} + H_{ion} & o & \Psi_{nmo} = \frac{a+ib}{ \sqrt{2}} & o & \Psi_{bmo} = \frac{a+b}{ \sqrt{2+2S}} \end{pmatrix} \nonumber$
8. We now calculate the kinetic (T), potential (V) and total energy (E) for each of the species involved in the mechanism.
In previous work we calculated the following values for the hydrogen atom. (The energy of the hydrogen ion by convention is 0).
$\begin{matrix} E_h = \frac{-1}{2} & T_h = \frac{1}{2} & V_h = -1 \end{matrix} \nonumber$
The energy of the non-bonding intermediate state is calculated as follows.
$E_{nb} = \int \frac{a-ib}{ \sqrt{2}} H \frac{a+ib}{ \sqrt{2}} d \tau = \frac{1}{2} \left( \int a H a d \tau + \int b H b d \tau \right) = Haa = Taa + Vaa \nonumber$
because
$Hbb = Haa \nonumber$
$\begin{matrix} \text{Total:} & Haa ( \alpha,~R) = -0.4594 & \text{Kinetic:} & Taa ( \alpha,~R) = 0.7663 & \text{Potential:} & Vaa ( \alpha,~R) = -1.2258 \ & & \text{Nuclear potential energy:} & \frac{1}{R} = 0.4992 \end{matrix} \nonumber$
In exercise 4 we calculated the energy contributions of the final molecular state of H2+.
$\begin{matrix} \text{Total:} & \text{Kinetic:} & \text{Potential:} \ E ( \alpha,~R) = =0.5865 & \frac{Taa ( \alpha,~R) + Tab ( \alpha,~R)}{1 + S( \alpha,~R)} = 0.5865 &\frac{Vaa ( \alpha,~R) + Vab ( \alpha,~R)}{1 + S( \alpha,~R)} = -1.1730 \end{matrix} \nonumber$
Nuclear potential energy:
$\frac{1}{R} = 0.4992 \nonumber$
We are now in a position to construct the following quantitative summary of the mechanism.
$\begin{pmatrix} o & \frac{ \text{Initial}}{ \text{Atomic State}} & o & \frac{ \text{Intermediate}}{ \text{Non Bonding State}} & o & \frac{ \text{Final}}{ \text{Molecular State}} & o \ o & H_{atom} + H_{ion} & o & \Psi_{nmo} = \frac{a+ib}{ \sqrt{2}} & o & \Psi_{bmo} = \frac{a+b}{ \sqrt{2+2S}} \ o & \alpha = 1 & \Delta & \frac{ \alpha = 1.238}{R= 2.003} & \Delta & \frac{ \alpha = 1.238}{R = 2.003} \ T & 0.50 & 0.2663& 0.7663 & -0.1798 & 0.5865 \ V & -1.0 & -0.2258 & -1.2258 & 0.0528 & -1.1730 \ V_{ne} & -1.0 & -0.7250 & -1.7250 & 0.0528 & -1.6722 \ V_{nn} & 0 & 0.4992 & 0.4992 & 0 & 0.4992 \ E & -0.50 & 0.0407 & -0.4595 & -0.1270 & -0.5865 \end{pmatrix} \nonumber$
Interpretation of the Mechanism
The first step is atomic and endoergic. It is atomic because there is no constructive interference between the atomic orbitals. It is endoergic because kinetic energy increases more than potential energy decreases. The increase in kinetic energy and the decrease in potential energy are due predominantly to atomic orbital contraction from α = 1 in the hydrogen atom to α = 1.238 in the hydrogen molecule ion.
The second step is exoergic and molecular. The atomic orbitals form a molecular orbital leading to charge redistribution relative to the non-bonding intermediate state. This redistribution causes a decrease in kinetic energy because the electron is now more delocalized, and an increase in potential energy because the electron is drawn away from the nuclear centers into the high-potential energy internuclear region.
Overall the kinetic energy increases and potential energy decreases on bond formation, and the virial theorem is satisfied. This is the source of the conclusion that covalent bonding is a potential energy effect. However, as this analysis shows, both kinetic and potential energy play important roles in covalent bond formation.
This mechanism clarifies an important issue. It is commonly thought, and frequently taught, that from a potential energy point of view the internuclear region is the most favorable place for electron density because it is midway between to nuclei. However, this mechanism clearly shows that the redistribution of charge into the internuclear region due to the interference term in the bonding electron density occurs with an increase in potential energy (0.0528 Eh ). From a purely electrostatic perspective the prefered location for an electron is in the nucleus.
The Bonding Electron Density
9. How much of the total electron density, Ψbmo2, actually contributes to covalent bonding? This question can be answered by comparing the electron density in the intermediate non-bonding state with that of the final molecular state. First we will demonstrate that Ψnmo2 = (a2 + b2)/2 and that this wave function is normalized because a and b are normalized.
$\begin{matrix} \Psi_{nmo} = \frac{a+ib}{ \sqrt{2}} & \left( \left| \frac{a+ib}{ \sqrt{2}} \right| \right)^2 \rightarrow \frac{1}{2} a^2 + \frac{1}{2} b^2 & \int \frac{a-ib}{ \sqrt{2}} \frac{a+ib}{ \sqrt{2}} d \tau = \frac{1}{2} \left( \int a^2 d \tau + \int b^2 d \tau \right) = 1 \end{matrix} \nonumber$
Ψnmo2 shows the "atomic" electron density - the electron density that would exist in the absence of constructive interference between the two atomic orbitals, 1sa and 1sb.
The non-bonding wave function is given in the appendix. Graph |Ψbmo|2 and |Ψnmo|2 along the bond axis and compare them.
10. The question posed in section 9 can now be answered. The bonding electron density can be defined as the charge redistribution that occurs in step 2 of the mechanism. Plot the difference between the total electron density and the non-bonding electron density along the bond axis. Interpret the plot. For example, where does the electron density in the internuclear region come from? What is the sign on the potential energy term for this transfer of electron density to the internuclear region? If this is an endoergic process, how is it funded?
$\rho_b ( \alpha,~x,~y) = \left( \left| \Psi_{bmo} ( \alpha,~x,~y,~z,~R) \right| \right)^2 - \left( \left| \Psi_{nmo} ( \alpha,~x,~y,~z,~R) \right| \right)^2 \nonumber$
This graph shows that electron density is being transferred from the nuclear centers (low potential energy) to the internuclear region (high potential energy). This explains why potential energy increases during step 2. The kinetic energy decreases more because this redistribution of charge also involves charge delocalization, so it is the larger decrease in kinetic energy that funds the transfer of charge to the internuclear region.
11. The electron density in the internuclear region is equal to a polynomial in the overlap integral S and is given by 0.738S(1 - S)(1 - 0.577S2). Use the value of S you calculated earlier to evaluate the amount of charge in the bonding region of H2+. Comment on this value. For example, how does its value compare with a literal interpretation of the Lewis electron pair model of chemical bonding that says that chemical bonds are formed by sharing electrons between two nuclei.
$0.738 S( \alpha,~R) (1-S( \alpha,~R)) \left( 1 - 0.577 S( \alpha,~R)^2 \right) = 16.1 \% \nonumber$
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In this analysis the molecular orbital for the hydrogen molecule ion is formed as a linear combination of scaled hydrogenic 1s orbitals centered on the nuclei, a and b.
$\begin{matrix} \Psi_{MO} = \frac{a+b}{ \sqrt{2+2S}} & \text{where} & a = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} (- \alpha,~r_a ) & b = \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha,~r_b) & S = \int ab d \tau \end{matrix} \nonumber$
The molecular energy operator in atomic units:
$H = \frac{-1}{2} \left[ \frac{d}{dr} \left( r^2 \frac{d}{dr} \blacksquare \right) \right] - \frac{1}{r_a} - \frac{1}{r_b} + \frac{1}{R} \nonumber$
The energy integral to be minimized by the variation method:
$E = \frac{ \int (a+b) H(a+b) d \tau}{2+2S} = \frac{Haa + Hab}{1+S} \nonumber$
When this integral is evaluated a two-parameter variational expression for the energy (highlighted below) is obtained.
Electron mass:
m = 1
Seed values for the variational parameter and internuclear separation:
$\begin{matrix} \alpha = 1 & R = .1 \end{matrix} \nonumber$
$E ( \alpha,~R) = \frac{ - \alpha^2}{2m} + \frac{ \frac{ \alpha^2}{m} - \alpha - \frac{1}{R} + \frac{1}{R} (1 + \alpha R) \text{exp} (-2 \alpha R) + \alpha \left( \frac{ \alpha}{m} - 2 \right) (1 + \alpha R) \text{exp} (- \alpha R)}{1 + \text{exp} ( - \alpha R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} + \frac{1}{R} \nonumber$
Minimization of the energy of the hydrogen molecule ion follows. There are two variational parameters, the orbital scale factor and the internuclear distance.
$\begin{matrix} \text{Given} & \frac{d}{d \alpha} E( \alpha,~R) = 0 & \frac{d}{dR} E ( \alpha,~R) = 0 & \begin{pmatrix} \alpha \ R \end{pmatrix} = \text{Find} ( \alpha,~R) & \begin{pmatrix} \alpha \ R \end{pmatrix} = \begin{pmatrix} 1.23803 \ 2.0033 \end{pmatrix} & E ( \alpha,~R) = -0.58651 \end{matrix} \nonumber$
The calculation yields a stable molecule ion as is shown here.
$\begin{matrix} ~ & \text{Hydrogen Molecule Ion} = & \text{Hydrogen Atom} + & \text{Hydrogen Ion} \ \text{Theory} & -0.5865 & -0.5000 & 0 & \text{Bond Energy} = 0.0865 \ \text{Experiment} & -0.6029 & -0.5000 & 0 & \text{Bond Energy} = 0.1029 \end{matrix} \nonumber$
The experimental ground state energy is -0.6029 Eh. The error in this calculation is calculate two ways: error in bond energy and error in total ground state energy.
$\begin{matrix} \frac{.1029 - .0865}{.1029} = 15.9378 \% & \left| \frac{E ( \alpha,~R)+ .6029}{-.6029} \right| = 2.71911 \% \end{matrix} \nonumber$
It is instructive to calculate the kinetic and potential energy contributions to the total energy.
$E = \int \Psi_{MO} (T+V) \Psi_{MO} d \tau = Taa + Tab + Vaa + Vab \nonumber$
$\begin{matrix} Taa ( \alpha,~R) = \frac{ \frac{ \alpha^2}{2m}}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} & Tab ( \alpha,~R) = \frac{ \frac{ \alpha^2}{2m} \text{exp} ( - \alpha,~R) \left( 1 + \alpha R - \frac{ \alpha^2 R^2}{3} \right)}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} \ Vaa ( \alpha,~R) = \frac{ \left( \frac{1}{R} + \alpha \right) \text{exp} (-2 \alpha R) - \alpha}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} & Taa ( \alpha,~R) = \frac{ \frac{-1}{3} \text{exp} ( \alpha,~R) \frac{3 \alpha R + 5 \alpha^2 R^2 - 3}{R}}{1 + \text{exp} ( - \alpha,~R) \left( 1 + \alpha R + \frac{ \alpha^2 R^2}{3} \right)} \end{matrix} \nonumber$
First we establish that these terms are correct by showing that they sum to the correct ground state energy calculated earlier.
$Taa ( \alpha,~R) + Tab ( \alpha,~R) + Vaa ( \alpha,~R) + Vab ( \alpha,~R) = -0.58651 \nonumber$
Next we establish that the virial theorem is obeyed. Quantum mechanical calculations that violate the appropriate virial theorem are not valid. For atomic and molecular systems the virial theorem is: <E> = -<T> = <V>/2.
$\begin{matrix} \text{Electron Kinetic Energy} & \text{Total Potential Energy} & \text{Virial Theorem Satisfied} \ Taa ( \alpha,~R) + Tab ( \alpha,~R) = 0.58651 & Vaa ( \alpha,~R) + Vab ( \alpha,~R) = -1.17301 & \frac{|Vaa ( \alpha,~R) + Vab ( \alpha,~R)|}{Taa ( \alpha,~R) + Tab ( \alpha,~R)} = 2 \end{matrix} \nonumber$
Next we separate the nuclear potential energy into its two components: electron-nucleus and nucleus-nucleus.
$\begin{matrix} \text{Electron Kinetic Energy} & \text{Electron-Nucleus Potential Energy} & \text{Nucleus-Nucleus Potential Energy} \ T = Taa ( \alpha,~R) + Tab ( \alpha,~R) & Ven = Vaa ( \alpha,~R) + Vab ( \alpha,~R) - \frac{1}{R} & Vnn = \frac{1}{R} \ T = 0.58651 & Ven = -1.67219 & Vnm = 0.49918 \ \frac{T}{T + |Ven|+Vnn} = 21.27 \% & \frac{|Ven|}{T + |Ven| + Vnn} = 60.63 \% & \frac{Vnn}{T + |Ven|+Vnn} = 18.10 \% \end{matrix} \nonumber$
In light of these calculations the following comments on the covalent bond in the H2+ are made. Ven is the largest and only negative term, and might be thought of as the glue holding the H2+ molecule together. However, a ground state (stable molecule) requires "energetic" opposition to the attractive Ven, otherwise molecular collapse occurs. This opposition is provided by T and Vnn. It might surprise those who think that chemical bonding is simply an electrostatic phenomenon, that nuclear repulsion makes a smaller contribution to molecular stability than electron kinetic energy.
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The covalent bond is a very challenging concept which few chemistry textbooks treat adequately. The simplest example of a covalent bond is the one‐electron bond in the hydrogen molecule ion. It plays the same role in the study of chemical bonding that the hydrogen atom plays in the study of atomic structure. The most penetrating, extensive, and perhaps controversial, analysis of the H2+ bond was provided by Ruedenberg over forty years ago (Rev. Mod. Phys. 1962, 34, 326‐352). The goal of this tutorial is quite modest in comparison; here an attempt is made to capture several of the basic features of the covalent bond with a simple, and yet plausible, one‐dimensional model.
This tutorial uses a one‐dimensional, double‐well potential (double Posch‐Teller potential) to model the H2+ covalent bond. Schrödingerʹs equation is solved numerically for the two lowest energy states using Mathcadʹs ordinary differential equation solver, and the numerical results are displayed graphically and interpreted. The calculations are carried out in atomic units: e = me = h/2π = 1.
As a numerical model, this approach is not based on the molecular orbital method (LCAO ‐ linear combination of atomic orbitals), but as we will see its results are consistent with the LCAO‐MO model for chemical bonding.
ODE Numerical Integration Algorithm
The Posch‐Teller parameters were determined by adjusting V0 so that it gave the correct H2+ ground state energy (‐0.6029 Eh) at the equilibrium bond length (2.00 a0).
$\begin{matrix} \text{Parameters:} & m = 1 & x_{max} = 8 & V_o = -1.11223 & \alpha = 1.5 & d = 1 \end{matrix} \nonumber$
$\begin{matrix} \text{Potential energy:} & V(x) = V_o \left[ \frac{1}{ \cosh [ \alpha (x-d)]^2} + \frac{1}{ \cosh [ \alpha (x+d)]^2} \right] \end{matrix} \nonumber$
Solve Schrödingerʹs equation numerically, and normalize and display the solution:
$\begin{matrix} \text{Given} & \frac{-1}{2m} \frac{d^2}{dx^2} \Psi (x) + V(x) \Psi (x) = E \Psi (x) & \Psi \left( - x_{max} \right) = 0 & \Psi ' \left( - x_{max} \right) = 0.1 \end{matrix} \ & \Psi = \text{Odesolve} \left( x,~x_{max} \right) & N = \frac{1}{ \sqrt{ \int_{-x_{max}}^{x_{max}} \Psi (x)^2 dx}} & \Psi (x) = N \Psi(x) \end{matrix} \nonumber$
Enter energy: E = =0.6029
This result for the ground state correctly shows that the electron density is delocalized over both nuclear centers forming a molecular state. It also shows that the electron density is greatest at the nuclear centers where the electron‐nucleus potential energy is lowest, and dips somewhat in the ʺbond regionʺ where the potential energy is much higher. It also shows that tunneling is occurring because the electron is present in regions where its total energy is lower than the local potential energy. Inspection of the right‐hand figure above shows that tunneling occurs for the following situations: x < ‐1.55, x > 1.55, and ‐0.396 < x < 0.396. The probability that tunneling is occuring is calculated below.
$\int_{-1.56}^{x_{max}} \Psi (x)^2 dx + \int_{-0.396}^{0.396} \Psi (x)^2 dx + \int_{1.56}^{x_{max}} \Psi (x)^2 dx = 0.229 \nonumber$
The next allowed state with energy E = ‐0.2222 Eh is clearly anti‐bonding because of the node in the internuclear region. The model also shows that this anti‐bonding state is unstable relative to a hydrogen atom and hydrogen ion which collectively have an energy of ‐0.500 Eh.
3.15: Localized and Delocalized Molecular Orbitals
The generation of localized molecular orbitals (LMOs) from canonical or delocalized molecular orbitals (DMOs) will be illustrated by modeling the π-electrons of 1,3-butadiene as particles in a one-dimensional box (PIB). Solving Schrödinger's equation for this application yields the familiar sine function for the canonical orbitals. The n = 1 and n = 2 states are occupied by two π-electrons each; Ψ and Ψ2 for both states are graphed below.
$\begin{matrix} \Psi (n,~x) = \sqrt{2} \sin (n \pi x) & x = 0, .01 .. 1 \end{matrix} \nonumber$
These canonical, delocalized wave functions present a problem for chemists who are accoustomed to localizing the π-electrons of butadiene between carbons 1 and 2, and carbons 3 and 4 with the Lewis structure shown below. In contradiction to this view, the DMOs clearly distribute the π-electron density over the entire carbon backbone
H2C=CH-CH=CH2
The chemist's localized-electron-pair model can be recovered to some extent from the delocalized, canonical wave functions by invoking the quantum mechanical superposition principle. The superposition principle teaches that any linear combination of the canonical solutions is also a valid wave function as long as they are only used to interpret physical properties that depend on the total electron density. The canonical DMOs are required for single-electron phenomena such as electronic spectroscopy and ionization phenomena.
Thus, it is easy to show that simply adding (Ψp) and subtracting (Ψm) the canonical wave functions yields LMOs in this simple example involving a particle-in-a-box model for the π-electrons of butadiene. Ψp shifts electron density to the left closer to the bond region between carbon atoms 1 and 2, while Ψm shows the electron density closer to the bond region between carbon atoms 3 and 4. Naturally, for more complicated molecular situations a more sophisticated protocol is required to generate LMOs.
$\begin{matrix} \Psi_p (x) = \frac{1}{ \sqrt{2}} ( \Psi (1,~x) + \Psi (2,~x)) & \Psi_m (x) = \frac{1}{ \sqrt{2}} ( \Psi (1,~x) - \Psi (2,~x)) \end{matrix} \nonumber$
For example, raising the level of theory from PIB to a low-level ab initio quantum mechanical basis set (STO-3G) yields the following HOMO-1 and HOMO for butadiene. These, of course, are the two occupied π molecular orbitals that are equivalent to the n = 1 and n =2 states for the PIB model. Just as for the PIB model the HOMO-1 has no nodes, while the HOMO has one node.
As noted above there are a variety of protocols in use to provide LMOs from the canonical DMOs. A method in wide-spread use forms LMOs from linear combinations of the DMOs by maximizing intra-orbital electron repulsion. This protocol yields the following LMOs for the π electrons in butadiene.
For a more detailed discussion of this subject and useful references to the pedagogical and primary literature consult the following reference. R. Bruce Martin, "Localized and Spectroscopic Orbitals: Squirrel Ears on Water," Journal of Chemical Education, 65(8), 668-670 (1988).
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Chemists use a variety of models to describe the electronic structure of molecules. The oldest and most rudimentary model which is still in widespread use, and is taught at all levels of the chemistry curriculum, is the localized electron pair model proposed by G. N. Lewis in 1916. This pre-quantum approach to chemical bonding was stimulated in part by the observation that half the elements in the periodic table have an odd number of electrons, but very few molecules do.
Today we know that the importance of the electron pair is explained by the Pauli Exclusion Principle. Electrons, as fermions, cannot have the same set of quantum numbers. Therefore, if the three spatial quantum numbers are the same the spin quantum number must be different, and the spin quantum number has only two allowed values - plus or minus 1/2. Thus only two electrons can occupy an orbital - a region of space.
Molecular orbital theory, a successor to the Lewis model, is based on the Exclusion Principle and the postulates of quantum mechanics. Thus it retains the concept of the electron pair but, due to the wave nature of quantum theory, it abandons spatial localization. This is perhaps the most challenging concept that quantum mechanics offers chemists, the delocalization of the electron pair over the entire molecule.
To emphasize the difference between the Lewis and MO approaches the bonding in water will be considered. The traditional Lewis structure for water is shown below. There are two equivalent bonding electron pairs and two equivalent non-bonding electron pairs.
Therfore, according to the Lewis localized electron pair bonding model there should be two bands in the experimental photoelectron spectrum of equal weight. However, the actual spectrum shows four distinct bands, indicating that there are four types of valence electrons in the water molecule, not two.
The quantum mechanical treatment of bonding is based primarily on the LCAO-MO approach where a molecular orbital is formed as a linear combination of atomic orbitals. A quantum mechanical calculation for water yields the following valence electron molecular orbitals. Note that there are four distinctly different molecular orbitals in agreement with the spectroscopic data mentioned above. Note also that the electron density is distributed over the molecule as a whole.
The canonical molecular orbitals are arranged on a relative energy scale below with comments.
Highest Occupied Molecular Orbital - HOMO: Non-bonding
HOMO-1: Also non-bonding with regard to oxygen and hydrogen
HOMO-2: Bonding - charge density present between oxygen and hydrogen.
HOMO-3: Bonding-charge density present between oxygen and hydrogen.
The fact that the delocalized molecular orbitals are in agreement with experimental data but the localized electron pair model is not presents a challenge for the chemist who is accustomed to placing confidence in the value of the Lewis model. However, the quantum mechanical superposition principle allows one to construct localized molecular orbitals (LMOs) from the canonical delocalized molecular orbitals (DMOs). In other words, quantum mechanics itself provides the justification for placing confidence in the utility of the Lewis electron-pair model.
There are a variety of computational protocols for generating LMOs from DMOs. The following LMOs were calculated using a method developed by Ruedenberg in which intra-orbital electron repulsions are maximized. It is clear that the four LMOs generated by this protocol bear a strong relationship to the simple localized electron pair model of Lewis for water shown above.
Bonding LMOs
Non-Bonding LMOs
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.16%3A_Two_Perspectives_on_the_Bonding_in_Water.txt
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Chemists use a variety of models to describe the electronic structure of molecules. The oldest and most rudimentary model which is still in widespread use, and is taught at all levels of the chemistry curriculum, is the localized electron pair model proposed by G. N. Lewis in 1916. This pre-quantum approach to chemical bonding was stimulated in part by the observation that half the elements in the periodic table have an odd number of electrons, but very few molecules do.
For example ammonia, NH3, has four atoms each with an odd number of valence electrons (5,1,1,1), giving the molecule the magic number of eight valence electrons or four pairs in the Lewis formulation. In the non-geometrical representation shown below, Lewis has the nitrogen sharing a pair of electrons with each hydrogen, and keeping a lone-pair for itself.
In a proper three-dimensional representation the ammonia molecule has C3v symmetry and looks like a camera tripod with the lone pair occupying the position of the camera.
However there is a serious problem with this simple model for the electronic structure of ammonia, and that is that it suggests that there are two types of valence electrons - three equivalent bonding pairs and one non-bonding pair. Unfortunately, there are three bands in ammonia's photoelectron spectrum, indicating that there are actually three types of valence electrons.
Quantum mechanical calculations based on the molecular orbital approach to chemical bonding (MO - LCAO; molecular orbitals formed as a linear combination of atomic orbitals) yield the following molecular orbitals which are in agreement with the experimental spectroscopic data.
Highest Occupied Molecular Orbital - HOMO
This MO is clearly non-bonding and represents the lone-pair electrons on the nitrogen.
The next three MOs, a degenerate pair and a single ground-state MO, represent the three bonding pairs of electrons. However they are not localized between the nitrogen and hydrogens as suggested by the simple Lewis structure shown above. They are indeed molecular orbitals, delocalizing the bonding electron density over the whole molecule.
HOMO-1
HOMO-2
HOMO-3
Thus the delocalized canonical molecular orbitals provided by ab initio quantum mechanics do not directly support the concept of the localized electron pair model of Lewis that has proven so useful to chemists. However, the quantum mechanical superposition principle allows one to construct localized molecular orbitals (LMOs) from the canonical delocalized molecular orbitals (DMOs).
In other words, using appropriate protocols linear combinations of the canonical MOs shown above yield the LMOs shown below which are consistent with the rudimentary Lewis structure shown above.
There are three localized bonding MOs like this, one for each N - H bond.
And one non-bonding MO like this.
3.18: A Molecular Orbital Approach to Bonding in Methane
Most general chemistry textbooks invoke sp3 hybridization to explain the bonding in the tetrahedral methane (CH4) molecule. The idea (valence bond theory, VBT) is that good overlap between the atomic orbitals centered on carbon and hydrogen leads to strong bonds. The problem is that the carbon 2s and 2p valence orbitals don’t point in the tetrahedral directions occupied by the hydrogen 1s valence orbitals. Linus Pauling solved this problem in 1931 by mathematically hybridizing carbon’s 2s and 2p orbitals so that the resulting hybrids (linear combinations) point toward the hydrogen 1s orbitals.
A typical representation of the valence bond approach to methane bonding is shown in the following graphic taken from the 5th edition of McMurray and Fay’s General Chemistry text.
It might be assumed that the tetrahedral geometry of methane requires sp3 hybridization of the carbon 2s and 2p valence atomic orbitals. However, this is not the case because there is a perfectly valid alternative bonding model, molecular orbital theory, which has solid empirical support.
Molecular orbital theory (MOT) for methane forms bonding molecular orbitals involving linear combinations of the unhybridized carbon 2s and 2p valence orbitals with the hydrogen 1s orbitals as shown in the graphic below. The plus and minus signs signify phase, not electrical charge.
A molecular orbital diagram showing both the bonding and anti‐bonding molecular energy levels is provided below. (McQuarrie & Simon, Physical Chemistry: A Molecular Approach, p. 388)
Methane has eight valence electrons, so according to the aufbau and Pauli exclusion principles the two lowest energy molecular orbitals (2a1 and 1t2) are fully occupied with electrons. This is consistent with the experimental valence electron ionization spectrum for methane shown below. (McQuarrie & Simon, Physical Chemistry: A Molecular Approach, p. 388)
The VBT model appears to predict only one valence electron ionization energy for methane, because there is only one kind of bonding electron – an electron in a carbon sp3 hybridized orbital which overlaps with a tetrahedrally located hydrogen 1s orbital. To see how VBT accounts for two ionization energies see the following reference: Journal of Chemical Education, 89, 570‐572 (2012).
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Lithium hydride is a white crystalline solid with the face-centered cubic crystal structure (see lattice shown below). The model for LiH(s) proposed in this study constists of the following elements:
1. The bonding in LiH(s) is completely ionic. The lattice sites are occupied by the spherical, two-electron ions, Li+ and H-.
2. The electrons of Li+ and H- occupy hydrogenic 1s atomic orbitals with adjustable scale factors α and β, respectively. Expressed in atomic units the wavefunctions have the form,
Ψ(1,2) = 1s(1)1s(2) = (α3 /π)exp[-α(r1 + r2)]
The scale factor determines how rapidly the wavefunction (and, therefore, the electron density) diminishes as the distance from the nucleus increases. α and β are, therefore, inversely related to the atomic radius. The larger α and β, the smaller the ionic radii are.
3. The average distance of an electron from the nucleus, <r>, in a scaled 1s orbital is 1.5/α. Therefore, it seems reasonable to take 2, or 3/α as the effective ionic radius in the solid. It is easy to show that 94% of the charge is contained within this radius. (See Appendix)
4. Van der Waals interations between the electron clouds of the ions and the quantum mechanical zero-point energy of the lattice are neglected.
To check the validity of this model the lattice energy of LiH(s) will be calculated and compared to the value obtained by a Born-Haber analysis. The lattice energy is defined as the energy required to bring about the following process,
LiH(s) ----> Li+ (g) + H-(g)
The determination of the lattice energy on the basis of the proposed model, therefore, proceeds by calculating the ground state energies of Li+(g) and H-(g). and subtracting from them the ground state energy of LiH(s). Since terms for the kinetic energy of the ions are not included, the calculations refer to absolute zero.
Li+(g) and H-(g)
The calculations for the ground-state energies of Li+ (g) and H-(g) are similar to that of He. The energy operators consist of five terms: kinetic energy operators for each of the electrons, electron-nuclear potential energy operators for each of the electrons, and an electron-electron potential energy operator.
$\begin{matrix} H_{Li} = - \frac{1}{2r_1} \frac{d^2}{dr_1^2} r_1 - \frac{1}{2r_2} \frac{d^2}{dr_2^2} r_2 - \frac{3}{r_1} - \frac{3}{r_2} + \frac{1}{r_{12}} \ H_{H} = - \frac{1}{2r_1} \frac{d^2}{dr_1^2} r_1 - \frac{1}{2r_2} r_2 - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_{12}} \end{matrix} \nonumber$
When the trial wavefunction and the appropriate energy operator is used in the variational integral,
$E = \int_0^{ \infty} \Psi (1,~2) H \Psi (1,~2) d \tau_1 d \tau_2 \nonumber$
the following expression result (see Appendix for details):
$\begin{matrix} E_{Li} = \alpha^2 - 6 \alpha + \frac{5}{8} \alpha & E_H = \beta^2 - 2 \beta + \frac{5}{8} \beta \end{matrix} \nonumber$
Minimization of the energy with respect to the scale factors to obtain the ground state energies of the gas-phase ions is the next step.
Calculation of the energies of the gas phase ions:
Seed value for the cation scale factor: α = 3
Calculate the energy and radius of the gase phase cation: $E_{Li} ( \alpha ) = \alpha^2 - 5.375 \alpha$
$\begin{matrix} \alpha = \text{Minimize} \left( E_{Li},~ \alpha \right) & E_{Li} ( \alpha ) = -7.2227 & E_{Li} = E_{Li} ( \alpha ) & R_{Li} ( \alpha ) & R_{Li} = \frac{3}{ \alpha} & R_{Li} = 1.1163 \end{matrix} \nonumber$
Seed value for the anion scale factor: β = 1
Calculate the energy and radius of the gas phase anion: $E_{Li} ( \beta ) = \beta^2 - 1.375 \beta$
$\begin{matrix} \beta = \text{Minimize} \left( E_H,~ \beta \right) & \beta = 0.6875 & E_H ( \beta) = -0.4727 & E_H = E_H ( \beta ) & R_H = \frac{3}{ \beta} & R_H = 4.3636 \end{matrix} \nonumber$
Lithium hydride solid - LiH(s)
As noted above, LiH has the face-centered cubic structure shown below.
The ground state energy of LiH(s) consists of three terms: the internal energy of Li+, the internal energy of H-, and the coulombic interaction energy of the ions occupying the lattice sites.
$E_{LiH} = E_{Li} + E_H + E_{coul} \nonumber$
where
$\begin{matrix} E_{coul} = - \frac{1.748}{R_c + R_a} & \text{for } \frac{R_c}{R_a} >= .414 & E_{coul} = - \frac{1.748}{ \sqrt{2} R_a} & \text{for } \frac{R_c}{R_a} < 0.414 \end{matrix} \nonumber$
Here 1.748 is the Madelung constant for the face-centered cubic structure for singly charged ions. Rc and Ra are the radii of the cation and anion. (Rc + Ra) is the inter-ionic separation for situations (Rc/Ra >= .414) in which there is cation-anion contact, while 1.414Ra is the inter-ionic separation for those circumstances (Rc/Ra < .414) in which there is only anion-anion contact. On the basis of assumption 3 of the model, Rc and Ra are replaced by 3/α and 3/β, the effective ionic radii of the cation and the anion. The coulombic contribution now has the form
$\begin{matrix} E_{coul} = - \frac{1.478}{ \frac{3}{ \alpha} + \frac{3}{ \beta}} & \text{for } \frac{ \beta}{ \alpha} >= .414 & E_{coul} = - \frac{1.748}{ \frac{ \sqrt{2} 3}{ \beta}} & \text{for} \frac{ \beta}{ \alpha} < .414 \end{matrix} \nonumber$
Minimization of the energy of the solid simultaneously with respect to α and β is outlined below.
Energy of the solid assuming anion-cation contact. $f( \alpha,~ \beta) = \alpha^2 - 5.375 \alpha + \beta^2 - 1.375 \beta - \frac{1.748}{ \frac{3}{ \alpha} + \frac{3}{ \beta}}$
Energy of the solid assuming anion-anion contact and that the cation rattles in the octahedral hole. $g( \alpha,~ \beta) = \alpha^2 - 5.375 \alpha + \beta^2 - 1.375 \beta - \frac{1.748 \beta}{3 \sqrt{2}}$
Composite expression for the energy of the solid using a conditional statement. $E_{LiH} ( \alpha,~ \beta) = \text{if} \left( \frac{ \beta}{ \alpha} \geq .414,~f( \alpha,~ \beta),~ g( \alpha,~ \beta) \right)$
Minimization of the energy of LiH with respect to the parameters α and β.
$\begin{pmatrix} \alpha \ \beta \end{pmatrix} = \text{Minimize} \left( E_{Li},~ \alpha,~ \beta \right) \nonumber$
$\begin{matrix} \alpha = 2.6875 & R_c = \frac{3}{ \alpha} & R_c = 1.1163 & \text{R}_c \text{ (experimental) = 1.134} \ \beta = 0.8935 & R_a = \frac{3}{ \beta} & R_a = 3.3576 & \text{R}_a \text{ (experimental) = 3.931} \end{matrix} \nonumber$
$\begin{matrix} E_{LiH} ( \alpha,~ \beta) = -8.0210 & E_{LiH} = E_{LiH} ( \alpha,~ \beta) \end{matrix} \nonumber$
Comparison of gas-phase and solid-state ion energies (see Appendix for interpretation):
$\begin{matrix} \text{Cation:} & E_{Lis} = \alpha^2 - 5.375 \alpha & E_{Lis} = -7.227 & E_{Li} = -7.227 & \text{Cation energy doesn't change.} \ \text{Anion:} & E_{Hs} = \beta^2 - 1.375 \beta & E_{Hs} = -0.4032 & E_H = -0.4727 & \text{Anion energy increases.} \end{matrix} \nonumber$
$\begin{matrix} \text{Coulomb energy in solid state:} & E_{LiH} - E_{Lis} - E_{Hs} = -0.3681 \ \text{The calculated lattice energy for LiH(s):} & U_{Lattice} = E_{Li} + E_H - E_{LiH} & U_{Lattice} = 0.3257 \end{matrix} \nonumber$
This result in atomic units is equivalent to a lattice energy expressed in SI units of 856 kJ/mol. A Born-Haber analysis (see below) yields a lattice energy of 912 kJ/mol. Thus, the calculated result of the proposed model is in error by only 6%. The errors for the solid-state ionic radii are 1.6% (cation) and 14.6% (anion). Given the simplicity of the model these comparisons with experimental data are encouraging. For further details on this model see the reference cited below.
$\text{LiH(s)} \xrightarrow{ - \Delta H^o _{form} = 90.4 kJ} \text{Li(s)} + \frac{1}{2} \text{H}_2 \text{(g)} \xrightarrow[ \frac{1}{2} BDE = 218 kJ]{ \Delta H_{sub} = 155 kJ} \text{Li(g) + H(g)} \xrightarrow[EA = -72kJ]{IE = 520 kJ} \text{Li}^+ \text{(g)} + \text{H}^- \text{(g)} \nonumber$
F. Rioux, "Simple Calculation of the Lattice Energy of Lithium Hydride," Journal of Chemical Education 54, 555 (1977).
Appendix:
$\int_0^{ \frac{3}{ \alpha}} \left( \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha,~R) \right) 4 \pi r^2 dr = 93.8 \% \nonumber$
The table below provides a summary of the lattice energy calculation carried out in this tutorial.
$\begin{pmatrix} \text{Property} & \text{Gas Phase} & \text{Solid State} \ \text{Cation, } \alpha & 2.6875 & 2.6875 \ \text{Cation Radius} & 1.1163 & 1.163 \ \text{Cation Energy} & -7.2227 & -7.2227 \ \text{Anion, } \beta & 0.6875 & 0.8935 \ \text{Anion Radius} & 4.364 & 3.3576 \ \text{Anion Energy} & -0.4727 & -0.4302 \ \frac{ \text{Inter Ion}}{ \text{Coulomb Energy}} & x & -0.3681 \ \text{Total Energy} & -7.6953 & -8.0210 \ \text{Lattice Energy} & x & 0.3257 \end{pmatrix} \nonumber$
From the table it is clear that in the formation of LiH solid, the hydride anion contracts significantly from its gas-phase size. This increases its energy (-0.4302+0.4727=0.0425). The increase in anion energy is more than offset by the attractive inter-ion coulombic energy (-0.3681). In other words, the anion suffers a modest increase in energy by shrinking in size so that it can be on-average closer to the cation, thereby increasing the coulombic attraction between the ions and leading to a stable ionic solid.
Most of the integrals required in the analysis above are now evaluated.
Previous memory of α and β values is cleared: $\begin{matrix} \alpha = \alpha & \beta = \beta \end{matrix}$
Trial one-electron wavefunction:
$\Psi (r,~ \beta = \sqrt{ \frac{ \beta^3}{ \pi}} \text{exp} ( - \beta,~r) \nonumber$
Demonstrate that is is normalized:
$\int_0 ^{ \infty} \Psi (r,~ \beta )^2 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow 1 \nonumber$
Calculate the average value of the electron's distance from the nucleus:
$R( \beta ) = \int_)^{ \infty} \Psi (r,~ \beta) r \Psi (r,~ \beta ) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{3}{2 \beta} \nonumber$
Calculate the average value of the kinetic energy of the electron:
$T( \beta ) = \int_0 ^{ \infty} \Psi (r,~ \beta) - \frac{1}{2r} \frac{d^2}{dr^2} (r \Psi (r,~ \beta )) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{ \beta^2}{2} \nonumber$
Calculate the average value of the electron-nucleus potential energy:
$V( \beta, ~Z) = \int_0^{ \infty} \Psi (r,~ \beta) - \frac{Z}{r} \Psi (r,~ \beta) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow -Z \beta \nonumber$
Calculate the average value of the electron-electron potential energy in two steps:
1. The electrostatic potential at r due to electron 1 is:
$\begin{array}{c|c} \Phi (r, ~ \beta _ = \frac{1}{r} \int_0^r \Psi (x,~ \beta)^2 4 \pi x^2 dx + \int_r^{ \infty} \frac{ \Psi (x,~ \beta )^2 4 \pi x^2}{x} dx & ^{ \text{assume,~} \beta >0}_{ \text{simplify}} \rightarrow - \frac{e^{-2 \beta r} + \beta r e^{-2 \beta r} - 1}{r} \end{array} \nonumber$
2. The electrostatic interaction between the two electrons is:
$\begin{array}{c|c} V_{EE} ( \beta ) = \int_0^{ \infty} \Phi (r,~ \beta) \Psi (r,~ \beta)^2 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{5 \beta}{8} \end{array} \nonumber$
To summarize, the trial wavefunction chosen for two electron systems lead to the following expression for the energy.
$E (Z,~ \beta) = \beta^2 - 2Z \beta + \frac{5}{8} \beta = \beta^2 - 2 \beta \left( Z - \frac{5}{16} \right) \nonumber$
Minimization of the energy with respect to the variational parameter β yields: $\beta = Z - \frac{5}{16}$
Ground state energy:
$E(Z) = - \left( Z - \frac{5}{16} \right)^2 \nonumber$
Ionic radius:
$R_Z = \frac{3}{Z - \frac{5}{16}} \nonumber$
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Lithium hydride is a white crystalline solid with the face-centered cubic crystal structure (see lattice shown below). The model for LiH(s) proposed in this study constists of the following elements:
1. The bonding in LiH(s) is completely ionic. The lattice sites are occupied by the spherical, two-electron ions, Li+ and H-.
2. The electrons of Li+ and H- occupy hydrogenic 1s atomic orbitals with adjustable scale factor α. Expressed in atomic units the wavefunctions have the form,
Ψ(1,2) = 1s(1)1s(2) = (α3/π)exp[-α(r1 + r2)]
The scale factor determines how rapidly the wavefunction (and, therefore, the electron density) diminishes as the distance from the nucleus increases. α and β are, therefore, inversely related to the atomic radius. The larger α and β, the smaller the ionic radii are.
3. The average distance of an electron from the nucleus, <r>, in a scaled 1s orbital is 1.5/α. Therefore, it seems reasonable to take 2, or 3/α as the effective ionic radius in the solid. It is easy to show that 94% of the charge is contained within this radius. (See Appendix)
4. It is also assumed that the Li+ and H- ions in the solid have the same α value, energy and radius as they do in the gas phase. Under this assumption the lattice energy is the negative of the inter-ion coulombic energy in the solid state.
To check the validity of this model the calculated lattice energy of LiH(s) will be compared to the value obtained from a Born-Haber analysis. The lattice energy is defined as the energy required to bring about the following process,
LiH(s) ----> Li+ (g) + H-(g)
The determination of the lattice energy on the basis of the proposed model, therefore, proceeds by calculating the ground state energies of Li+(g) and H-(g). and subtracting from them the ground state energy of LiH(s). Since terms for the kinetic energy of the ions are not included, the calculations refer to absolute zero.
Li+(g) and H-(g)
We begin with variational calculations for the ground-state energies of Li+ (g) and H- (g). These calculations will yield the ionic radii which will subsequently be used to calculate the LiH lattice energy.
The energy operators consist of five terms: kinetic energy operators for each of the electrons, electron-nuclear potential energy operators for each of the electrons, and an electron-electron potential energy operator.
$\begin{matrix} H_{Li} = - \frac{1}{2r_1} \frac{d^2}{dr_1^2} r_1 - \frac{1}{2r_2} \frac{d^2}{dr_2^2} r_2 - \frac{3}{r_1} - \frac{3}{r_2} + \frac{1}{r_{12}} \ H_{H} = - \frac{1}{2r_1} \frac{d^2}{dr_1^2} r_1 - \frac{1}{2r_2} r_2 - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_{12}} \end{matrix} \nonumber$
When the trial wavefunction and the appropriate energy operator is used in the variational integral,
$E = \int_0^{ \infty} \Psi (1,~2) H \Psi (1,~2) d \tau_1 d \tau_2 \nonumber$
the following expression result (see the appendix for computational details):
$\begin{matrix} E_{Li} = \alpha^2 - 6 \alpha + \frac{5}{8} \alpha & E_H = \beta^2 - 2 \beta + \frac{5}{8} \beta \end{matrix} \nonumber$
Minimization of the energy with respect to the scale factors to obtain the ground state energies and radii of the ions is shown below.
Calculation of the energies of the gas phase ions:
Seed value for the cation scale factor: α = 3
Calculate the energy and radius of the gase phase cation: $E_{Li} ( \alpha ) = \alpha^2 - 5.375 \alpha$
$\begin{matrix} \alpha = \text{Minimize} \left( E_{Li},~ \alpha \right) & E_{Li} ( \alpha ) = -7.2227 \ R_{Li} = \frac{3}{ \alpha} & R_{Li} = 1.1163 \end{matrix} \nonumber$
Energy and radius of the gas phase anion: $E_{Li} ( \alpha ) = \beta^2 - 1.375 \beta$
$\begin{matrix} \alpha = \text{Minimize} \left( E_H,~ \alpha \right) & \alpha = 0.6875 & E_H ( \alpha) = -0.4727 \ R_H = \frac{3}{ \alpha} & R_H = 4.3636 \end{matrix} \nonumber$
Lithium hydride solid - LiH(s)
As noted above, LiH has the face-centered cubic structure shown below.
The ground state energy of LiH(s) consists of three terms: the internal energy of Li+, the internal energy of H-, and the coulombic interaction energy of the ions occupying the lattice sites.
$E_{LiH} = E_{Li} + E_H + E_{coul} \nonumber$
Assuming a face-centered structure (Madelund constant = 1.748) with cation-anion and anion-anion contact, the coulombic term is as shown below.
$E_{Li} (2.6875) + E_H (0.6875) - \frac{1.748}{R_{Li} + R_H} = -8.0143 \nonumber$
Since it has been assumed that the Li+ (g) and H- (g) ions are the same in the gas phase and the solid state, the lattice energy is the negative of Ecoul.
$\begin{matrix} U_{Lattice} = \frac{1.748}{R_{Li} + R_H} & U_{Lattice} = 0.319 \end{matrix} \nonumber$
This result in atomic units is equivalent to a lattice energy expressed in SI units of 838 kJ/mol. The Born-Haber analysis shown below yields a lattice energy of 912 kJ/mol. Thus, the calculated result of the proposed model is in error by 8%.
$\text{LiH(s)} \xrightarrow{ - \Delta H^o _{form} = 90.4 kJ} \text{Li(s)} + \frac{1}{2} \text{H}_2 \text{(g)} \xrightarrow[ \frac{1}{2} BDE = 218 kJ]{ \Delta H_{sub} = 155 kJ} \text{Li(g) + H(g)} \xrightarrow[EA = -72kJ]{IE = 520 kJ} \text{Li}^+ \text{(g)} + \text{H}^- \text{(g)} \nonumber$
Improved results can be obtained by allowing the gas phase ions to change size on the formation of the solid under the influence of the inter-ion coulombic interaction, and by abandoning the simplifying restriction of cation-anion and anion-anion contact. See the reference below for an outline of such a calculation.
F. Rioux, "Simple Calculation of the Lattice Energy of Lithium Hydride," Journal of Chemical Education 54, 555 (1977).
Appendix:
Because this is a live Mathcad document and the scale factor, α, has been used above, β will be used for the calculations outlined below.
Trial one-electron wavefunction:
$\Psi (r,~ \beta ) = \sqrt{ \frac{ \beta^3}{ \pi}} \text{exp} ( - \beta,~r) \nonumber$
Demonstrate that is is normalized:
$\int_0 ^{ \infty} \Psi (r,~ \beta )^2 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow 1 \nonumber$
Calculate the average value of the electron's distance from the nucleus:
$R( \beta ) = \int_0^{ \infty} \Psi (r,~ \beta) r \Psi (r,~ \beta ) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{3}{2 \beta} \nonumber$
Demonstrate 94% of electron density is contained within 2<r>: $\int_0 ^{ \frac{3}{ \beta}} \Psi (r,~ \beta)^2 4 \pi r^2 dr \text{ assume, } \beta >0 \rightarrow (-25)e^{-6} + 2 = 93.8 \% \nonumber \] Calculate the average value of the kinetic energy of the electron: $T( \beta ) = \int_0 ^{ \infty} \Psi (r,~ \beta) - \frac{1}{2r} \frac{d^2}{dr^2} (r \Psi (r,~ \beta )) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{1}{2} \beta^2 \nonumber$ Calculate the average value of the electron-nucleus potential energy: $V( \beta, ~Z) = \int_0^{ \infty} \Psi (r,~ \beta) - \frac{Z}{r} \Psi (r,~ \beta) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow (- \beta ) Z \nonumber$ Calculate the average value of the electron-electron potential energy in two steps: 1. The electrostatic potential at r due to electron 1 is: $\begin{array}{c|c} \Phi (r, ~ \beta _ = \frac{1}{r} \int_0^r \Psi (x,~ \beta)^2 4 \pi x^2 dx + \int_r^{ \infty} \frac{ \Psi (x,~ \beta )^2 4 \pi x^2}{x} dx & ^{ \text{assume,~} \beta >0}_{ \text{simplify}} \rightarrow \frac{- \left[ e^{-2 \beta r} + \beta r e^{-2 \beta r} - 1 \right]}{r} \end{array} \nonumber$ 2. The electrostatic interaction between the two electrons is: $\begin{array}{c|c} V_{EE} ( \beta ) = \int_0^{ \infty} \Phi (r,~ \beta) \Psi (r,~ \beta)^2 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{5 \beta}{8} \end{array} \nonumber$ To summarize, the trial wavefunction chosen for two electron systems lead to the following expression for the energy. $E (Z,~ \beta) = \beta^2 - 2Z \beta + \frac{5}{8} \beta = \beta^2 - 2 \beta \left( Z - \frac{5}{16} \right) \nonumber$ Minimization of the energy with respect to the variational parameter β yields: \( \beta = Z - \frac{5}{16}$
Ground state energy:
$E(Z) = - \left( Z - \frac{5}{16} \right)^2 \nonumber$
Ionic radius:
$R_Z = \frac{3}{Z - \frac{5}{16}} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.20%3A_An_Even_Simpler_LiH_Lattice_Energy_Calculation.txt
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The charge cloud model for atomic and molecular systems was developed in the early 1950s by George Kimball and his graduate students Gertrude Neumark and Lee Kleiss. Unfortunately their research was only published through University Microfilms at the University of Michigan.
I took an interest in the charge cloud model for pedagogical reasons in the 70s and 80s. I believed it could be used to introduce students to the quantum mechanical explanation for atomic and molecular stability and structure using only the most rudimentary mathematical tools ‐ algebra, geometry and basic calculus.
I will present in skeletal form the calculations that appear in the following publications. Please consult these references for more details on the calculations such as comparisons with available experimental data.
ʺThe Stability of the Hydrogen Atom,ʺ Frank Rioux, J. Chem. Educ. 50, 550 (1973).
ʺCharge cloud study of atomic and molecular structure,ʺ Frank Rioux and Peter Kroger, Am. J. Phys. 44, 56 (1976).
ʺAn Ionic Model for Metallic Bonding,ʺ Frank Rioux, J. Chem. Educ. 62, 383 (1985).
A thorough and insightful historical review of the charge cloud model appeared recently in the Journal of Chemical Education.
ʺThe Kimball Free‐Cloud Model: A Failed Innovation in Chemical Education?,ʺ William B. Jensen, J. Chem. Educ. 91, 1106 (2014).
The basic building block of the charge cloud model is a spherical electron charge cloud of radius R and uniform charge distribution. This is the wave function of the electron. Its use leads to the following expressions for electron kinetic energy, electron‐nucleus potential energy when the nucleus is in the center of the charge cloud, and electron‐electron potential energy for two concentric interpenetrating charge clouds. All charge cloud calculations satisfy the virial theorem. The highlighted calculations do not predict stable species. The calculations are carried out in atomic units: h/2π = me = e = ao = 4πεo = 1.
$\begin{matrix} T = \frac{9}{8 R^2} & V_{ne} = \frac{-3Z}{2R} & V_{ee} = \frac{6}{5R} \end{matrix} \nonumber$
Hydrogen atom: point charge proton surrounded by a spherical electron charge cloud of radius R and even charge distribution throughout.
$\begin{matrix} E_H (R) = \frac{9}{8R^2} - \frac{3}{2R} & \begin{array}{c|c} R = \frac{d}{dR} E_H (R) = 0 & _{ \text{float, } 5}^{ \text{solve, R}} \rightarrow 1.5 \end{array} & E_H (R) = -0.5 \end{matrix} \nonumber$
Hydride anion: point charge proton surrounded by two spherical electron charge clouds of radius R and even charge distribution throughout. It is not stable on the basis of the charge cloud model.
$\begin{matrix} \text{Reset R:} & E_{H'} (R) = \frac{9}{4R^2} + \frac{6}{5R} - \frac{3}{R} & \begin{array}{c|c} R = \frac{d}{dR} E_{H'} (R) = 0 & _{ \text{float, } 5}^{ \text{solve, R}} \rightarrow 3.5 \end{array} & E_{H'} (R) = -0.36 \end{matrix} \nonumber$
Positronium: two oppositely charged concentric interpenetrating charge clouds. The correct ground state energy is ‐0.25 au.
$\begin{matrix} \text{R = R} & E_{p} (R) = \frac{9}{4R^2} - \frac{6}{5R} & \begin{array}{c|c} R = \frac{d}{dR} E_{p} (R) = 0 & _{ \text{float, } 5}^{ \text{solve, R}} \rightarrow 3.75 \end{array} & E_{H'} (R) = -0.16 \end{matrix} \nonumber$
Helium ion: spherical electron charge cloud with +2 nuclear point charge at the center.
$\begin{matrix} \text{R = R} & E_{He'} (R) = \frac{9}{8R^2} - \frac{3}{R} & \begin{array}{c|c} R = \frac{d}{dR} E_{He'} (R) = 0 & _{ \text{float, 3}}^{ \text{solve, R}} \rightarrow 0.75 \end{array} & E_{He'} (R) = -2 \end{matrix} \nonumber$
Helium atom: two concentric electron charge clouds with a +2 nuclear point charge at the center.
$\begin{matrix} \text{R = R} & E_{He} (R) = \frac{9}{4R^2} - \frac{6}{R} + \frac{6}{5R} & \begin{array}{c|c} R = \frac{d}{dR} E_{He} (R) = 0 & _{ \text{float, 3}}^{ \text{solve, R}} \rightarrow 0.937 \end{array} & E_{He} (R) = -2.56 \end{matrix} \nonumber$
First ionization energy: $E_{He'} (0.75) - E_{He} (R) = 0.56$
Hydrogen molecule ion: a spherical electron charge cloud with two protons positioned at +/‐ r from the center. The electron‐proton and proton‐proton potential energies for this configuration are:
$\begin{matrix} V_{en} = \frac{-3}{2R} + \frac{r^2}{2R^3} & V_{nn} = \frac{1}{2r} \end{matrix} \nonumber$
$E_{H2'} (R,~r) = \frac{9}{8R^2} - \frac{3}{R} + \frac{r^2}{R^3} + \frac{1}{2r} \nonumber$
$\begin{matrix} R = 1 & r = .2 & \begin{pmatrix} R \ r \end{pmatrix} = \text{Minimize} \left( E_{H2'},~R,~r \right) & \begin{pmatrix} R \ r \end{pmatrix} = \begin{pmatrix} 1.243 \ 0.783 \end{pmatrix} & E_{H2'} (R,~r) = -0.728 \end{matrix} \nonumber$
Bond energy: $2E_H (1.5) - E_{H2} (R,~r) = 0.21$
Hydrogen molecule: two concentric electron charge clouds with protons positioned at +/‐ r from the center.
$E_{H2} (R,~r) = \frac{9}{4R^2} - \frac{6}{R} + \frac{2r^2}{R^3} + \frac{6}{5R} + \frac{1}{2r} \nonumber$
$\begin{matrix} \left( E_{H2},~ R,~r \right) & \begin{pmatrix} R \ r \end{pmatrix}= \begin{pmatrix} 1.364 \ 0.682 \end{pmatrix} & E_{H2} (R,~r) = -1.21 \end{matrix} \nonumber$
Bond energy:
$2E_H (1.5) - E_{H2} (R,~r) = 0.21 \nonumber$
LiH(s): This simple ionic solid has a face‐centered cubic lattice structure shown below. Itʹs lattice sites are occupied by the two‐electron ions, Li+(1s2) and H (1s2). Since the ions are isoelectronic it is clear that the small spheres must be the lithium cation with nuclear charge +3, and the large spheres are the hydride anion with nuclear charge +1.
The total energy of LiH consists of three terms: the energy of the lithium cation, the energy of the hydride anion and the coulombic potential energy between the ions. As indicated below the cation‐anion potential energy depends on the relative size of ions.
$\begin{matrix} E_{cation} = \frac{9}{4R_c^2} - \frac{3Z_c}{R_c} + \frac{6}{5R_c} & E_{anion} = \frac{9}{4R_a^2} - \frac{3Z_a}{R_a} + \frac{6}{5R_a} \end{matrix} \nonumber$
$\begin{matrix} \frac{R_c}{R_a} \geq 0.414 & E_{coul} = \frac{-1.748}{R_a + R_c} & \frac{R_c}{R_a} < 0.414 & E_{coul} = \frac{-1.748}{ \sqrt{2} R_a} \end{matrix} \nonumber$
Nuclear charges: $\begin{matrix} Z_a = 1 & Z_c = 3 \end{matrix}$
Seed values for the ionic radii: $\begin{matrix} R_a = .5 & R_c = .5 \end{matrix}$
$\begin{array}{c|c} E_{LiH} \left( R_c,~R_a \right) = & \frac{9}{4R_c^2} - \frac{3Z_c}{R_c} + \frac{6}{5R_c} + \frac{9}{4R_a^2} - \frac{3Z_a}{R_a} + \frac{6}{5 R_a} - \frac{1.748}{R_a + R_c} ~ \text{if } \frac{R_c}{R_a} \geq 0.414 \ ~ & \frac{9}{4R_c^2} - \frac{3Z_c}{R_c} + \frac{6}{5R_c} + \frac{9}{4R_a^2} - \frac{3Z_a}{R_a} + \frac{6}{5 R_a} - \frac{1.748}{R_a + R_c} ~ \text{otherwise} \end{array} \nonumber$
$\begin{matrix} \begin{pmatrix} R_c \ R_a \end{pmatrix} = \text{Minimize} \left( E_{LiH},~ R_c,~R_a \right) & \begin{pmatrix} R_c \ R_a \end{pmatrix} = \begin{pmatrix} 0.577 \ 1.482 \end{pmatrix} & E_{LiH} \left( R_c,~R_a \right) = -7.784 \end{matrix} \nonumber$
Calculation of the energies of the gas-phase ions:
$\begin{matrix} E_c \left( R_c \right) = \frac{9}{4 R_c^2} - \frac{3Z_c}{R_c} + \frac{6}{5R_c} & R_c = \text{Minimize} \left( E_c,~R_c \right) & R_c = 0.577 & E_c \left( R_c \right) = -6.76 \ E_c \left( R_c \right) = \frac{9}{4 R_a^2} - \frac{3Z_a}{R_a} + \frac{6}{5R_a} & R_a = \text{Minimize} \left( E_a,~R_a \right) & R_a = 2.500 & E_a \left( R_a \right) = -0.360 \end{matrix} \nonumber$
Lattice energy calculation: LiH (s) --> Li+ (g) + H- (g)
$E_c \left( R_c \right) + E_a \left( R_a \right) - E_{LiH} \left( R_c,~R_a \right) = 0.494 \nonumber$
Li(s): Lithium metal is modeled as an ionic solid with the same face‐centered structure as used for LiH. The small spheres in the figure above are again the lithium cations and the large spheres are the electride anions, spherical electron charge clouds of radius Ra.
$\begin{array}{c|c} E_{LiH} \left( R_c,~R_a \right) = & \frac{9}{4R_c^2} - \frac{3Z_c}{R_c} + \frac{6}{5R_c} + \frac{9}{8R_a^2} - \frac{1.748}{R_a + R_c} ~ \text{if } \frac{R_c}{R_a} \geq 0.414 \ ~ & \frac{9}{4R_c^2} - \frac{3Z_c}{R_c} + \frac{6}{5R_c} + \frac{9}{8R_a^2} - \frac{1.748}{ \sqrt{2} R_a} ~ \text{otherwise} \end{array} \nonumber$
$\begin{matrix} \begin{pmatrix} R_c \ R_a \end{pmatrix} = \text{Minimize} \left( E_{Li},~ R_c,~R_a \right) & \begin{pmatrix} R_c \ R_a \end{pmatrix} = \begin{pmatrix} 0.577 \ 1.82 \end{pmatrix} & E_{Li} \left( R_c,~R_a \right) = -7.1 \end{matrix} \nonumber$
Lattice energy calculation for lithium metal assuming that the energy of a gas‐phase electron is zero: Li(s) ‐‐‐> Li+(g) + e (g)
$E_c \left( R_c \right) + 0 = E_{Li} \left( R_c,~R_a \right) = 0.34 \nonumber$
Calculation of the enthalpy of formation of LiH(s) using the results of previous calculations: Li(s) + 1/2 H2(g) ‐‐‐> LiH(s)
$-7.784 - \left( -7.1 - \frac{121}{2} \right) = -0.079 \nonumber$
Methane: Two ionic models for methane are proposed in order to show that VSEPRʹs explanation for the greater stability of tetrahedral methane is spurious. Gillespieʹs Valence Shell Electron Pair Repulsion (VSEPR) theory [J. Chem. Educ. 40, 295 (1963)] holds that the configuration of valence electron pairs surrounding a central nonvalence nuclear kernel is determined by minimization of electron‐electron potential energy.
In the ionic models considered here the +4 carbon kernel (nucleus and nonvalence electrons) is treated as a point charge which interacts electrostatically with four unpolarized hydride anions of radius R in both the square planar (D4h) and tetrahedral (Td) arrangement. In Td methane the carbon kernel occupies the tetrahedral hole and is not visible in the figure below. The energy of the hydride anion as a function of charge cloud radius was calculated previously.
D4h: In the energy expression for square planar geometry the first term in brackets multiplied by four is the hydride ion energy contribution, the next two are the repulsive interactions between the hydrides and the final term is the attractive hydride‐kernel potential energy.
$E_{D4h} (R) = 4 \left( \frac{9}{4R^2} + \frac{6}{5R} - \frac{3}{R} \right) + \frac{2}{R} + \frac{1}{ \sqrt{2} R} - \frac{16}{ \sqrt{2} R} \nonumber$
$\begin{matrix} R = R & \begin{array}{c|c} R = \frac{d}{dR} E_{D4h} (R) = 0 & _{ \text{float, 5}} ^{ \text{solve, R}} \rightarrow 1.1388 \end{array} & E_{D4h} (R) = -6.94 \end{matrix} \nonumber$
$\begin{matrix} V_{hh} = \frac{2}{R} + \frac{1}{ \sqrt{2} R} = 2.377 & V_{hk} = - \frac{16}{ \sqrt{2} R} = -9.935 \end{matrix} \nonumber$
Td: In the energy expression for tetrahedral geometry the repulsive interaction between the hydrides and the attractive hydride‐kernel potential energy follow the hydride ions energy contribution.
$E_{Td} (R) = 4 \left( \frac{9}{4R^2} + \frac{6}{5R} - \frac{3}{R} \right) - \frac{16}{ \sqrt{ \frac{3}{2}} R} \nonumber$
$\begin{matrix} R = R & \begin{array}{c|c} R = \frac{d}{dR} E_{Td} (R) = 0 & _{ \text{float, 5}} ^{ \text{solve, R}} \rightarrow 1.0426 \end{array} & E_{Td} (R) = -8.279 \end{matrix} \nonumber$
$\begin{matrix} V_{hh} = \frac{3}{R} = 2.877 & V_{hk} = - \frac{16}{ \sqrt{2} R} = -10.851 \end{matrix} \nonumber$
These calculations show that indeed tetrahedral methane is the energetically preferred geometry and that the reason for this is the stronger hydride‐kernel attractive interaction, which is due to its smaller charge cloud radius and the close packing of the charge clouds around the nuclear kernel. The repulsive interaction between the hydride anions is actually higher in tetrahedral methane. In other words, molecular geometry is driven by maximizing the attractive electron‐nuclear interactions and not by minimizing the electron‐electron repulsive interactions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.21%3A_Charge_Cloud_Models_for_Some_Simple_Atomic_Molecular_and_Solid_Systems.txt
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During the last three decades the emphasis on teaching VSEPR theory in the general chemistry sequence has increased dramatically. In fact in the area of molecular structure it might be described as the theory of choice at the introductory level of the chemistry curriculum. However given long-term reservations in some quarters (see quotes below) regarding its validity, and given the tightness of the general chemistry curriculum the question of whether or not its use is justified should be a matter of discussion. This is an important issue because VSEPR is generally taught at the expense of more comprehensive and important bonding theories such as molecular orbital theory. I think it is important for the community of undergraduate chemistry teachers to discuss the question of whether or not VSEPR belongs in the chemistry curriculum.
Recently I analyzed the VSEPR model quantitatively using a simple quantum mechanical model for bonding created by Henry Bent (1) in the 1960s called the Tangent Spheres Model (TSM). The purpose of this paper is to share the results of my analysis with the general chemistry faculty in order to facilitate a discussion of the role VSEPR should play in the general chemistry curriculum. TSM is a rudimentary quantum mechanical model that yields quantitative predictions for molecular geometries with relative ease. Because of its simplicity as an electronic structure model, the conclusions drawn from an analysis of its results need to be validated by higher level quantum mechanical calculations. In other words the TSM can point us in the direction of further study.
Before proceeding to the TSM calculations, I would like to review a sample of the concerns that have been voiced about teaching and using VSEPR in the recent literature. As can be seen by the dates of these citations, there have been periodic expressions of caution about the use and interpretation of VSEPR since its first appearance in the chemistry curriculum.
Recent articles in the literature have championed the use of the valence shell electron pair repulsion, VSEPR, approach to predicting molecular structure. Recently, the authors of general chemistry textbooks have grasped this idea and incorporated it into the curriculum. This approach leads to the impression that electron-electron repulsions determine the geometries of molecules, because the model is presented in the form of an explanation and structures are used to confirm the model. The factors that influence the geometry of molecules are much more complicated than the VSEPR model leads one to believe. Drago, JCE 50, 244 (1973).
In teaching VSEPR theory, as in teaching any other simple model, the student should understand that such a simple approach neglects many other factors in bonding. Its emphasis on interelectronic repulsions rather than on bond formation contrasts sharply with other models in which replusions are considered only to refine structural details. VSEPR theory like most simple models, produces mostly correct predictions. However, such a simple theory can easily be misused or overused. It should be stressed that although VSEPR theory is an easy-to-use tool, such a simple tool - by the very nature of its simplicity - may not be as accurate as necessary. McKenna and McKenna, JCE 61, 771 (1984).
On the whole, the VSEPR method predicts the geometries of main group compounds and complexes rather well. This is not the same thing as saying that it provides a correct explanation of molecular geometry. Indeed, in the opinion of some, its status and use is just that of prediction, that of an aid to getting the right answer. One is forced to remember a difficult lesson: a model that leads to a correct prediction is not necessarily a correct model. Kettle, Physical Inorganic Chemistry, University Science Books, 1996, Appendix 2.
Among my concerns is the fact that much of the caution that is voiced in these quotations appears to be absent in our general chemistry textbooks. If it isn't stated explicitly, the impression one is left with after reading textbook presentations of VSEPR is that it is not just a predictive devise, but an explanation for the equilibrium geometries that molecules achieve. I will argue below that it does not have explanatory success because it is concerned only with electron-electron repulsions and they represent a relatively minor contribution to the total molecular energy.
The Tangent Spheres Model
The Tangent Spheres Model assumes electron pairs are spheres of radius R (an adjustable or variational parameter) and that the electron density within the spheres is uniform. As described, the TSM might be viewed as a primitive precursor of the current density functional theory (DFT). The Pauli Exclusion Principle is incorporated into the model by assuming that spheres are hard, that is that they occupy space to the exclusion of other spherical electron pairs. Quantum mechanics enters TSM through a 1/R2 dependence for electron kinetic energy. All potential energy contributions are calculated classically, as they are in any ab initioquantum mechanical calculation. In other words, quantum mechanics always manifests itself in the kinetic energy term. The consequence of de Broglie's hypothesis of wave-particle duality for matter is a non-classical expression for the kinetic energy of material objects such as the electron.
To take a specific example, VSEPR predicts tetrahedral geometry for methane because this geometry puts the electron pairs farthest apart and therefore minimizes valence electron repulsions. To challenge this approach the valence energy of methane was calculated assuming both square planar and tetrahedral geometry using the Tangent Spheres method. To build square planar or tetrahedral methane using TSM requires five spheres. A small sphere represents the nuclear kernel - the nucleus and non-valence electrons and carries a net +4 charge. Four larger spheres contain the four pairs of valence electrons each with a proton imbedded in the center. Each valence sphere, therefore, carries a net charge of -1. The arrangement of the spheres for the two geometries is shown schematically in the accompanying Mathcad worksheet showing the computational details.
As we shall see below, TSM also predicts tetrahedral geometry for methane, but clearly identifies electron-nuclear attractions as the driving force in determining the equilibrium geometry. The entire TSM calculation is appended as a Mathcad worksheet as mentioned above. For computational details click here.
In the TSM approach the total molecular energy for methane is partitioned into the following contributions:
• Electron kinetic energy: 9/(8R2)
• Intra-pair electron-electron repulsion: 6/(5R)
• Inter-pair electron-electron repulsion: Coulomb's Law using point charges
• Nuclear-nuclear repulsion: Coulomb's Law using point charges
• Hydrogen - hydrogen
• Carbon - hydrogen
• Electron-nuclear attraction
• For carbon: Coulomb's Law using point charges
• For hydrogen: -3/(2R)
The results are summarized below in tabular form. The various energy contributions are reported in hartrees ( 1 hartree = 27.2 eV = 2.6255 MJ/mol). The results in the table show that the TSM calculations satisfy the virial theorem. That is total energy is half the potential energy and the kinetic energy is the negative of the total energy.
Methane Square Planar Tetrahedral Difference
Electron kinetic energy 6.94 8.28 1.34
Electron-electron repulsion, total 13.73 16.11 2.38
Intra-pair electron repulsion 4.22 4.60 0.38
Inter-pair electron repulsion 9.51 11.51 2.00
Nuclear-nuclear repulsion, total 12.31 15.41 3.1
Nuclear-nuclear repulsion: H-H 2.38 2.88 0.50
Nuclear-nuclear repulsion: C-H 9.94 12.53 2.60
Electron-nuclear attraction, total -39.92 -48.08 -8.16
Electron-carbon attraction -19.87 -25.06 -5.19
Electron-hydrogen attraction: molecular -9.51 -11.51 -2.00
Electron-hydrogen attraction: atomic -10.54 -11.51 -0.97
Total energy -6.94 -8.28 -1.34
Sphere radius 1.14 1.04 -0.10
In ab initio quantum mechanical geometry optimizations the total energy is minimized, thus, the VSEPR model seems suspect in concentrating on a single energy contribution to predict molecular geometry. The rudimentary quantum mechanical results provided in the table seem to support this suspicion. Nothing the VSEPR theory claims to be important is supported by the calculations based on TSM. For example, in going from square planar to tetrahedral geometry:
• inter-pair electronic repulsions increase rather than decrease
• the valence electrons get closer together rather than farther apart (see R in table)
• the inter-pair electron repulsion term is much smaller than the valence electron attraction for the carbon nucleus
In fact it can be stated that on the basis of the TSM analysis it appears that the major driving force for molecular geometry appears to be maximizing electron-nuclear attractions, not minimizing electron-electron repulsions. TSM suggest, therefore, that the acronym should be VSEPNA (valence shell electron pair nuclear attraction). According to TSM that geometry is best which allows electrons to get closest to the nucleus. This is, of course, a packing problem. In this example, the tetrahedral hole is clearly smaller than the square planar (octahedral) hole, allowing the negatively charged valence electrons to get closer to the positively charged nucleus. Or we could argue that the tetrahedral arrangement of valence electron pairs is close packing and the square planar arrangement isn't.
Reference:
1. Bent, H. A. J. Chem. Educ. 40, 446-452 (1963).
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In a previous tutorial [1] it was shown that the reason methane is tetrahedral (Td) rather than square planar (D4h) is because electron-nucleus attractions are greater for the tetrahedral geometry. VSEPR teaches (incorrectly) that minimization of electron-electron potential energy drives molecular geometry. However, the previous variational calculation showed that electron-electron repulsion is actually higher in Td methane than it is in the D4h molecule.
In what follows a simple electrostatic calculation will be presented which reaches the same conclusion as the more elaborate and rigorous variational calculation based on Henry Bent’s [2] Tangent Spheres Model (TSM) of chemical bonding and molecular geometry. This calculation was suggested to the author in a private communication from Henry Bent.
A simple ionic model is proposed for the bonding in methane. The +4 carbon kernel (nucleus and nonvalence electrons) is treated as a point charge which interacts electrostatically with four unpolarized hydride anions of unit diameter in either a square planar or tetrahedral arrangement. In tetrahedral methane the +4 kernel occupies the tetrahedral hole and is not visible in the diagram shown below.
Rudimentary geometrical considerations provide the necessary molecular parameters in the following table.
Molecular Parameter D4h Td
Hydride-Hydride Distance 4 @ 1 and 2 @ √2 6 @ 1
Hydride-Nucleus Distance 4 @ √2/2 4 @ √(3/8)
From an electrostatic perspective a hydride anion is equivalent to an electron. Therefore, the calculation of the potential energy will be expressed in terms of electron-electron potential energy and electronnucleus potential energy. Using the molecular parameters listed in the table above we calculate the potential energy contributions for both geometries as follows.
Square Planar Methane
$\begin{pmatrix} V_{ee} = 4 \frac{(-1)(-1)}{1} + 2 \frac{(-1)(-1)}{ \sqrt{2}} = 5.41 \ V_{en} = 4 \frac{(-1)(+4)}{ \frac{ \sqrt{2}}{2}} = -22.63 \end{pmatrix} \nonumber$
Tetrahedral Methane
$\begin{matrix} V_{ee} = 6 \frac{(-1)(-1)}{1} = 6.00 \ V_{en} = 4 \frac{(-1)(+4)}{ \sqrt{ \frac{3}{8}}} = -26.13 \end{matrix} \nonumber$
These results are summarized in the following table.
Energy Contribution D4h Td
Vee 5.41 6.00
Ven -22.63 -26.13
Vtot -17.22 -20.13
The results of this simple electrostatic calculation contradict the VSEPR model in two significant ways:
1. Electron-electron repulsions are greater for tetrahedral geometry than they are for square planar geometry.
2. Td is favored over D4h because of electron-nucleus attractions. In other words, electron-nuclear attractions are the most important energy contribution in determining molecular geometry.
The latter conclusion should not be surprising. There are four types of energy contributions in a molecule under the Born-Oppenheimer approximation: (1) electron kinetic energy; (2) electron-nucleus potential energy; (3) electron-electron potential energy; (4) nucleus-nucleus potential energy. Electronnucleus potential energy is the only attractive term, and electron-electron potential energy is the smallest of the “repulsive” terms. Clearly electron-nucleus attraction is the single most important term in determining molecular geometry.
Furthermore, this guarantees that electron-electron potential energy will track in the opposite direction. Electron domains get closer to nuclei in two ways: by adopting a close-packed geometry (Td versus D4h) and by shrinking in size. Both lead to larger electron-electron potential energy and lower (more negative) electron-nucleus potential energy.
For the reasons enumerated above, chemical educators should recall VSEPR. It is not a valid model for molecular geometry and takes up space in textbooks that would be better devoted to viable quantum mechanical models of molecular geometry such as TSM [1, 2] or molecular orbital theory. Even in those textbooks in which it is juxtaposed with more credible models it distracts attention from them because of its specious predictive methodology.
1. Rioux, F. http://www.users.csbsju.edu/~frioux/vsepr/NVSEPR.htm
2. Bent, H. A. J. Chem. Educ. 40, 446-452 (1963).
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.23%3A_A_Simple_Electrostatic_Critique_of_VSEPR.txt
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Gillespie's Valence Shell Electron Pair Repulsion (VSEPR) theory holds that the configuration of valence electron pairs surrounding a central nonvalence nuclear kernel is determined by minimization of electron-electron potential energy. The following summary statement by Gillespie is taken from J. Chem. Educ. 40, 295 (1963).
The theory proposes that the stereochemistry of an atom is determined primarily by the repulsive interactions between the electron pairs in a valence shell. The electrons in a valence shell are regarded as occupying essentially localized orbitals that are oriented in space around the nucleus and completed inner electron shells (nuclear kernel) so that their average distance apart is maximized. The tendency of electrons in a pair in a valence-shell to adopt an arrangement which maximizes their distance apart may be regarded as arising from repulsive interactions between the electron pairs.
However, quantum theory teaches that minimization of the total energy determines electronic structure. The following analysis will show that the VSEPR short-cut is specious; electron-nucleus attraction is the most important factor in determining electronic geometry and electron-electron repulsion actually increases in the transition from square planar to the more stable tetrahedral geometry in the system examined. If VSEPR gets the right answer, it does so for the wrong reason.
For the sake of mathematical simplicity the calculation of the total energy of an atomic configuration of charges will be used to refute the VSEPR theory that electron-electron repulsion determines geometry. Consider a +8 charge nuclear kernel interacting with four spherical two-electron charge clouds of uniform density and radius R in both square-planar and tetrahedral geometry. The kinetic energy of a uniform single electron charge cloud of radius R is 9/8R2. The intra-pair repulsion is 6/5R and the other electrostatic contributions to the total energy also go like +/- 1/R and are easily determined using elementary geometry. The large spheres in the figures shown below contain two electrons and the small sphere represents the +8 nuclear kernel which is assumed to have the same total energy in both geometries. Further computational detail can be found at http://www.users.csbsju.edu/~frioux/...loudModels.pdf.
For each geometry the total energy will be minimized with respect to the charge cloud radius using the Mathcad programming environment.
The calculations are carried out in atomic units: h/2π = me = e = ao = 4πεo = 1.
Square Planar Geometry
Total energy as a function of R: $E_D (R) = \frac{9}{R^2} + \frac{8}{R} + \frac{4}{ \sqrt{2} R} + 4 \frac{6}{5R} - \frac{64}{ \sqrt{2}R}$
Minimize energy with respect to R: $\begin{matrix} \begin{array}{c|c} R = \frac{d}{dR} E_D (R) = 0 & _{ \text{float, 4}} ^{ \text{solve, R}} \rightarrow 0.6076 \end{array} & E_D (R) = -24.381 \end{matrix}$
Kinetic energy: $\frac{9}{R^2} = 24.378$
Electron-Electron Repulsion: $\frac{8}{R} + \frac{4}{ \sqrt{2} R} + 4 \frac{6}{5R} = 25.722$
Electron-nucleus attraction: $- \frac{64}{ \sqrt{2} R} = -74.481$
Total potential energy: $25.722 - 74.481 = -48.759$
Average electron-nucleus distance: $\sqrt{2} R = 0.859$
Clear R memory: $R = R$
Tetrahedral Geometry
Total energy as a function of R: $E_T (R) = \frac{9}{R^2} + \frac{12}{R} + 4 \frac{6}{5R} - \frac{64}{ \sqrt{ \frac{3}{2}} R}$
Minimize energy with respect to R: $\begin{matrix} \begin{array}{c|c} R = \frac{d}{dR} E_T (R) = 0 & _{ \text{float, 4}} ^{ \text{solve, R}} \rightarrow 0.5077 \end{array} & E_T (R) = -34.920 \end{matrix}$
Kinetic energy: $\frac{9}{R^2} = 34.916$
Electron-electron repulsion: $\frac{12}{R} + 4 \frac{6}{5R} = 33.090$
Electron-nucleus attraction: $- \frac{64}{ \sqrt{ \frac{3}{2}} R} = -102.926$
Total potential energy: $33.090 - 102.926 = -69.836$
Average electron-nucleus distance: $\sqrt{ \frac{3}{2}} R= 0.622$
These calculations are now summarized in tabular format.
$\begin{pmatrix} \text{Energy Contribution} & \text{Square Planar} & \text{Tetrahedral} & \text{Change} \ \text{Kinetic Energy} & 24.378 & 34.916 & 10.538 \ \text{Electron electron potential energy} & 25.722 & 33.090 & 7.368 \ \text{Electron nucleus potential energy} & -74.481 & -102.926 & -28.445 \ \text{Total energy} & -24.381 & -34.920 & -10.539 \ \text{Electron nucleus distance} & 0.859 & 0.622 & -0.237 \end{pmatrix} \nonumber$
Of the three contributions to the total energy, electron-nucleus potential energy has the largest magnitude and is the only negative term. Kinetic energy and electron-electron potential energy have "repulsive" character and are of similar magnitudes for both geometries. They both increase in going from square-planar to tetrahedral geometry, but are overwhelmed by the very large decrease in electron-nuclear potential energy due to the closer proximity of the electrons to the nuclear kernel in the tetrahedral configuration shown in the bottom line of the table. The only conclusion that can be drawn from these calculations is that electron-nuclear attraction determines the configuration of valence electron density around a nonvalence nuclear kernel. And tetrahedral geometry is more stable because it is close-packed and brings the electrons closer to the nucleus than the square planar arrangement.
Further insight can be obtained by graphing the total energy as a function of R for both geometries.
$\begin{matrix} R = .2, .21 .. 5 & E_D (R) = \frac{9}{R^2} - \frac{29.626}{R} & E_T (R) = \frac{9}{R^2} - \frac{35.456}{R} \end{matrix} \nonumber$
Both the square planar and tetrahedral examples show that it is kinetic energy that is responsibility for the stability of matter. While potential energy is going to negative infinity like -1/R as R decreases, kinetic energy is going to positive infinity like 1/R2.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.24%3A_Another_Critique_of_VSEPR.txt
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Why do non‐bonding electron pairs occupy the equatorial rather than the axial positions in trigonal bipyramidal geometry? VSEPR teaches that it is because they have more room in the equatorial positions and this leads to lower electron‐electron repulsion. The Tangent Spheres Model (TSM) [Bent, H. A. J. Chem. Educ. 40, 446‐452 (1963)] calculation below shows that equatorial electrons indeed have a lower energy than the axial electrons, but not for the reason VSEPR postulates.
In the TSM approach electron pairs are represented as non‐penetrating spherical charge clouds of radius R and uniform charge density. These Pauli spheres are surogates for the hybridized atomic orbitals that are frequently invoked in VSEPR explanations. There are five such pairs (three equatorial and two axial) in trigonal bipyramidal geometry. The trigonal planar hole created by the equatorial electron pairs is occupied by the nuclear kernel (nucleus plus non‐valence electrons). The pockets created by the trigonal planar spheres are occupied by the two axial pairs of electrons. The equatorial electrons and one of the axial pairs is shown in the figure below.
In the calculations carried out here, the equatorial and axial electron spheres are allowed to have different radii. The nuclear kernel is treated as +10 point charge centered in the trigonal planar hole, and its electrostatic interaction with the valence electrons is its only contribution to the total energy. The contributions to the energy are identified in the table below. These Mathcad calculations are carried out using atomic units.
$\begin{bmatrix} \text{Energy Contribution} & \text{Trigonal Bipyramidal Geometry} \ \text{Electron Kinetic Energy} & 6 \frac{6}{8 R_e^2} + 4 \frac{9}{8 R_a^2} \ \text{Intra Pair Electron Repulsion} & 3 \frac{6}{5 R_e} + 2 \frac{6}{5 R_a} \ \text{Inter Pair Electron Repulsion} & 3 \frac{-2(-2)}{2R_e} + 6 \frac{-2(-2)}{R_e + R_a} + \frac{-2(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \end{bmatrix} \nonumber$
Seed values for the equatorial and axial radii: $\begin{matrix} R_e = .2 & R_a = R_e \end{matrix}$
Minimization of the energy with respect to the equatorial and axial radii:
$\begin{matrix} E \left( R_e,~R_a \right) = & 6 \frac{9}{8R_e^2} + 4 \frac{9}{8R_a^2} + 3 \frac{6}{5R_e} + 2 \frac{6}{5R_a} + 3 \frac{(-2)(-2)}{2R_e} + 6 \frac{(-2)(-2)}{R_e + R_a} ... \ & \frac{(-2)(-2)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} + 3 \frac{(-2)10}{ \frac{2}{3} \sqrt{3} R_e} + 2 \frac{(-2)10}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \end{matrix} \nonumber$
$\begin{matrix} \text{Given} & \frac{d}{dR_e} E \left( R_e,~R_a \right) = 0 & \frac{d}{dR_a} E \left( R_e,~R_a \right) = 0 \ \begin{pmatrix} R_e \ R_a \end{pmatrix} = \text{Find} \left( R_e,~R_a \right) & R_e = 0.317 ~ R_a = 1.057 & E \left( R_e,~R_a \right) = -71.389 \end{matrix} \nonumber$
In what follows the results of this calculation will be broken down into separate contributions for interpretive purposes. The first thing to note is that TSM predicts that the equatorial electrons are smaller than the axial electrons and, therefore, close to one another. This contradicts the VESPR idea that the equatorial position offers nonbonding electron pairs require more space.
The calculation immediately below shows that equatorial electrons have a lower energy than axial electrons. The question is why do they have a lower energy; is it because of reduced electron‐electron repulsion or some other factor contributing to the energy. The smaller radius of the equatorial electrons not only brings them closer to each other (increasing electron‐electron repulsion) it also brings them closer to the nuclear kernel (increasing nuclear‐electron attraction).
Individual equatorial electron energy:
$\begin{matrix} \varepsilon \left( R_e,~R_a \right) = \frac{9}{8 R_e^2} + \frac{6}{5R_e} + 4 \left( \frac{(-1)(-1)}{2R_e} \right) + 4 \left( \frac{(-1)(-1)}{R_e + R_a} \right) + \frac{(-1)(10)}{ \frac{2}{3} \sqrt{3} R_e} & \varepsilon \left( R_e,~R_a \right) = -3.111 \end{matrix} \nonumber$
Individual axial electron energy:
$\begin{matrix} \gamma \left( R_e,~R_a \right) = & \frac{9}{8R_a^2} + \frac{6}{5R_a} + 6 \frac{(-1)(-1)}{R_e + R_a} ... & & \gamma \left( R_e,~R_a \right) = -0.287 \ ~ & + \frac{(-1)(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} + \frac{(-1)(10)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \end{matrix} \nonumber$
To determine why the equatorial electrons have a lower energy than the axial electrons requires that the total equilibrium energy be broken down to its constituent parts. The relative importance of each energy type (kinetic, electron‐nucleus potential, electron‐electron potential) is calculated as the percentage of its magnitude to the sum of the magnitudes of all energy contributions.
Equatorial-equatorial electron repulsion:
$3 \frac{6}{5R_e} + 3 \frac{(-2)(-2)}{2R_e} = 30.327 \nonumber$
Axial-equatorial electron repulsion:
$6 \frac{(-2)(-2)}{R_e + R_a} = 17.469 \nonumber$
Axial-axial electron repulsion:
$2 \frac{6}{5R_a} + \frac{(-2)(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} = 3.78 \nonumber$
Total electron-electron repulsion:
$\begin{matrix} 30.327 + 17.469 + 3.78 = 51.576 & \colorbox{yellow}{16.3%} \end{matrix} \nonumber$
Equatorial electron-nuclear kernel attraction:
$3 \frac{ (-2)(10)}{ \frac{2}{3} \sqrt{3} R_e} = -164.151 \nonumber$
Axial electron-nuclear kernel attraction:
$2 \frac{(-2)(10)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} = -30.203 \nonumber$
Total electron-nuclear attraction:
$\begin{matrix} -164.151 - 30.203 = -194.354 & \colorbox{yellow}{61.2%} \end{matrix} \nonumber$
Equatorial electron kinetic energy:
$6 \frac{9}{8R_e^2} = 67.364 \nonumber$
Axial electron kinetic energy:
$4 \frac{9}{8R_a^2} = 4.025 \nonumber$
Total electron kinetic energy:
$\begin{matrix} 67.364 - 4.025 = 71.389 & \colorbox{yellow}{22.5%} \end{matrix} \nonumber$
Every detail of this calculation contradicts the VSEPR dogma. First, electron‐electron repulsion is the least important contribution to the total energy, significantly below electron kinetic energy. Second, equatorial‐equatorial electron repulsion is larger than equatorial‐axial and axial‐axial electron repulsion. Third, VSEPR completely ignores the most important contribution to the total energy in its prediction of molecular geometry ‐ electron‐nuclear potential energy. It also ignores the importance of kinetic energy in determining molecular stability and geometry.
The validity of this TSM calculation can be tested by other calculations in which protons are imbedded in the axial or equatorial electron pairs. Energy minimization with respect to the sphere radii (axial, equatorial, imbedded) should show a lower energy for the structure in which the axial electron pair has the imbedded proton.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.25%3A_Why_Nonbonding_Electrons_Occupy_the_Equatorial_Position_in_Trigonal_Bipyramidal_Geometry.txt
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Why do non‐bonding electron pairs occupy the equatorial rather than the axial positions in trigonal bipyramidal geometry? VSEPR teaches that it is because they have more room in the equatorial positions and this leads to lower electron‐electron repulsion. The modified Tangent Spheres Model (TSM) [see Bent, H. A. J. Chem. Educ. 40, 446‐452 (1963) for an introduction] calculation below shows that equatorial electrons indeed have a lower energy than the axial electrons, but not for the reason VSEPR postulates.
In this modified TSM approach an electron is represented as a particle in a sphere of radius R. The wave function for such an electron is the eigenfunction of the kinetic energy operator in spherical coordinates, with energy π2/(2R2) in atomic units, as is shown below.
$\begin{matrix} \Phi (r,~R) = \frac{1}{ \sqrt{2 \pi R}} \frac{ \sin \left( \frac{ \pi r}{R} \right)}{r} & \frac{ - \frac{1}{2r} \frac{d^2}{dr^2} \left( r \Phi (r,~R) \right)}{\Phi (r,~R)} \rightarrow \frac{1}{2} \frac{ \pi^2}{R^2} \end{matrix} \nonumber$
According to the Pauli Exclusion Principle each spherical wave function (orbital) can accomodate two electrons. These non‐interpenetrating Pauli spheres are surogates for the hybridized atomic orbitals that are frequently invoked in VSEPR explanations.
There are five such pairs (three equatorial and two axial) in trigonal bipyramidal geometry. The trigonal planar hole created by the equatorial electron pairs is occupied by the nuclear kernel (nucleus plus non‐valence electrons). The pockets created by the trigonal planar spheres are occupied by the two axial pairs of electrons. The equatorial electrons and one of the axial pairs is shown in the figure below.
In the calculations carried out here, the equatorial and axial electron spheres are allowed to have different radii. This means there are two variational parameters in the energy calculation, the radii of the axial and equatorial spherical electron clouds. The nuclear kernel is treated as +10 point charge centered in the trigonal planar hole, and its electrostatic interaction with the valence electrons is its only contribution to the total energy. The contributions to the energy are identified in the table below. These Mathcad calculations are carried out using atomic units.
$\begin{bmatrix} \text{Energy Contribution} & \text{Trigonal Bipyramidal Geometry} \ \text{Electron Kinetic Energy} & 6 \frac{ \pi^2}{2R_e^2} + 4 \frac{ \pi^2}{2R_a^2} \ \text{Intra Pair Electron Repulsion} & 3 \frac{1.786}{R_e} + 2 \frac{1.786}{R_a} \ \text{Inter Pair Electron Repulsion} & 3 \frac{(-2)(-2)}{2R_e} + 6 \frac{(-2)(-2)}{R_e + R_a} + \frac{(-2)(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \ \text{Electron Kernel Attraction} & 3 \frac{(-2)(10)}{ \frac{2}{3} \sqrt{3} R_e} + 2 \frac{(-2)(10)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \end{bmatrix} \nonumber$
Seed values for the equatorial and axial radii: $\begin{matrix} R_e = .2 & R_a = R_e \end{matrix}$
$\begin{matrix} E \left( R_e,~R_a \right) = & 6 \frac{ \pi^2}{2 R_e^2} + 4 \frac{ \pi^2}{2R_a^2} + 3 \frac{1.786}{R_e} + 2 \frac{1.786}{R_a} + 3 \frac{(-2)(-2)}{2R_e} + 6 \frac{(-2)(-2)}{R_e + R_a} ... \ & + \frac{(-2)(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} + 3 \frac{(-2)(10)}{ \frac{2}{3} \sqrt{3} R_e} + 2 \frac{(-2)(10)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \end{matrix} \nonumber$
$\begin{matrix} \text{Given} & \frac{d}{dR_e} E \left( R_e,~R_a \right) = 0 & \frac{d}{dR_a} E \left( R_e,~R_a \right) = 0 \ \begin{pmatrix} R_e \ R_a \end{pmatrix} = \text{Find} \left( R_e,~R_a \right) & R_e = 1.449 ~ R_a = 5.331 & E \left( R_e,~R_a \right) = -14.799 \end{matrix} \nonumber$
In what follows the results of this calculation will be broken down into separate contributions for interpretive purposes. The first thing to note is that TSM predicts that the equatorial electrons are smaller than the axial electrons and, therefore, close to one another. This contradicts the VESPR idea that the equatorial position offers nonbonding electron pairs require more space.
The calculation immediately below shows that equatorial electrons have a lower energy than axial electrons. The question is why do they have a lower energy; is it because of reduced electron‐electron repulsion or some other factor contributing to the energy. The smaller radius of the equatorial electrons not only brings them closer to each other (increasing electron‐electron repulsion) it also brings them closer to the nuclear kernel (increasing nuclear‐electron attraction).
Individual equatorial electron energy:
$\begin{matrix} \varepsilon \left( R_e,~R_a \right) = \frac{ \pi^2}{2R_e^2} + \frac{1.786}{R_e} + 4 \left( \frac{(-1)(-1)}{R_e} \right) + 4 \left( \frac{(-1)(-1)}{R_e + R_a} \right) + \frac{(-1)(10)}{ \frac{2}{3} \sqrt{3} R_e} & \varepsilon \left( R_e,~R_a \right) = -0.423 \end{matrix} \nonumber$
Individual axial electron energy:
$\begin{matrix} \gamma \left( R_e,~R_a \right) = & \frac{ \pi^2}{2R_a^2} + \frac{1.786}{R_a} + 6 \frac{(-1)(-1)}{R_e + R_a} ... & & \gamma \left( R_e,~R_a \right) = 0.024 \ ~ & + \frac{(-1)(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} + \frac{(-1)(10)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} \end{matrix} \nonumber$
To determine why the equatorial electrons have a lower energy than the axial electrons requires that the total equilibrium energy be broken down to its constituent parts. The relative importance of each energy type (kinetic, electron‐nucleus potential, electron‐electron potential) is calculated as the percentage of its magnitude to the sum of the magnitudes of all energy contributions.
Equatorial-equatorial electron repulsion:
$3 \frac{1.786}{R_e} + 3 \frac{(-2)(-2)}{2R_e} = 7.839 \nonumber$
Axial-equatorial electron repulsion:
$6 \frac{(-2)(-2)}{R_e + R_a} = 3.54 \nonumber$
Axial-axial electron repulsion:
$2 \frac{1.786}{R_a} + \frac{(-2)(-2)}{2 \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} = 0.974 \nonumber$
Total electron-electron repulsion:
$\begin{matrix} 7.839 + 3.54 + 0.974 = 12.353 & \colorbox{yellow}{17.9%} \end{matrix} \nonumber$
Equatorial electron-nuclear kernel attraction:
$3 \frac{(-2)(10)}{ \frac{2}{3} \sqrt{3} R_e} = -35.863 \nonumber$
Axial electron-nuclear kernel attraction:
$2 \frac{(-2)(10)}{ \sqrt{ \left( R_e + R_a \right)^2 - \frac{4}{3} R_e^2}} = -6.088 \nonumber$
Total electron-nuclear attraction:
$\begin{matrix} -35.863 - 6.088 = -41.951 & \colorbox{yellow}{60.7%} \end{matrix} \nonumber$
Equatorial electron kinetic energy:
$6 \frac{ \pi^2}{2R_e^2} = 14.104 \nonumber$
Axial electron kinetic energy:
$4 \frac{ \pi^2}{2R_a^2} = 0.695 \nonumber$
Total electron kinetic energy:
$\begin{matrix} 14.104 + 0.695 = 14.799 & \colorbox{yellow}{21.4%} \end{matrix} \nonumber$
Every detail of this calculation contradicts the VSEPR dogma. First, electron‐electron repulsion is the least important contribution to the total energy, significantly below electron kinetic energy. Second, equatorial‐equatorial electron repulsion is larger than equatorial‐axial and axial‐axial electron repulsion. Third, VSEPR completely ignores the most important contribution to the total energy in its prediction of molecular geometry ‐ electron‐nuclear potential energy. It also ignores the importance of kinetic energy in determining molecular stability and geometry.
Appendix: Calculation of intra‐pair electron repulsion for sphere of unit radius.
$\begin{matrix} R = 1 & \int_0^R \Psi (r,~R)^2 \left( \frac{1}{r} \int_0^r \Phi (x,~R)^2 4 \pi x^2 dx + \int_0^R \frac{ \phi (x,~R)^2 4 \pi x^2}{x} dx \right) 4 \pi r^2 dr = 1.786 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.26%3A_A_Modified_Tangent_Spheres_Model_Analysis_of_Trigonal_Bipyramidal_Geometry.txt
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Enter and solve the butadiene Huckel determinant for energy eigenvalues.
$\begin{array}{c|c} \left| \begin{pmatrix} \alpha - E & \beta & 0 & 0 \ \beta & \alpha - E & \beta & 0 \ 0 & \beta & \alpha - E & \beta \ 0 & 0 & \beta & \alpha - E \end{pmatrix} \right| = 0 & _{ \text{float, 4}}^{ \text{solve, E}} \rightarrow \begin{pmatrix} \alpha + .6180 \beta \ \alpha - 1.618 \beta \ \alpha + 1.618 \beta \ \alpha - .6180 \beta \end{pmatrix} \end{array} \nonumber$
Calculate the eigenvectors:
$\begin{array}{c|c} \text{eigenvecs} \left( \begin{pmatrix} \alpha - E & \beta & 0 & 0 \ \beta & \alpha - E & \beta & 0 \ 0 & \beta & \alpha - E & \beta \ 0 & 0 & \beta & \alpha - E \end{pmatrix} \right) = 0 & _{ \text{float, 4}}^{ \text{simplify}} \rightarrow \begin{pmatrix} .317 & .6013 & -.6013 & -.3717 \ .6014 & -.3716 & .3716 & .6014 \ .6014 & .3716 & .3716 & -.6014 \ .3717 & .6013 & .6013 & .3717 \end{pmatrix} \end{array} \nonumber$
Construct an energy level diagram and show the occupied levels.
$\begin{matrix} \text{Energy} & \text{Occupancy} & \text{Wave function coefficients} \ \alpha - 0.618 \beta & - & \begin{pmatrix} -.3717 & .6014 & -.6014 & .3717 \end{pmatrix} \ \alpha - 0.618 \beta & - & \begin{pmatrix} -.6014 & .3717 & .3717 & -.6014\end{pmatrix} \ \alpha + 0.618 \beta & _{-}xo _{-} & \begin{pmatrix} -.6013 & -.3716 & .3716 & .6013 \end{pmatrix} \ \alpha + 1.618 \beta & _{-}xo_{-} & \begin{pmatrix} .3717 & .6013 & .6013 & .3717 \end{pmatrix} \end{matrix} \nonumber$
Calculate the π-electron energy:
$E_ \pi = \left[ 2 \left( \alpha + 1.618 \beta \right) + 2 \left( \alpha + 0.618 \beta \right) \right] \rightarrow E_ \pi = 4 \alpha + 4.472 \beta \nonumber$
Calculate the delocalization energy:
$E_d = \left[ 4 \alpha + 4.472 \beta - 2 \left( 2 \alpha + 2 \beta \right) \right] \rightarrow E_d = .472 \beta \nonumber$
Calculate the wavelength of the photon required for the HOMO-LUMO transition.
$\begin{array}{c|c} \frac{hc}{ \lambda} = \left( \alpha - 0.618 \beta \right) - \left( \alpha + 0.618 \beta \right) & _{ \text{float, 3}} ^{ \text{solve, } \lambda} \rightarrow -.809 h \frac{c}{ \beta} \end{array} \nonumber$
3.28: A Numeric Huckel MO Calculation Using Mathcad
Enter the number of carbon atoms. $\text{Natoms} = 4$
Enter the number of occupied molecular orbitals. $\text{Nocc} = 2$
Enter the Huckel matrix.
$\begin{matrix} H = \begin{pmatrix} 0 & 1 & 0 & 0 \ 1 & 0 & 1 & 0 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 0 \end{pmatrix} & H = -H \end{matrix} \nonumber$
Calculate eigenvalues and eigenvectors:
$\begin{matrix} E = \text{eigenvals(H)} & \text{Display = rsort} \left( \text{stack} \left( E^T,~ \text{eigenvecs(H)} \right),~ 1 \right) \end{matrix} \nonumber$
Display eigenvalues and eigenvectors:
$\text{Display} = \begin{pmatrix} -1.618 & -0.618 & 0.618 & 1.618 \ 0.372 & -0.602 & 0.602 & -0.372 \ 0.602 & -0.372 & -0.372 & 0.602\ 0.602 & 0.372 & -0.372 & -0.602 \ 0.372 & 0.602 & 0.602 & 0.372 \end{pmatrix} \nonumber$
Display energy level diagram. $\begin{matrix} \text{E = sort(E)} & \text{i = 1 .. Natoms} \end{matrix}$
Calculate total π-electronic energy:
$\begin{matrix} E_ \pi = 2 \sum_{i = 1} ^{ \text{Nocc}} E_i & E_ \pi = -4.472 \end{matrix} \nonumber$
Calculate the delocalization energy:
$\begin{matrix} E_{deloc} = E_ \pi + 2 \text{Nocc} & E_{deloc} = -0.472 \end{matrix} \nonumber$
Calculate the delocalization energy per atom:
$\frac{ \text{E}_{deloc}}{ \text{Natoms}} = -0.118 \nonumber$
$C = \text{submatrix (Display, 2, Natoms + 1, 1, Natoms)} \nonumber$
Enter the number of the molecualr orbital to be plotted.
$\begin{matrix} r = 1 & s = 1 & 2 \sum_{i=1}^{ \text{Nocc}} \left[ \left( C^{<i>} \right)_r \left( C^{<i>} \right)_s \right] = 1 & \pi- \text{electron density on carbon 1} \ r = 1 & s = 2 & 2 \sum_{i=1}^{ \text{Nocc}} \left[ \left( C^{<i>} \right)_r \left( C^{<i>} \right)_s \right] = 0.894 & \pi- \text{bond order between carbons 1 and 2} \ r = 2 & s = 3 & 2 \sum_{i=1}^{ \text{Nocc}} \left[ \left( C^{<i>} \right)_r \left( C^{<i>} \right)_s \right] = 0.447 & \pi- \text{bond order between carbons 2 and 3} \ r = 3 & s = 4 & 2 \sum_{i=1}^{ \text{Nocc}} \left[ \left( C^{<i>} \right)_r \left( C^{<i>} \right)_s \right] = 1 & \pi- \text{bond order between carbons 3 and 4} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.27%3A_A_Symbolic_Huckel_MO_Calculation_Using_Mathcad.txt
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1. Number the carbons after inspection of the molecular structure and fill in data needed below.
$\begin{matrix} \text{Natoms = 60} & \text{the number of carbon atoms and } π- \text{electrons} \ \text{Nocc = 30} & \text{the number of occupied molecular occupied} \end{matrix} \nonumber$
2. Create a 60 x 60 null matrix.
$\begin{matrix} \text{i = 1 .. Natoms} & \text{j = 1 .. Natoms} & \text{H}_{i,~j} = 0 \end{matrix} \nonumber$
3. Form the Huckel matrix from the null matrix using the results of part 1.
$\begin{matrix} \text{k = 1 .. Natoms - 1} & \text{H}_{k,~k+1} = 1 \end{matrix} \nonumber$
$\begin{matrix} H_{1,~5} = 1 & H_{1,~9} = 1 & H_{2,~12} = 1 & H_{3,~15} = 1 & H_{4,~18} = 1 & H_{6,~20} = 1 & H_{7,~22} = 1 \ H_{8,~25} = 1 & H_{10,~26} = 1 & H_{11,~29} = 1 & H_{13,~30} = 1 & H_{14,~33} = 1 & H_{16,~34} = 1 & H_{17,~37} = 1 \ H_{19,~38} = 1 & H_{21,~40} = 1 & H_{23,~42} = 1 & H_{24,~44} = 1 & H_{27,~45} = 1 & H_{28,~47} = 1 & H_{31,~48} = 1 \ H_{32,~50} = 1 & H_{35,~51} = 1 & H_{36,~53} = 1 & H_{39,~54} = 1 & H_{41,~55} = 1 & H_{43,~57} = 1 & H_{46,~58} = 1 \ H_{49,~59} = 1 & H_{52,~60} = 1 & H_{56,~60} = 1 & H_{j,~i} = H_{i,~j} & H = -H\end{matrix} \nonumber$
4. Calculate the eigenvalues and eigenvectors. It is not feasible to look at the eigenvectors as a group because that would require displaying a 60 x 60 matrix. The eigenvalues will be displayed later.
$\begin{matrix} \text{E = eigenvals(H)} & \text{C = submatrix} \left( \text{rsort} \left( \text{stack} \left( \text{E}^T,~ \text{eigenvecs(H)} \right), ~ 1 \right),~2 \text{Natoms +1, 1, Natoms} \right) \end{matrix} \nonumber$
5. Use the eigenvectors to calculate selected π-electron densities and π-bond orders. If r = s you are calculating the π-electron density on carbon r. If r is not equal to s you are calculating the r-s π-bond order. Several examples are given below. These calculations can be used to show that all the carbons have the same π-electron density and that there are only two physically meaningful π-bond orders.
$\begin{matrix} r = 1 & s = 1 & 2 \sum_{i = 1}^{ \text{Nocc}} \left( C^{<i>} \right)_r,~ \left( C^{<i>} \right)_s = 1 & \pi- \text{electron density on carbon 1} \ r = 1 & s = 2 & 2 \sum_{i = 1}^{ \text{Nocc}} \left( C^{<i>} \right)_r,~ \left( C^{<i>} \right)_s = 0.476 & \pi- \text{bond order between carbons 1 and 2} \ r = 1 & s = 5 & 2 \sum_{i = 1}^{ \text{Nocc}} \left( C^{<i>} \right)_r,~ \left( C^{<i>} \right)_s = 0.476 & \pi- \text{bond order between carbons 1 and 5} \ r = 1 & s = 9 & 2 \sum_{i = 1}^{ \text{Nocc}} \left( C^{<i>} \right)_r,~ \left( C^{<i>} \right)_s = 0.601 & \pi- \text{bond order between carbons 1 and 9} \end{matrix} \nonumber$
Note that the π-bond order is higher for bonds that fuse two hexagons compared to bonds that fuse a hexagon and a pentagon.
6. Display the energy eigenvalues: $\text{E = sort(E)}$
$\begin{matrix} i = 1 .. 15 & E_i & i = 16 .. 30 & E_i & i = 31 .. 45 & E_i & i = 46 .. 60 & E_i \ ~ & \begin{array}{|c|} \hline -3 \ \hline -2.757 \ \hline -2.757 \ \hline -2.757 \ \hline -2.303 \ \hline -2.303 \ \hline -2.303 \ \hline -2.303 \ \hline -2.303 \ \hline -1.82 \ \hline -1.82 \ \hline -1.82 \ \hline -1.562 \ \hline -1.562 \ \hline -1.562 \ \hline \end{array} & ~ & \begin{array}{|c|} \hline -1.562 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -1 \ \hline -0.618 \ \hline -0.618 \ \hline -0.618 \ \hline -0.618 \ \hline -0.618 \ \hline \end{array} & ~ & \begin{array}{|c|} \hline 0.139 \ \hline 0.139 \ \hline 0.139 \ \hline 0.382 \ \hline 0.382 \ \hline 0.382 \ \hline 1.303 \ \hline 1.303 \ \hline 1.303 \ \hline 1.303 \ \hline 1.303 \ \hline 1.438 \ \hline 1.438 \ \hline 1.438 \ \hline 1.618 \ \hline \end{array} & ~ & \begin{array}{|c|} \hline 1.618 \ \hline 1.618 \ \hline 1.618 \ \hline 1.618 \ \hline 2 \ \hline 2 \ \hline 2 \ \hline 2 \ \hline 2.562 \ \hline 2.562 \ \hline 2.562 \ \hline 2.562 \ \hline 2.618 \ \hline 2.618 \ \hline 2.618 \ \hline \end{array} \end{matrix} \nonumber$
The energy eigenvalues can also be displayed graphically as follows: $i = 1 .. 60$
Attention should be drawn to the nine-fold degenerate state at E = -1. C60 belongs to the icosahedral point group and the largest degeneracy permitted is 5-fold. Thus, this state is an example of the accidental degeneracy of a 5-fold degenerate state and a 4-fold degenerate state.
7. It is easy to show that this manifold of energy states is in agreement with the basic facts about C60. For example, it is diamagnetic and a non-conductor. Placing the 60 π-electrons is the lowest available energy states completely fills all the bonding molecular orbitals. The HOMO has an energy of -.618 |β| and contains ten paired electrons giving a diamagnetic and non-conducting electronic structure. The LUMO is three-fold degenerate and not far away in energy. This is consistent with the fact that C60 has a high electron affinity and forms ionic compounds with alkali metals such as potassium. For example, K3C60 (three unpaired electrons in the LUMO) is a conductor and becomes a superconductor at low temperatures, while K6C60 (six paired electrons in the LUMO) is a non-conductor.
The Huckel calculation can be extended with a simple example of modeling. Since the results of the following calculation require the assignment of a value for β, the results should not be taken too seriously. The HOMO-LUMO gap is .757 |β|. Giving |β| a typical value 2.5 eV enables one to calculate the wavelength of light required to promote an electron from the HOMO to the LUMO.
$\begin{matrix} eV = 1.6021777 (10)^{-19} \text{joule} & h = 6.62608 (10)^{-34} \text{joule sec} & c = 2.99792458 (10)^8 \frac{m}{sec} \ \beta = 2.5 eV & nm = 10^{-9} m & \lambda = 1 nm \ \text{Given} & \frac{hc}{ \lambda} = .757 | \beta | & \text{Find}( \lambda ) = 655 nm \end{matrix} \nonumber$
This result is just barely within the visible region and not in particularly good agreement with a known visible transition around 400 nm. However, it should be noted that the HOMO-LUMO transition is formally forbidden. The HOMO-LUMO+1 is allowed and the energy gap is |β| which would give an optical transition at 496 nm, in better agreement with the experimental value.
There are a number of very weak features in the visible spectrum between 440 and 670 nm any of which may be attributable to the forbidden HOMO-LUMO transtion. However, it is also true that small amounts of other fullerenes may be contributing to this part of the visible spectrum.
8. Calculate the π-electron stabilization energy per carbon atom. This is calculated by summing the energies of the occupied orbitals and multiplying by two. Subtract from this the π-electron binding energy of the equivalent number of ethylene molecules and divide by the number of carbon atoms, which is the same as the number of π-electrons.
$\begin{matrix} \Delta E_{ \pi} = \frac{2 \sum_{i - 1}^{ \text{Nocc}} \left| E_i \right| - \text{Natoms}}{ \text{Natoms}} & \Delta E_{ \pi} = 0.553 \end{matrix} \nonumber$
Recall that the π-electron stabilization energy per carbon atom for benzene is 0.333.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.29%3A_A_Numerical_Huckel_MO_Calculation_on_C60.txt
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As is well-known C60 resembles an American soccerball, containing 12 pentagons and 20 hexagons (see Figure 1). Removing the leather, but keeping the seams, leaves 60 vertices for the carbon atoms and 90 covalent bonds between them, 60 single bonds and 30 double bonds.
However, we should note that the actual bonding in C60 is somewhat more complex. The reason for this is that sp2 hybrid orbitals all lie in the same plane, but inspection of any model of buckminsterfullerene clearly shows that the environment at each carbon is not planar. So the hybridization at each carbon cannot be pure sp2, and current research indicates that the actual hybridization is sp2.3. Since we are trying to develop a simple, but serviceable, model of the bonding and electronic structure of C60, we will assume sp2 hybridization as a first approximation. It might be a good idea at this point to review what Zumdahl has to say about models on pages 362 to 364 in his textbook.
In chapter 8 we spent a fair amount of time drawing resonance structures for molecules and ions that had more than one correct Lewis structure. C60 is such a molecule and, as a matter of fact, it has 12,500 resonance structures! Try drawing at least two resonance structures, one in which all double bonds are in the hexagons, and one in which the pentagons have double bonds.
Associated with the concept of resonance is the idea of delocalization of electrons. For example, for a molecule with two resonance structures of equal plausibility (review formal charge arguments) we would say that the molecule's electronic structure is not correctly represented by either one or the other, but an average of the two (see Zumdahl's discussion of resonance in section 8.12). The fact that more than one Lewis structure can be written for a molecule is an indication of weakness in the Lewis' localized electron pair model of chemical bonding, because the existence of resonance structures actually delocalizes the r-electrons over the molecule as a whole. Benzene is, perhaps, the best example of this. Rather than writing structures A or B below, we write C and think Figure 9.48b in Zumdahl.
Since C60 has 12,500 resonance structures involving its 60 r-electrons we may profitably consider the r-electrons to be particles delocalized over a sphere. The solution of Schrödinger's equation for a particle on a sphere is well known. Since a particle on a sphere is restricted to move in two directions, in spherical polar coordinates its position can be described by two angles, θ and ϕ. Thus, two spatial quantum numbers are required and they are the old familiar 1 and m1 of Chapter 7 of Zumdahl. Since the particles in question are electrons, the spin quantum number, ms, is also required. The rules for the quantum numbers are as stated in Chapter 7: 1 can have integer values starting with 0; m1 can have integer values ranging from -1 to +1; ms, can have the values of ±1/2.
Using the rules for the 1 and m1 quantum numbers makes it possible to construct an energy level diagram for an electron on a sphere. This is shown in Figure 4.
This figure shows 10 unpaired electrons (Hund's Rule) in the highest occupied energy level. This would suggest that C60 is paramagnetic and perhaps a conductor of electricity. In fact C60 is diamagnetic (no unpaired electrons) and does not conduct electricity. Thus, this initial model for the bonding and electronic structure of C60 is not correct.
A way out of this difficulty is to recall that C60 is not a sphere, but a truncated iscosohedron. The symmetry of the truncated icosohedron is high, but not as high as a perfect sphere. The main consequence of this is that the degeneracies of some of the higher energy levels are split by the lower iscosahedral symmetry. You will see more of this in CHEM 234 when we study transition metal complexes and crystal field theory. Figure 6 shows what happens to the energy levels under icosahedral symmetry and how they are filled with 60 electrons.
Note that this energy level diagram is in agreement with the basic facts about C60. It is a non-conductor and diamagnetic. In addition, it suggests that C60 might have a fairly high electron affinity because it has a low-lying unoccupied energy level with room for six electrons. This is consistent with the fact that K3C60 is known and is a conductor and passes into a super-conducting state at low temperature. The electro-positive potassium atoms supply three electrons which half-fill the lowest unoccupied energy level. K6C60 is also known and is a non-conductor. This too is consistent with the model as the potassiums supply six valence electrons completely filling the C60's lowest unoccupied energy level.
In Chapter 10 we will see that the crystal structures of solid C60, K3C60, and K6C60 can be described using the same concepts as are used to describe the more familiar structures of NaCl, CsCl2, and CaF2. For now we can say, for example, that solid buckminsterfullerene consists of a face-centered cubic packing of spherical C60 molecules (see Figures 10.15 and 10.17 in Zumdahl). K3C60 can be described as face-centered cubic C603- ions with K+ ions in all the tetrahedral and octahedral holes (see Figure 10.33 in Zumdahl for an illustration of tetrahedral and octahedral holes). You should confirm that this structure gives the correct stoichiometric ratios for the ions. By contrast K6C60 is body centered C603- ions with four K+ ions in each of the six cube faces as shown in Figure 7 below. Again you should confirm that this structure gives the correct ratio of the stoichiometric coefficients.
While it is not possible to go into detail at this point, it is also the case that the model presented here for the bonding and electronic structure of buckminsterfullerene is consistent with the absorption spectra of C60 in the ultraviolet, visible, and infrared regions of the electromagnetic spectrum.
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The recent discovery of a new allotropic form of carbon1 and its production in macroscopic amounts2 has generated a tremendous amount of research activity in chemistry, physics and material science.3 Among the areas of current interest are the electronic properties of the fullerenes and this note describes a simple model for the electronic structure of Coo that is consistent with recent experimental findings. It is based on the most elementary principles of quantum mechanics and group theory.
After the soccer-ball structure for C60 was first suggested in 1985 it became important to obtain as much independent supporting evidence as possible. This came in the areas of nmr, IR, and Raman spectroscopy. The nmr spectrum showed a single resonance indicating 60 equivalent carbon atoms in the structure. The IR spectrum was found to have four lines, while Raman spectroscopy yielded ten lines. The second half of this paper will use group theory to demonstrate that a C60 molecule with a soccer-ball structure must have four IR active and ten Raman active vibrational modes.
Electronic Structure
As is well-known C60 is a carbon cage consisting of 20 hexagons and 12 pentagons and resembles a soccerball. Removing the leather, but keeping the seams, leaves 60 vertices for the carbon atoms and 90 covalent bonds between them. Actually C60 has spheroidal geometry and belongs to the truncated icosahedral symmetry group, lh. Curly and Smalley, codiscoverers of buckminsterfullerene, have described it as the "roundest molecule that can possibly exist"4 so the model presented here assumes, initially, that Coo is a perfect sphere. Each carbon is sigma bonded to three other carbons using three of its four valence electrons to form these bonds. The remaining electron is considered to be delocalized on the surface of the sphere created by the 60 atom carbon cage.
The quantum mechanical behavior of an individual electron restricted to the surface of a sphere is well-known.5 Solving Schrödinger's equation for a particle on a sphere yields the spherical harmonic wavefunctions, pictures of which can be found in Atkins' physical chemistry textbook. The energy levels associated with the spherical harmonic states are a function of the radius of the sphere and the angular momentum quantum number.
$E_L = \frac{h^2}{8 \pi^2 m_e R^2} L(L+1) \nonumber$
Just as the quantum mechanical solution for the one-electron hydrogen atom can be adapted for qualitative treatments of the electronic structure of multi-electron atoms, so the energy level diagram for the electron moving on a sphere can be used to describe the electronic structure of C60. The energy level diagram shown in Figure 1 provides a qualitative description of the electronic structure obtained when one applies the aufbau principle, the Pauli exclusion principle, and Hund' s rule to the addition of 60 electrons to the available spherical harmonic energy levels.
As it stands this is not a satisfactory picture because pristine C60 is an insulator and has no un-paired electrons. This difficulty is resolved by recalling that C60 is not a perfect sphere, but has the lower symmetry of the icosahedral group. Invoking icosahedral symmetry at this point splits the degeneracies of all levels above L = 2. However, initially it is only necessary to examine what happens to the highest occupied level, L = 5, since all other levels are completely filled.
Using traditional group theoretical methods6, it can be shown that the L = 5 spherical harmonics transform under the rotations of the icosahedral symmetry group as shown in the last row of the icosahedral character table shown below. The behavior of the spherical harmonics under the rotations of the icosahedral group is given by
$\chi \left( c_{ \alpha} \right) = \frac{ \sin \left( L + \frac{1}{2} \right) \alpha}{ \sin \frac{ \alpha}{2}} \nonumber$
Ih E 12C5 12C52 20C3 15C2
A 1 1 1 1 1
T1 3 1.618 -.618 0 -1
T2 3 -.618 1.618 0 -1
G 4 -1 -1 1 0
H 5 0 0 -1 1
L=5 11 1 1 -1 -1
It is easy to show that this reducible representation is a linear combination of the fivefold degenerate Hu, an the two three-fold degenergate, T1u and T2u irreducible representations of the icosahedral group. Group theory doesn't predict the order of the levels, but Figure 2 shows that if the five-fold degenerate level is placed lowest, an energy level diagram that captures the essentials of the known electronic structure of C60 is obtained.7,8 This assignment is consistent with HOMO, LUMO, and LUMO+ 1 levels of the Huckel molecular orbital calculation on C60.8,9 In addition, if the splittings of the L = 3 and L = 4 states are also examined in the manner outlined above, the complete energy level diagram for the π-electrons of C60 shown in Figure 3 is obtained. This set of π-electron levels is qualitatively consistent with the results of an ab initio calculation based on the pseudopotentiallocal density method.10
The HOMO-LUMO energy gap is known to be 1.5 eV. However, the HOMO -> LUMO electronic transition is optically forbidden. Because Hu level is full, (Hu)10, the ground electronic state has Ag symmetry. The first excited state (Hu)9(T1u)1 produces the reducible representation shown in the table below. This is obtained by taking the direct product of the Hu and T1u irreducible representations (HuxT1u). The reducible representation of the second excited state, (Hu)9(T1g)1 is also shown in the table. It is important to note that both excited states are 15-fold degenerate.
E 12C5 12C52 20C3 15C2 i 12S10 12S102 20S6 15
σ
HuT1u 15 0 0 0 -15 15 0 0 0 15
HuT1g 15 0 0 0 -15 -15 0 0 0 15
Employing the usual methods it is not difficult to show that the first and second excited electronic configurations contain the following irreducible representations.
$\begin{matrix} \left( H_u \right)^9 \left(T_{1u} \right)^1 \rightarrow T_{1g} + T_{2g} + G_g + H_g \ \left( H_u \right)^9 \left( T_{1g} \right)^1 \rightarrow T_{1u} + T_{2u} + G_u + H_u \end{matrix} \nonumber$
Thus, it can be seen that the excited electronic configurations each give rise to four excited states. In order for a transition to any of these states from the ground state to be allowed, the transition probability integral ∫ΨiμΨfdτ must be non-zero. For this integral to be non-zero the direct product of the irreducible representations for the ground electronic state, the electric dipole operator, and the excited electronic state must contain the totally symmetric irreducible representation Ag. Because the ground state itself has Ag symmetry and the electric dipole operator has T1u symmetry, only an excited state with T1u symmetry will lead to a direct product which contains the Ag irreducible representation.
Inspection of the irreducible representations for the first and second exited electronic configurations shows that only the second excited electronic configuration contains the T1u irreducible representation. Thus, while the HOMO → LUMO transition is forbidden, the HOMO → LUMO+ 1 transition is allowed. This result is consistent with the visible spectrum of the free molecule.9
A further comment on the magnitude of the HOMO-LUMO gap might be made at this point. The energies of the spherical harmonic states shown in Figure 1 were calculated using equation (1) and a value of 710 pm for the diameter of the carbon cage. At the L = 5 level the energy difference between adjacent states is 3.1 and 3.6 eV. While the model doesn't provide a detailed quantitative analysis of.the splitting of the L = 5 level, using reasonable assumptions one can obtain a value for the HOMO-LUMO gap that is "in the ball park. "
In summary, this analysis provides a simple interpretation of the fact that C60 is an insulator. The model also provides low-lying, un-occupied orbitals to form conduction bands and receive electrons from donors such as potassium. Furthermore, the fact that the LUMO is triply degenerate is consistent with the experimental evidence that K3C60 is a conductor and K6C60 is an insulator.7 While this simple model is not a rival to the more sophisticated molecular orbital or band theory calculations, it does provide the non-specialist with an appealing and simple alternative.
Vibrational Spectroscopy
To analyze the vibrational modes of C60 using group theory it is necessary to determine how the 180 degrees of freedom of the C60 molecule transform under the symmetry operations of the Ih group. This is actually quite easily done because the rotations and the inversion symmetry operation move all the carbon atoms and, therefore, have characters of O. The identity operation leaves all carbons unmoved for a character of 180, while the 15 planes of symmetry contain four carbon atoms each and can be shown to have a character of 4. The model of C60 on the first page shows one of these planes. It is perpendicular to the plane of the paper and clearly contains four carbon atoms, two at the top and two at the bottom.
This makes it very easy to decompose the reducible representation, Γtot, into its irreducible representations by the usual methods. This is summarized in the table on the next two pages. For simplicity only the E and cr symmetry operations are shown, but it must be remembered that there are 120 symmetry operations total. For example, the occurrence of Hg is calculated as follows:
$\Lambda_{tot} \cdot H_g = \frac{[(1)(180)(5)+(15)(4)(1)]}{120} = 8 \nonumber$
After translation and rotation are subtracted from the total, 174 vibrational degrees of freedom remain. However, group theory shows that many vibrational modes are degenerate. In fact, as the table below shows, there are only 46 distinct vibrational frequencies.
$\Lambda_{vib} = 2A_g + 3T_{1g} + 4T_{2g} + 6G_g + 8H_g + A_u + 4T_{1u} + 5T_{2u} + 6G_u + 7H_u \nonumber$
Of these, the table indicates that only the four triply degenerate T1u modes are IR active while ten vibrational modes (2Ag + 8Hg) are Raman active. This analysis is in complete agreement with the experimental measurements11 and was considered to be crucial evidence in support of the proposed soccer-ball structure for C60. Several of the rotational axes for C60 are shown below to illustrate that they do move all atoms.
Ih E 15σ Occurrence h = 120
Γxyz
180 4
Ag 1 1 2 Raman active
T1g 3 -1 4 Rx, Ry, Rz
T2g 3 -1 4
Gg 4 0 6
Hg 5 1 8 Raman active
Au 1 -1 1
T1u 3 1 5 Tx, Ty, Tz /IR
T2u 3 1 5
Gu 4 0 6
Hu 5 -1 7
Literature cited:
1. Kroto, H. W.; Heath, J. R.; O'Brien, S. C.; Curl, R. F.; Smalley, R. E. Nature 1985, 318, 162-164.
2. Kratschmer, W.; Lamb, L. D.; Fostiropolous, K.; and Huffman, D. R., Nature 1990, 347, 354.
3. Accounts of Chemical Research 1992, 25 , number 3.
4. Curl, R.F.; Smalley, R. E. Sci. Am. 1991, 10, 54-63.
5. Atkins, P. W. Physical Chemistry, 1990 (W. H. Freeman and Company, New York, 4th edition, Chapter 12, pages 336-338).
6. Atkins, P. W. Molecular Quantum Mechanics, 1983 (Oxford University Press, Oxford, New York, 2nd Edition, Chapter 7, pages 163-167).
7. Weaver, J. H., et al. Phys. Rev. Lett. 1991, 66, 1741-1744.
8. Haddon, R. C., Accounts of Chemical·Research 1992, 25, 127-133.
9. Hebard, A. F., Physics Today 1992, 11, 26-32.
10. Martins, J. L.; Troullier, N.; Weaver, J. H., Chem. Phys. Lett. 1991, 180, 457-60.
11. Chung, F.; Sternberg, S., American Scientist 1993, 81, 56-71.
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High level quantum mechanical calculations reveal that phenanthrene is more stable than anthracene, and that the enhanced stability has its origin in the energy of the π‐electron density. This tutorial shows that even a simple Huckel molecular orbital calculation predicts the greater stability of phenanthrene.
Huckel calculations can be carried out in several ways; perhaps the most economical and revealing is the numeric method which is based on a connectivity matrix. Anthracene and phenanthrene, shown below, have 14 carbon atoms, so the connectivity matrix is 14 x 14. The only non‐zero matrix elements contain the integer 1 indicating adjacent (bonded) carbons. For example, using the numbering scheme shown below carbon 1 is bonded to carbons 2 and 14 in both molecules, and this connectivity (bonding) is registered in the first row of the respective Huckel matirices shown below.
The Huckel connectivity matrices for anthracene and phenanthrene are,
$\begin{matrix} Ha = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} & Hp = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
The energy eigenvalues are the negatives of the eigenvalues of the connectivity matrices. The energy eigenvalues, π−electron energy and the π−delocalization energy are calculated for each molecule below. There are seven doubly occupied molecular orbitals for each molecule and the energy of a localized π electron pair (ethene) is ‐2. Thus the π‐electron energy is twice the sum of the occupied energy levels. The delocalization energy is the π‐electron energy minus the energy of an equivalent number of localized π‐electron pairs. See the appendix for the calculation of the ethene (localized) π‐electron energy.
Anthracene
$\begin{matrix} \text{Ha = -Ha} & \text{Ea = eigenvals(Ha)} & \text{Ea = sort(Ea)} \end{matrix} \nonumber$
$Ea^T = \begin{pmatrix} -2.414 & -2 & -1.414 & -1.414 & -1 & -1 & -0.414 & 0.414 & 1 & 1 & 1.414 & 1.414 & 2 & 2.414 \end{pmatrix} \nonumber$
$\begin{matrix} E_{ \pi a} = 2 \sum_{i = 1}^7 Ea_i & E_{ \pi a} = -19.314 & Ea_{deloc} = E_{ \pi a} -7(-2) & Ea_{deloc} = -5.314 \end{matrix} \nonumber$
Phenanthrene
$\begin{matrix} \text{Hp = -Hp} & \text{Ep = eigenvals(Hp)} & \text{Ep = sort(Ep)} \end{matrix} \nonumber$
$Ep^T = \begin{pmatrix} -2.435 & -1.951 & -1.516 & -1.306 & -1.142 & -0.769 & -0.605 & 0.605 & 0.769 & 1.142 & 1.306 & 1.516 & 1.951 & 2.414 \end{pmatrix} \nonumber$
$\begin{matrix} E_{ \pi p} = 2 \sum_{i = 1}^7 Ep_i & E_{ \pi p} = -19.448 & Ep_{deloc} = E_{ \pi p} -7(-2) & Ep_{deloc} = -5.448 \end{matrix} \nonumber$
Summary
$\begin{pmatrix} \text{Molecule} & E_{ \pi} & E_{deloc} \ \text{Anthracene} & -19.314 & -5.314 \ \text{Phenanthrene} & -19.448 & -5.448 \end{pmatrix} \nonumber$
According to the Huckel model for π‐electron energy phenanthrene is more stable than anthracene. This result is in agreement with higher level quantum mechanical calculations and the enthalpies of formation of anthracene and phenanthrene, 227 and 201 kJ/mol, respectively.
Appendix
Ethene is the bench mark for a localized pair of π electrons. As shown below, the eigenvalues for the Huckel connectivity matrix for ethene are ‐1 and +1. Thus a localized pair of π electrons has an energy of ‐2.
$\begin{matrix} He = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & \text{He = -He} & \text{Ee = eigenvals(He)} & \text{Ee = sort(Ee)} & Ee^T = \begin{pmatrix} -1 & 1 \end{pmatrix} \end{matrix} \nonumber$
3.33: A Numerical Huckel Calculation on C10H8 Isomers
The numeric version of the Huckel molecular orbital theory (HMOT) is based on a connectivity matrix which records which atoms are bonded to each other. The eigenvalues of this matrix provide the HMOT energy levels and the HMOT wave functions.
The purpose of this exercise is to see what HMOT has to say about the relative stability of the C10H8 isomers shown below.
Napthalene calculation: $\begin{matrix} \text{Natoms = 10} & \text{Nocc = 5} \end{matrix}$
$H = \begin{pmatrix} 0&1&0&0&0&1&0&0&0&1 \ 1&0&1&0&0&0&0&0&0&0 \ 0&1&0&1&0&0&0&0&0&0 \ 0&0&1&0&1&0&0&0&0&0 \ 0&0&0&1&0&1&0&0&0&0 \ 1&0&0&0&1&0&1&0&0&0 \ 0&0&0&0&0&1&0&1&0&0 \ 0&0&0&0&0&0&1&0&1&0 \ 0&0&0&0&0&0&0&1&0&1 \ 1&0&0&0&0&0&0&0&1&0 \end{pmatrix} \nonumber$
Calculate the eigenvalues and eigenvectors: $\begin{matrix} \text{E = eigenvals(H)} & \text{E = sort(E)} \end{matrix}$
$E^T = \begin{pmatrix} -2.303 & -1.618 & -1.303 & -1 & -0.618 & 0.618 & 1 & 1.303 & 1.618 & 2.303 \end{pmatrix} \nonumber$
Calculate total π-electronic energy: $\begin{matrix} E_{ \pi} = 2 \sum_{i = 1}^{ \text{Nocc}} E_i & E_{ \pi} = -13.683 \end{matrix}$
Calculate the delocalization energy: $\begin{matrix} E_{deloc} = E_{ \pi} + 2 \text{Nocc} & E_{deloc} = -3.683 \end{matrix}$
Calculate the delocalization energy per atom: $\frac{E_{deloc}}{Natoms} = -0.368$
Fulvalene calculation:
$H = \begin{pmatrix} 0&1&0&0&1&0&0&0&0&0 \ 1&0&1&0&0&0&0&0&0&0 \ 0&1&0&1&0&0&0&0&0&0 \ 0&0&1&0&1&0&0&0&0&0 \ 1&0&0&1&0&1&0&0&0&0 \ 0&0&0&0&1&0&1&0&0&1 \ 0&0&0&0&0&1&0&1&0&0 \ 0&0&0&0&0&0&1&0&1&0 \ 0&0&0&0&0&0&0&1&0&1 \ 0&0&0&0&0&1&0&0&1&0 \end{pmatrix} \nonumber$
Calculate the eigenvalues and eigenvectors: $\begin{matrix} \text{E = eigenvals(H)} & \text{E = sort(E)} \end{matrix}$
$E^T = \begin{pmatrix} -2.115 & -1.618 & -1.618 & -1.303 & 0.254 & 0.618 & 0.618 & 1 & 1.861 & 2.303 \end{pmatrix} \nonumber$
Calculate the total π-electronic energy: $\begin{matrix} E_{ \pi} = 2 \sum_{i = 1}^{ \text{Nocc}} E_i & E_{ \pi} = -12.799 \end{matrix}$
Calculate the delocalization energy: $\begin{matrix} E_{ \text{deloc}} = E_{ \pi} + 2 \text{Nocc} & E_{ deloc} = -2.799 \end{matrix}$
Calculate the delocalization energy per atom: $\frac{E_{deloc}}{Natoms} = -0.28$
Azulene calculation:
$H = \begin{pmatrix} 0&1&0&0&1&0&0&0&0&1 \ 1&0&1&0&0&0&0&0&0&0 \ 0&1&0&1&0&0&0&0&0&0 \ 0&0&1&0&1&0&0&0&0&0 \ 1&0&0&1&0&1&0&0&0&0 \ 0&0&0&0&1&0&1&0&0&0 \ 0&0&0&0&0&1&0&1&0&0 \ 0&0&0&0&0&0&1&0&1&0 \ 0&0&0&0&0&0&0&1&0&1 \ 1&0&0&0&0&0&0&0&1&0 \end{pmatrix} \nonumber$
Calculate the eigenvalues and eigenvectors: $\begin{matrix} \text{E = eigenvals(H)} & \text{E = sort(E)} \end{matrix}$
Calculate the total π-electronic energy: $\begin{matrix} E_{ \pi} = 2 \sum_{i = 1}^{ \text{Nocc}} E_i & E_{ \pi} = -13.364 \end{matrix}$
Calculate the delocalization energy per atom: $\frac{E_{deloc}}{Natoms} = -0.336$
The HMOT calculations indicate that napthalene has the largest delocalization energy and fulvalene has the smallest delocalization energy.
$\begin{pmatrix} \text{Molecule} & E_{ \pi} & E_{deloc} & \frac{E_{deloc}}{atom} \ \text{Napthalene} & -13.683 & -3.683 & -0.368 \ \text{Azulene} & -13.364 & -3.364 & -0.336 \ \text{Fulvalene} & -12.799 & -2.799 & -0.280 \end{pmatrix} \nonumber$
The enthalpies of formation are available for napthalene and azulene and are 150 and 280 kJ/mol, respectively. Thus the rudimentary Huckel calculation is consistent with the thermodynamic data.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.32%3A_A_Numerical_Huckel_Calculation_on_Anthracene_and_Phenanthrene.txt
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Calculate the wavefunctions and energies of the σ orbitals in the HF molecule, taking β = -1.0 eV. The values of the Coulomb integrals αH and αF are taken as the negatives of the orbital ionization energies of the atoms.
The molecular orbital is formed as a linear combination of the valence orbitals of H and F.
$\Psi = C_h \Psi_h + C_f \Psi_f \nonumber$
where
$\Psi_h = 1s(h) and \Psi_f = 2p_z (f) \nonumber$
The variational integral for this calculation is,
$E = \frac{ \int_0^{ \infty} \left( C_h \Psi_h + C_f \Psi_f \right) H \left( C_h \Psi_h + C_h \Psi_f \right) d \tau}{ \int_0^{ \infty} \left( C_h \Psi_h + C_f \Psi_f \right)^2 d \tau} \nonumber$
Minimization of the variational energy integral simultaneously with respect to Ch and Cf yields two linear homogeneous equations (see McQuarrie and Simon, section 7-2, for details).
$\begin{matrix} C_h \left[ H_{hh} - ES_{hh} \right] + C_f \left[ H_{hf} - ES_{hf} \right] = 0 \ C_h \left[ H_{hf} - ES_{hf} \right] + C_f \left[ H_{ff} - ES_{ff} \right] = 0 \end{matrix} \nonumber$
where the integral and their parameterizations are presented below.
$\begin{matrix} H_{hh} = \int_0^{ \infty} \Psi_h H \Psi_h d \tau = -13.6 & S_{hh} = \int_0^{ \infty} \Psi_h^2 d \tau = 1 \ H_{ff} = \int_0^{ \infty} \Psi_f H \Psi_f d \tau = \alpha_f = -18.6 & S_{ff} = \int_0 ^{ \infty} \Psi_f^2 d \tau = 1 \ H_{hf} = \int_0^{ \infty} \Psi_h H \Psi_f d \tau = \beta = -1 & Shf = \int_0^{ \infty} \Psi_h \Psi_f d \tau = 0 \end{matrix} \nonumber$
In matrix form equations (4) can be written as,
$\begin{pmatrix} \alpha_h - E & \beta \ \beta & \alpha_f - E \end{pmatrix} \begin{pmatrix} C_h \ C_f \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \nonumber$
or, as an eigenvalue equation,
$\begin{pmatrix} \alpha_h & \beta \ \beta & \alpha_f \end{pmatrix} \begin{pmatrix} C_h \ C_f \end{pmatrix} = \begin{pmatrix} E & 0 \ 0 & E \end{pmatrix} \begin{pmatrix} C_h \ C_f \end{pmatrix} \nonumber$
Mathcad can now be used to find the eigenvalues and eigenvectors of (6). First we must give it the values for all the parameters.
$\begin{matrix} \alpha_f = -18.6 & \alpha_h = -13.6 & \beta = -1 \end{matrix} \nonumber$
Define the variational matrix shown on the left side of equation (6):
$H = \begin{pmatrix} \alpha_h & \beta \ \beta & \alpha_f \end{pmatrix} \nonumber$
Find the eigenvalues:
$\begin{matrix} \text{E = sort(eigenvals(H))} & \text{E}_0 = -18.79 & E_1 = -13.41 \end{matrix} \nonumber$
Now find the eigenvectors:
$\begin{matrix} \text{The bonding MO is:} & \Psi_h = \text{eigenvec} \left( H,~E_0 \right) & \Psi_b = \begin{pmatrix} 0.19 \ 0.98 \end{pmatrix} & \overrightarrow{ \left( \Psi_b^2 \right)} = \begin{pmatrix} 0.04 \ 0.96 \end{pmatrix} \ \text{The anti-bonding MO is:} & \Psi_h = \text{eigenvec} \left( H,~E_1 \right) & \Psi_b = \begin{pmatrix} -0.98 \ 0.19 \end{pmatrix} & \overrightarrow{ \left( \Psi_a^2 \right)} = \begin{pmatrix} 0.96 \ 0.04 \end{pmatrix} \end{matrix} \nonumber$
Squaring the coefficients of the wavefunction we find that an electron in the bonding molecular orbital spends 96% of its time on the fluorine atom and 4% of its time on the hydrogen atom. Because fluorine has a kernel charge (nuclear charge minus two non-valence electrons) of +7 and gets credit, according to the appended MO diagram for six non-bonding electrons, plus 96% of the two bonding electrons, the net charge on flourine can be shown to be -.92. This value indicates that the H-F bond is quite polar, which is consistent with the high electronegativity of the flourine atom.
On the Pauling scale the electronegativities of H and F are 2.1 and 4.0, respectively. The partial charge on H is calculated as: nuclear charge - non-valence electrons - unshared valence electrons - bonding electrons times χH/(χH + χF). As shown below these formal charge calculations do not give fluorine such a high negative charge.
$\begin{matrix} Z_H = 1 & _F = 9 & \chi_H = 2.1 & \chi_F = 4.0 \ \delta_H = 1-0-0-2 \frac{ \chi_H}{ \chi_H + \chi_F} & \delta_H = 0.31 & \delta_F = 9 - 2 - 6 - 2 \frac{ \chi_F}{ \chi_H + \chi_F} & \delta_F = -0.31 \end{matrix} \nonumber$
Mathcad offers another way of performing the variational calculation. The variational integral given in equation (3) can also be written in the following form in which the integrals have been replaced by the experimental parameters used to replace them.
$E = \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2 C_h C_f \beta}{C_h^2 + C_f^2 + 2C_h C_f S_{hf}} \nonumber$
This expression can be minimized simultaneously with respect to Ch and Cf in a Given/Find solve block. As is usually the case, Mathcad requires seed values or actual values for all the variables that appear in the solve block. There are two solutions, the ground state and the excited state. That is, the bonding molecular orbital and the anti-bonding molecular orbital. Within the solve block we have the equation for the energy, its first derivative with respect to Ch and Cf each set equal to zero, and the normalization requirement. To find the ground state we chose initial values of the coefficients that favor the fluorine 2pz orbital.
$\begin{matrix} C_h = 0.1 & C_f = .9 & \alpha_h = -13.6 & \alpha_f = -18.6 & \beta = - 7 S_{hf} = 0 & E = -10 \end{matrix} \nonumber$
Given
$\begin{matrix} E = \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2C_h C_f \beta}{C_h^2 + C_f^2 2C_h S_{hf}} & \frac{d}{dC_h} \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2C_h C_f \beta}{C_h^2 + C_f^2 2C_h S_{hf}} = 0 \ \frac{d}{dC_f} \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2C_h C_f \beta}{C_h^2 + C_f^2 2C_h S_{hf}} = 0 & C_h^2 + C_f^2 + 2C_h C_f S_{hf} = 1 \end{matrix} \nonumber$
Given these constraints, we ask Mathcad to find a solution. The solution yields wavefunction coefficients and the ground-state energy.
$\text{Find} \left( C_h,~C_f,~E \right) = \begin{pmatrix} 0.19 \ 0.98 \ -18.79 \end{pmatrix} \nonumber$
To find the anti-bonding state, we choose initial coefficients that favor the hydrogen 1s orbital.
$\begin{matrix} C_h = .9 & C_f = 0.1 & \alpha = -13.6 & \alpha_f = -18.6 & \beta = -1 & S_{hf} = 0 & E = -10 \end{matrix} \nonumber$
Given
$\begin{matrix} E = \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2C_h C_f \beta}{C_h^2 + C_f^2 2C_h S_{hf}} & \frac{d}{dC_h} \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2C_h C_f \beta}{C_h^2 + C_f^2 2C_h S_{hf}} = 0 \ \frac{d}{dC_f} \frac{C_h^2 \alpha_h + C_f^2 \alpha_f + 2C_h C_f \beta}{C_h^2 + C_f^2 2C_h S_{hf}} = 0 & C_h^2 + C_f^2 + 2C_h C_f S_{hf} = 1 \end{matrix} \nonumber$
Given these constraints, we ask Mathcad to find a solution.
$\text{Find} \left( C_h,~F_f,~E \right) = \begin{pmatrix} 0.98 \ -0.19 \ -13.41 \end{pmatrix} \nonumber$
There is yet another way that Mathcad can do this calculation that requires that the symbolic processor be loaded. Equation (5) in matrix form has a non-trivial solution only if the determinant of the coefficient matrix is equal to zero.
$\begin{pmatrix} \alpha_h - E & \beta \ \beta & \alpha_f - E \end{pmatrix}$ has determinant $\alpha_h - \alpha_f - \alpha_h E - E \alpha_f + E^2 - \beta^2$ which has solution(s)
$\begin{pmatrix} \frac{1}{2} \alpha_h + \frac{1}{2} \alpha_f + \frac{1}{2} \sqrt{ \alpha_h^2 - 2 \alpha_h \alpha_f + \alpha_f^2 + 4 \beta^2} \ \frac{1}{2} \alpha_h + \frac{1}{2} \alpha_f + \frac{1}{2} \sqrt{ \alpha_h^2 - 2 \alpha_h \alpha_f + \alpha_f^2 + 4 \beta^2} \end{pmatrix} = \begin{pmatrix} -13.41 \ -18.79 \end{pmatrix} \nonumber$
Find the bonding molecular orbital:
E = -18.793
Given $\left( \alpha_h - E \right) C_h + \beta C_f = 0$ and $\begin{matrix} C_h^2 + C_f^2 + 2C_h C_f S_{hf} =1 & \text{Find} \left( C_h,~C_f \right) = \begin{pmatrix} 0.19 \ 0.98 \end{pmatrix} \end{matrix}$
Find the anti-bonding molecular orbital:
E = -13.407
Given $\left( \alpha_h - E \right) C_h + \beta C_f = 0$ and $\begin{matrix} C_h^2 + C_f^2 + 2C_h C_f S_{hf} =1 & \text{Find} \left( C_h,~C_f \right) = \begin{pmatrix} 0.98 \ -0.19 \end{pmatrix} \end{matrix}$
The molecular orbital calculation can be summarized graphically as shown below.
Reference: See page 428 of Atkins and de Paula, Physical Chemistry, 7th edition.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.34%3A_Semi-empirical_Molecular_Orbital_Calculation_on_HF.txt
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The bonding in XeF2 can be interpreted in terms the three-center four-electron bond. In linear XeF2 the molecular orbital can be considered to be a linear combination of the fluorine 2p orbitals and the central xenon 5p.
$\Psi = C_{F1} \Psi_{F1} + C_{Xe} \Psi_{Xe} + C_{F2} \Psi_{F2} \nonumber$
Minimization of the variational integral
$E = \frac{ \int \left( C_{F1} \Psi_{F1} + C_{Xe} \Psi_{Xe} + C_{F2} \Psi_{F2} \right) H \left( C_{F1} \Psi_{F1} + C_{Xe} \Psi_{Xe} + C_{F2} \Psi_{F2} \right) d \tau}{ \int \left( C_{F1} \Psi_{F1} + C_{Xe} \Psi_{Xe} + C_{F2} \Psi_{F2} \right)^2 d \tau} \nonumber$
yields the following 3 x 3 Huckel matrix
$H = \begin{pmatrix} \alpha_F & \beta & 0 \ \beta & \alpha_{Xe} & \beta \ 0 & \beta & \alpha_F \end{pmatrix} \nonumber$
All overlap integrals are zero. The Coulomb integrals are parameterized as the negative of the valence orbital ionization energies (-12.13 eV for the Xe 5p orbital, -17.42 eV for the F 2p). The non-zero resonance integral is given a value of -2.0 eV.
$\begin{matrix} \alpha_{F1} = \alpha_F = \int \Psi_{F1} H \Psi_{F1} d \tau = -17.42 & \alpha_{F2} = \alpha_F = \int \Psi_{F2} H \Psi_{F2} d \tau = -17.42 \ \alpha_{Xe} = \int \Psi_{Xe} H \Psi_{Xe} d \tau = -12.13 & \beta= \int \Psi_{Xe} H \Psi_F d \tau = -2 \end{matrix} \nonumber$
Mathcad can now be used to find the eigenvalues and eigenvectors of H. First we must give it the values for all of the parameters.
$\begin{matrix} \alpha_F = -17.42 & \alpha_{Xe} = -12.13 & \beta = -2.00 \end{matrix} \nonumber$
Define the variational matrix shown on the left side of equation (6):
$H = \begin{pmatrix} \alpha_F & \beta & 0 \ \beta & \alpha_{Xe} & \beta \ 0 & \beta & \alpha_F \end{pmatrix} \nonumber$
Find the eigenvalues: $\begin{matrix} \text{E = sort(eigenvals(H))} & E_0 = -18.647 & E_1 = -17.42 & E_2 = -10.903 \end{matrix}$
Now find the eigenvectors:
$\begin{matrix} \text{The bonding MO is:} & \Psi_b = \text{eigenvec} \left( H,~E_0 \right) & \Psi_b = \begin{pmatrix} 0.649 \ 0.389 \ 0.649 \end{pmatrix} & \overrightarrow{ \left( \Psi_b \right)^2} = \begin{pmatrix} 0.421 \ 0.158 \ 0.421 \end{pmatrix} \ \text{The non-bonding MO is:} & \Psi_{nb} = \text{eigenvec} \left( H,~E_1 \right) & \Psi_{nb} = \begin{pmatrix} -0.707 \ 0 \ 0.707 \end{pmatrix} & \overrightarrow{ \left( \Psi_{nb} \right)^2} = \begin{pmatrix} 0.5 \ 0 \ 0.5 \end{pmatrix} \ \text{The anti-bonding MO is:} & \Psi_a = \text{eigenvec} \left( H,~E_2 \right) & \Psi_b = \begin{pmatrix} -0.282 \ 0.917 \ -0.282 \end{pmatrix} & \overrightarrow{ \left( \Psi_a \right)^2} = \begin{pmatrix} 0.079 \ 0.842 \ 0.079 \end{pmatrix} \end{matrix} \nonumber$
The molecular orbital diagram for this system is shown below.
Is the molecule stable? The diagram shows two electrons each in the bonding and non-bonding orbitals for a total energy of -72.14 eV. The energy of the isolated atoms is 2(-12.13 eV) + 2(-17.42 eV) = - 59.1 eV. Thus, the molecule is more stable than the isolated atoms according to this crude semi-empirical model.
The partial charges are calculated next. The fluorine atoms have a kernel charge of +7. They get credit for 6 non-bonding atomic electrons, 42.1% of the two electrons in the bonding MO, and 50% of the two electrons in the non-bonding MO. This gives a partial charge of (7 - 6 - 2(.421) - 2(.50)) -0.842.
The xenon atom has a kernel charge of +8. It gets credit for 6 non-bonding atomic electrons, 15.8% of the two electrons in the bonding MO, and no credit for the two electrons in non-bonding MO. This yields a partial charge of (8 - 6 - 2(.158)) +1.684. This value is reasonable because xenon is bonded to the most electronegative element in the periodic table.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.35%3A_Semi-empirical_Molecular_Orbital_Calculation_on_XeF2.txt
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The purpose of this tutorial is to use a one‐dimensional Posch‐Teller potential to model the electronic behavior of metals. Schrödingerʹs equation is integrated numerically for the first six energy states, producing two bands of bound states separated by a significant band gap. The integration algorithm is taken from J. C. Hansen, J. Chem. Educ. Software, 8C2, 1996.)
$\begin{matrix} \text{Set parameters:} & n = 200 & xmin = -6 & xmax = 6 & \Delta = \frac{xmax - xmin}{n-1} \ \mu = 1 & Vo = -10 & \alpha = 2 & d = 2 & a = 1 \end{matrix} \nonumber$
Calculate the position vector, the potential energy matrix, and the kinetic energy matrix. Then combine them into a total energy matrix.
$\begin{matrix} i = 1 .. n & j = 1 .. n & x_i = xmin + (i-1) \Delta \end{matrix} \nonumber$
$\begin{matrix} V_{i,~j} = \text{if} \left[ \right] & T_{i,~j} = \text{if} \left[ i = j,~ \frac{ \pi^2}{6 \mu \Delta^2},~ \frac{(-1)^{i-j}}{(i-j)^2 \mu \Delta^2} \right] \end{matrix} \nonumber$
Total energy matrix: H = T + V
Calculate eigenvalues: $\text{E = sort(eigenvals(H))}$
Display selected eigenvalues: m = 1 .. 6
$E_m = \begin{array}{|c|} \hline -6.5283 \ \hline -6.4484 \ \hline -6.3942 \ \hline -1.8562 \ \hline -1.3274 \ \hline -0.6173 \ \hline \end{array} \nonumber$
Calculate eigenvectors: $\begin{matrix} k = 1 .. 6 & \Psi (k) = \text{eigenvec} \left( H,~E_k \right) \end{matrix}$
Display the potential energy and energy eigenvalues to show band formation:
Display the potential energy and electron density distributions offset by an arbitrary amount for purposes of clarity:
Discussion of the model: The metal cations occupy the potential energy minima. The electron density distributions show that the allowed valence electron states are delocalized over the metal structure. Delocalization is achived (especially for the lower energy band) by quantum tunneling through the internal energy barriers representing the inter‐nuclear regions.
Bunching of the allowed energy states into two ʺbandsʺ separated by a significant band gap occurs because the states have different electron densities at the locations of the potential energy minima and maxima, in addition to the fact that kinetic energy increases with increasing quantum number. This is illustrated by the following table in which the expectation values for kinetic and potential energy are calculated. Note that the kinetic energy always increases with increasing quantum number, but within a band the potential energy actually decreases slightly.
$\begin{pmatrix} \text{"Kinetic Energy"} & \text{"Potential Energy"} & \text{"Total Energy"} \ \Psi (1)^T T \Psi (1) & \Psi (1)^T V \Psi (1) & E_1 \ \Psi (2)^T T \Psi (2) & \Psi (2)^T V \Psi (2) & E_2 \ \Psi (3)^T T \Psi (3) & \Psi (3)^T V \Psi (3) & E_3 \ \Psi (4)^T T \Psi (4) & \Psi (4)^T V \Psi (4) & E_4 \ \Psi (5)^T T \Psi (5) & \Psi (5)^T V \Psi (5) & E_5 \ \Psi (6)^T T \Psi (6) & \Psi (6)^T V \Psi (6) & E_6 \ \end{pmatrix} = \begin{pmatrix} \text{"Kinetic Energy"} & \text{"Potential Energy"} & \text{"Total Energy"} \ 1.1917 & -7.7201 & -6.5283 \ 1.3766 & -7.8251 & -6.4484 \ 1.5634 & -7.9576 & -6.3942 \ 1.924 & -3.7855 & -1.8562 \ 2.4726 & -3.8000 & -1.3274 \ 3.2362 & -3.8535 & -0.6173 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.36%3A_Posch-Teller_Potential_Model_for_Metals.txt
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Thumbnail: White light is dispersed by a prism into the colors of the visible spectrum. (CC BY-SA 3.0; D-Kuru).
04: Spectroscopy
$\begin{matrix} \text{Planck's constant:} & h = 6.62608 10^{-34} & \text{Speed of light:} & c = 2.9979 10^{8} \frac{m}{sec} \end{matrix} \nonumber$
$\begin{matrix} \text{Conversion factors:} & nm = 10^{-9} m & pm = 10^{-12}m & aJ = 10^{-18} \text{Joule} \end{matrix} \nonumber$
$\begin{matrix} \text{Energy of a photon:} & E_{photon} = h \nu = \frac{hc}{ \lambda} \end{matrix} \nonumber$
Energy of the hydrogen atom: where n is a quantum number and can have integer values.
$E_{atom} = \frac{-2.178 aJ}{n^2} \nonumber$
Emission Spectroscopy
In emission spectroscopy a photon is created as the electron undergoes a transition from a higher to a lower energy state. Energy conservation requires
$E_{atom} ~^{initial} = E_{atom} ~^{final} + E_{photon} \nonumber$
Example: Calculate the frequency, wavelength, and energy of the photon emitted when an electron undergoes a transition from the n=2 to the n=1 state.
$\begin{matrix} n_i = 2 & n_f = 1 & \begin{array}{c|c} \nu = \frac{-2.178 aJ}{n_i^2} = \frac{-2.178 aJ}{n_f^2} + h \nu ~ & _{float,~4} ^{solve,~ \nu} \rightarrow \frac{0.2465e16}{sec} \end{array} \end{matrix} \nonumber$
$\begin{matrix} \text{Calculate the wavelength of the photon:} & \lambda = \frac{c}{ \nu} & \lambda = 121.619 nm \end{matrix} \nonumber$
$\begin{matrix} \text{Calculate the energy of the photon:} & h \nu = 1.633 aJ \end{matrix} \nonumber$
Absorption Spectroscopy
In absorption spectroscopy a photon is absorbed and an electron is promoted to a higher energy level. Energy conservation requires
$E_{atom} ~^{initial} = E_{atom} ~^{final} + E_{photon} \nonumber$
Example: Calculate the frequency, wavelength, and energy of the photon required to promote the electron from the n=1 to the n=3 level.
$\begin{matrix} \nu = \nu & n_i = 1 & n_f = 3 & \begin{array}{c|c} \nu = \frac{-2.178 aJ}{n_i^2} + h \nu = \frac{-2.178 aJ}{n_f^2} ~ & _{float,~4} ^{solve,~ \nu} \rightarrow \frac{0.2922e16}{sec} \end{array} \end{matrix} \nonumber$
$\begin{matrix} \text{Calculate the wavelength of the photon:} & \lambda = \frac{c}{ \nu} & \lambda = 102.598 nm \end{matrix} \nonumber$
$\begin{matrix} \text{Calculate the energy of the photon:} & h \nu = 1.936 aJ \end{matrix} \nonumber$
4.02: A Particle-in-a-Box Model for Color Centers
Electrons trapped (see figure below) in anion vacancies of alkali halide salts absorb visible radiation and cause the typically white solids to appear colored.* A simple explanation of this phenomenon is that due to the wave nature of matter (the basic postulate of quantum theory), the energy of a trapped particle, such as an electron, is quantized. This permits a simple interpretation of the absorption spectra (and, therefore, the color) of F‐centers. It is assumed that a photon of visible light promotes the trapped electron from its ground state to its first excited state.
This simplest model for the electron under these conditions is to assume that it behaves like a particle in a cubic box. The allowed energy levels of an electron in a cubic box are,
$E = \frac{h^2}{8 ma^2} \left[ \left( n_x \right)^2 + \left( n_y \right)^2 + \left( n_z \right)^2 \right] \nonumber$
where, h is Planckʹs constant, m is the electron mass, a is the box dimension, and nx, ny, and nz are quantum numbers.
On the basis of this simple model the first allowed transition is given by
$\Delta E = \frac{hc}{ \lambda} = E_f - E_i = \frac{h^2}{8ma^2} \left[ \left(2^2 + 1^2 + 1^2 \right) - \left( 1^2 + 1^2 + 1^2 \right) \right] \nonumber$
where c is the speed of light and λ is the wavelength of the photon of light absorbed. This expression can be re‐written as follows:
$\begin{matrix} \lambda = \frac{8mc}{3h} a^2 = 1099 a^2 & \text{(where } λ \text{ and a are given in nanometers)} \end{matrix} \nonumber$
As equation (3) shows, the model predicts that the wavelength maximum is directly proportional to the box dimension squared.
Reasonable agreement between experimental absorbance maxima and the model can be obtained by assuming that the electron is restricted to a cube the size of one unit cell. This is shown below by doing a linear regression analysis on equation (3) assuming that the exponent of a is an adjustable parameter. In other words, it is assumed that the wavelength maximum is directly proportional to an, where a is the lattice constant and n is the parameter to be determinded.
The data below are the lattice constant and wavelength maximum for the following alkali halides: LiF, NaF, LiCl, KF, NaCl, NaBr, KCl, NaI, RbCl, KBr, RbBr, KI, and RbI.
i = 1 .. 13
$\begin{matrix} a_i = & \lambda_i = \ \begin{array}{|c|} \hline .402 \ \hline .462 \ \hline .514 \ \hline .534 \ \hline .562 \ \hline .596 \ \hline .628 \ \hline .646 \ \hline .654 \ \hline .658 \ \hline .686 \ \hline .706 \ \hline .732 \ \hline \end{array} & \begin{array}{|c|} \hline 250 \ \hline 341 \ \hline 385 \ \hline 455 \ \hline 458 \ \hline 540 \ \hline 556 \ \hline 588 \ \hline 609 \ \hline 625 \ \hline 694 \ \hline 689 \ \hline 756 \ \hline \end{array} \end{matrix} \nonumber$
Equation (3) is transformed into a linear function by taking the logarithm of both sides of the equation. This yields: ln(λ) = n ln(a) + ln(1099). This is, of course, the familiar: y = slope*x + intercept.
$\begin{matrix} y_i = \ln( \lambda_i ) & x_i = \ln (a_i) \ \text{a = slope(x, y)} & \text{b = slope(x, y)} & \text{c = corr(x, y)} \ \text{Slope:} & a = 1.7943 & \text{Intercept:} & b = 7.1841 \ \text{Correlation coefficient:} & c = 9.9499 \times 10^{-1} \end{matrix} \nonumber$
The data and the fit to the data are displayed below. The fit is good and the optimum value of n is reasonably close to the theoretical value of 2. Ways to increase the sophistication of the model are discussed below.
It is clear from the figure above that the simple model employed here accounts for the trend in the data. More sophisticated models would assume that the electron occupied a finite potential well. In other words that some tunneling into the potential barrier would be permitted. In addition to cubic wells, it would not be difficult to perform an analysis assuming that the trapped electrons occupy finite spherical potential wells.
*The F‐center (electron trapped in an anion vacancy) can be formed by the ionization of the halide anion as follows: X‐ + hν ‐‐‐‐‐‐‐> X + e‐. The neutral X drifts away and the negatively charged electron takes its place as the anion in the crystal structure.
Reference: G. P. Hughes, ʺColor Centers: An example of a particle trapped in a finite potential wellʺ, American Journal of Physics 45, 948, (1977).
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Using the particle-in-a-box (PIB) approximation or Hückel molecular orbital theory (HMOT) to model the π-electrons of various dyes is a standard lecture/lab exercise in the traditional undergraduate physical chemistry curriculum (1-8). In Volume III of his celebrated The Feynman Lectures on Physics, Richard Feynman offers a different approach by considering such dyes as examples of two-state electronic systems (9). Other examples of two-state systems that Feynman presents are photon polarization, the ammonia inversion, bonding in the hydrogen molecular ion and the hydrogen molecule, spin ½ systems, the strong nuclear force, and other more esoteric nuclear phenomena.
The PIB and HMOT models provide a manifold of quantized π-electron energy levels which are then populated with electrons. A HOMO-LUMO electronic transition provides the mechanism for the absorption of a photon of light and the origin of the dye color. These models each have one parameter: box length for PIB and the resonance integral, β, for HMOT.
Both models also start with Lewis structures, as does the Feynman approach. Two equivalent resonance structures can be written for the symmetric cyanine dyes, the subject of this tutorial.
$R_2 \ddot{N} [CH=CH]_n CH= ~^+NR_2 \leftrightarrow R_2N^+ =CH[CH=CH]_n \ddot{N}R_2 \nonumber$
The resonance structures are used to determine the box length for the PIB model and to construct the energy matrix for the HMOT calculation.
Feynman, however, considers the resonance structures as base states in a simple quantum mechanical analysis.
$\begin{matrix} |1 \rangle = | R_2 \ddot{N} [CH=CH]_n CH= ~^+NR_2 \rangle & |2 \rangle = | R_2N^+ = CH[CH=CH]_n \ddot{N}R_2 \rangle \end{matrix} \nonumber$
The energy matrix for these base states is shown below. The presence of off-diagonal elements indicates that these base states are not stationary states of the energy operator.
$\begin{bmatrix} E^0 & A \ A & E^0 \end{bmatrix} \nonumber$
Here E0 and A are negative quantities, which are generally approximated by an appeal to experimental data.
$\begin{matrix} \langle 1 | \hat{H} | 1 \rangle = \langle 2 | \hat{H} | 2 \rangle = E^0 & \langle 1 | \hat{H} | 2 \rangle = \langle 2 | \hat{H} | 1 \rangle = A \end{matrix} \nonumber$
The following quantum mechanical superpositions of the base states diagonalize the energy matrix and are, therefore, stationary states.
$| \Psi_{elec}^{ \pm} \rangle = \frac{1}{ \sqrt{2}} \left[ \left| R_2 \ddot{N} [CH=CH]_n CH= ~^+NR_2 \left\rangle \pm \right| R_2N^+ = CH[CH=CH]_n \ddot{N}R_2 \right\rangle \right] = \frac{1}{ \sqrt{2}} \left[ |1 \rangle \pm |2 \rangle \right] \nonumber$
$\begin{bmatrix} E^0 + A & 0 \ 0 & E^0 - A \end{bmatrix} \nonumber$
$\begin{matrix} \langle \Psi^+_{elec} | \hat{H} | \Psi^+_{elec} \rangle = E^0 + A & \langle \Psi^-_{elec} | \hat{H} | \Psi^-_{elec} \rangle = E^0 - A & \langle \Psi^-_{elec} | \hat{H} | \Psi^+_{elec} \rangle = \langle \Psi^+_{elec} | \hat{H} | \Psi^-_{elec} \rangle = 0 \end{matrix} \nonumber$
The in-phase superposition represents the ground electric state, while the anti-symmetric superposition represents the excited electronic state.
It must be stressed that a superposition is not a mixture. The dye is not a 50-50 mixture of two equivalent structures, nor is it rapidly oscillating back and forth between two equivalent structures. According to quantum mechanics the dye is simultaneously in both electronic base states.
To proceed to the issue of experimental validation, we need to recognize that the values of the matrix elements will depend on n, the length of the dye chromophore. This can be accomplished by adding subscripts in the energy matrix shown above.
$\begin{bmatrix} E_n^0 + A_n & 0 \ 0 & E^0_n - A_n \end{bmatrix} \nonumber$
This leads to the following picture of the absorption of visible light by the dyes.
$\begin{matrix} | \Psi_{elec}^- \rangle_n ----- E_n^0 - A_n \ \uparrow \ | \Psi_{elec}^+ \rangle_n ----- E_n^0 + A_n \end{matrix} \nonumber$
The wavelength of the photon required for this electronic excitation is,
$\lambda_n = - \frac{hc}{2A_n} \nonumber$
It is clear from this result that two experimental photon wavelengths are required to calibrate the model. These can be used to calculate the photon wavelength of a third dye. In general this would be formulated as follows.
$\frac{ \lambda_{n+1}}{ \lambda_n} = \frac{A_n}{A_{n+1}} \xrightarrow[implies]{constant~ratio} \lambda_{n+2} = \lambda_{n+1} \left( \frac{ \lambda_{n+1}}{ \lambda_n} \right) \nonumber$
Due to rotational and vibrational fine structure, and other complications, the visible spectra of dyes exhibit broad electronic absorption bands. The theoretically calculated wavelength is therefore compared to the longest wavelength absorption maximum, λmax, in the UV-VIS spectrum. The table below compares calculated photon wavelengths with experimental λmax values for a series of symmetric cyanine dyes (8), and shows that Feynman’s two-state model gives modest agreement with experimental results. This generally is also the level of success achieved by the PIB and HMOT models.
Symmetric Cyanine Dye Results
n 1 2 3 4
Experimental λmax/nm 313 416 519 625
Theoretical Prediction 553 647
The interaction of the dye chromophores with electromagnetic radiation is a complicated process, so simple models such as PIB, HMOT and Feynman’s two-state approach which isolate the electronic event from other degrees of freedom can only be expected to give “ball park” quantitative agreement with experimental results. However, conceptually we believe that these simple models capture the essence of the photon absorption process. From a pedagogical perspective it is valuable to have three complementary models for the same physical phenomenon.
Literature cited:
1. Gerkin, R. E. J. Chem. Educ. 1965, 42, 490-491.
2. Farrell, J. J. J. Chem. Educ. 1985, 62, 351-352.
3. Shoemaker, D. P.; Garland, C. W.; Nibler, J. W. Experiments in Physical Chemistry, 5th ed.; McGraw-Hill: New York, 1989, pp 440-445.
4. Sime, R. J. Physical Chemistry: Methods, Techniques, and Experiments; Saunders: Philidelphia, 1990; pp 687-694.
5. Moog, R. S. J. Chem. Educ. 1991, 68, 506-508.
6. Bahnick, D. A. J. Chem. Educ. 1994, 71, 171-173.
7. Anderson, B. D. J. Chem. Educ. 1997, 74, 985.
8. Autschbach, J. J. Chem. Educ. 2007, 84, 1840-1845.
9. Feynman, R. P.; Leighton, R. B.; Sands, M. The Feynman Lectures on Physics; Addison-Wesley Publishing Co.: Reading MA, 1965; p. 10-12.
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Abstract:
The umbrella inversion in ammonia is modeled as a harmonic potential with an internal Gaussian barrier to inversion. The allowed vibrational eigenstates are calculated by numerical integration of Schrödinger's equation. The significance of the inversion in the operation of the maser is stressed.
Ammonia has four IR active vibrational modes occurring at 3444, 3337, 1627, and 950 cm-1. The umbrella-like bending mode at 950 cm-1 is modeled by a harmonic potential well with a Gaussian internal barrier. [See for example: Swallen and Ibers, J. Chem. Phys. 36, 1914 (1962).]
$V = \frac{1}{2} kx^2 + b \text{exp} (-c x^2) \nonumber$
The presence of the Gaussian barrier within the dominant harmonic potential causes a bunching between adjacent symmetric (+) and anti-symmetric (-) states. All states are raised in energy by the presence of the barrier, but the (-) states are elevated less than the (+) states because they have a node in the barrier and the (+) states do not.
Below Schrödinger's equation is integrated numerically for the first four energy states and the results compared with experimental spectroscopic data. (Integration algorithm taken form J. C. Hansen, J. Chem. Educ. Software, 8C2, 1996.)
Set parameters (atomic units are used):
$\begin{matrix} n = 99 & xmin = -2 & xmax = 2 & \Delta = \frac{xmax = xmin}{n} \ \mu = 4668 & k = .07598 & b = .05684& c = 1.36696 \end{matrix} \nonumber$
Calculate position vector, the potential energy matrix, and the kinetic energy matrix. Then combine them into a total energy matrix.
$\begin{matrix} i = 0 .. n & j = 0 .. n & x_i = xmin + (i) \Delta \end{matrix} \nonumber$
$\begin{matrix} V_{i,~j} = \text{if} \left[ i = j,~ \left[ \frac{1}{2} k(x_i)^2 + b \text{exp} \left[ -c (x_i)^2 \right] \right],~0 \right] & T_{i,~j} = \text{if} \left[ i-j,~ \frac{ \pi^2}{6 \mu \Delta^2},~ \frac{(-1)^{i-j}}{(i-j)^2 \mu \Delta^2} \right] \end{matrix} \nonumber$
Hamiltonian matrix: H = T + V
Find eigenvalues: E = sort(eigenvals(H))
Selected eigenvales: m = 0 .. 3
$E_m = \begin{array}{|c|} \hline 0.04996046 \ \hline 0.04996492 \ \hline 0.05418694 \ \hline 0.05435966 \ \hline \end{array} \nonumber$
Find selected eigenevectors: $\Psi (m) = \text{eigenvec}(H, ~E_m)$
Plot selected eigenfunctions: Vpoti = Vi, i
The selection rule for transitions between the manifold of energy levels obtained with this potential is (+ --> -) or (- --> +). A comparison of the first four calculated energy levels with experimental spectroscopic data shows that the Gaussian potential is a good model for the ammonia inversion.
According to the selection rule, there are four allowed transitions between these energy states. Two transitions (E0 ---> E1 and E2 ---> E3) occur in the microwave region at 0.79 and 36 cm-1. The E1 ---> E0 transition is, of course, the basis for the ammonia maser. The other allowed transitions (E0 ---> E3 and E1 ---> E2) appear in the infra-red region at 968.3 and 932.5 cm-1. The E0 ---> E2 and E1 ---> E3 transitions are forbidden. A comparison between the experimental and calculated frequencies for the allowed transitions is given below.
$\begin{matrix} \text{Transition} & \text{Theory/cm}^{-1} & \text{Experiment/cm}^{-1} \ E_0 \rightarrow E_1 & (E_1 - E_0) 2.1947463 \times 10^5 = 0.98 & 0.79 \ E_2 \rightarrow E_3 & (E_3 - E_2) 2.1947463 \times 10^5 = 37.9 & 36.0 \ E_1 \rightarrow E_2 & (E_1 - E_2) 2.1947463 \times 10^5 = 926.6 & 932.5 \ E_0 \rightarrow E_3 & (E_13- E_0) 2.1947463 \times 10^5 = 965.5 & 968.3 \end{matrix} \nonumber$
The criteria for allowed transitions are that the photon must satisfy the Bohr frequency condition and that the time-dependent superposition involving the initial and final states must exhibit asymmetric oscillating dipole character. The animation provided in the Appendix allows one to confirm the allowed and forbidden transitions listed above.
The harmonic potential with internal Gaussian function representing the barrier to inversion is shown below:
The nitrogen is to the left of the plane of hydrogens in the left-hand well, NH3, and to the right in the right-hand well H3N. It is clear from the plots above that the v = 0 and v = 1 wave functions are linear superpositions of these states:
$\begin{matrix} | \Psi \rangle_0 = \frac{1}{ \sqrt{2}} \left[ | \ddot{N} H_3 \rangle + |H_3 \ddot{N} \rangle \right] & | \Psi \rangle_1 = \frac{1}{ \sqrt{2}} \left[ | \ddot{N} H_3 \rangle + |H_3 \ddot{N} \rangle \right] \end{matrix} \nonumber$
These states are equally populated at room temperature because the energy difference separating them is less than 1 cm-1. However they can be separated electrostatically because they have different dipole moments. The |Ψ>1 state is isolated in a resonant cavity and the v =1 ----> v = 0 transition is induced by 24 GHz micro-waves (see "The Maser", J. P. Gordon, Scientific American, Dec. 1958, page 42).
A few additional comments are in order, because by and large most of the time the ammonia inversion phenomenon is interpreted using classical concepts. In his Scientific American article Gordon juxtaposes the (incorrect) classical explanation with the (correct) quantum mechanical explanation.
For example, regarding molecular vibration Gordon says "The molecule does not vibrate; the broken circles (shown in figures on page 44) merely indicate that the atoms are regarded as simultaneously occupying a number of positions." In other words, each vibrational degree of freedom is represented by a wave function which is a weighted superposition of all possible positions. The molecule is in a stationary state; movement only occurs when the vibrational state is perturbed by electromagnetic radiation.
Regarding the inversion phenomenon Gordon writes the following. "On the classical theory we picture the nitrogen atom flipping back and forth at a characteristic frequency of about 24,000 million vibrations per second, or 24,000 megacycles per second. At any given instant the nitrogen atom is on one side of the hydrogens or on the other. From the quantum point of view the nitrogen has at a given time a certain probability of being on either side - in a sense it is partly on both sides. Moreover, the molecule as a whole has two distinct energy states (see calculations above). The difference in energy between the states equals the energy of a photon with frequency of 24,000 megacycles per second."
In other words, the quantum view states that the so-called inversion frequency is actually the frequency of the photon required to induce a transition between the v = 0 and v = 1 states calculated above.
There is at least one other major misconception regarding the inversion phenomenon -- this is the concept of inversion doubling. Harris and Bertolucci, in their otherwise excellent monograph "Symmetry and Spectroscopy", write "The quantum mechanical consequence of a mutiple-minimum potential well is that there are two states for every one state of the single-minimum state." This certainly seems to imply that you are getting two for the price of one, that the internal barrier is creating new vibrational states.
Actually all that is going on is a bunching of existing states. Sketch the harmonic oscillator eigenfunctions and then compare them with the eigenfunctions for harmonic oscillator with Gaussian internal barrier. The presence of the barrier raises the energy of all states, but raises the energy of the symmetric states more than the antisymmetric states because the antisymmetric states have a node (zero probability) in the barrier and the symmetric states don't have a node.
In summary, the ammonia maser is a stunning illustration of quantum mechanics in action. We should teach the maser phenomenon from the quantum mechanical point of view: classical models are more easily digestable, but unfortunately they are incorrect.
Appendix
The selection rule stated above can be illustrated with an animation of the time-dependent superposition involving the two states involved in a transition. McMillin [J. Chem. Ed. 55, 7 (1978)] has described an appealing model for "quantum jumps" that is referred to as the fluctuating dipole mechanism. The animation shown below is based on McMillin's dipole mechanism. Several other "quantum jump" tutorials on this page are also based on McMillin's mechanism.
The v = 0 to v = 1 transition is allowed because the time-dependent superposition involving these states exhibits oscillating dipole character. The ammonia molecule oscillates asymmetrically in time within the left and right potential wells providing a mechanism for coupling with the oscillating electromagnetic field as long as the Bohr frequency condition is satisfied.
The v = 0 to v = 2 transition is forbidden because the time-dependent superposition involving these states does not exhibit oscillating dipole character. The ammonia molecule oscillates symmetrically in time within the left and right potential well and does not provide a mechanism for coupling with the external electromagnetic field.
Select quantum numbers of initial and final states:
$\begin{matrix} vi = 0 & vf = 1 \end{matrix} \nonumber$
$\begin{matrix} \text{Energy of transition:} & \Delta E = E_{vf} - E_{vi} & \Delta E = 4.45734266 \times 10^{-6} & t = \frac{ \text{FRAME}}{ 10^{-8}} \end{matrix} \nonumber$
Time-dependent superposition of initial and final states:
$\Phi_i = \frac{ \Psi(vi)_i exp(-i E_{vi}t) + \Psi (vf)_i exp(-i E_{vf}t)}{ \sqrt{2}} \nonumber$
To animate click on Tools, Animation, Record and follow the instructions in the dialog box. Recommended setting: From 0 To 40 in 10 Frames/Sec.
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Schrödinger's equation is solved numerically for a symmetric double well potential: $V = bx^4 - cx^2$, which resembles the double-well potential used to model the ammonia inversion in the previous tutorial. The integration algorithm is taken from J. C. Hansen, J. Chem. Educ. Software, 8C2, 1996.
Set parameters:
$\begin{matrix} \text{Increments:} & n = 100 & \text{Integration limits:} & xmin = 4 & \Delta = \frac{xmax - xmin}{n-1} \ \text{Effective mass:} & \mu = 1 & \text{Constants:} & b = 1 & c = 6 \end{matrix} \nonumber$
Calculate position vector, the potential energy matrix, and the kinetic energy matrix. Then combine them into a total energy matrix.
$\begin{matrix} i = 1 .. n & j = 1 .. n & x_i = xmin + (i-1) \Delta \end{matrix} \nonumber$
$\begin{matrix} V_{i,~j} = \text{if} \left[ i = j,~b (x_i)^4 - c (x_j)^2,~0 \right] & T_{i,~j} = \text{if} \left[ i=j,~ \frac{ \pi^2}{6 \mu \Delta^2},~ \frac{(-1)^{i-j}}{(i-j)^2 \mu \Delta} \right] \end{matrix} \nonumber$
Hamiltonian matrix:
$H = T + V \nonumber$
Calculate eigenvalues: E = sort(eigenvals(H))
Display selected eigenvalues: m = 1 .. 6
$E_m = \begin{array}{|c|} \hline -6.6427 \ \hline -6.6406 \ \hline -2.4512 \ \hline -2.3156 \ \hline 0.4156 \ \hline 1.6785 \ \hline \end{array} \nonumber$
Calculate selected eigenvectors: k = 1 .. 4 Ψ(k) = eigenvec(H, Ek)
Display the results graphically:
The numerical results show that the energy eigenvalues are paired due to the presence of the central barrier because the odd energy states have a node at the center of the internal potential barrier. The canonical solutions to Schrödinger's equation (stationary states) show that the wave functions are delocalized over the entire potential well and that tunneling is occurring because of the probability of being in a barrier of greater potential energy than that of the total eigenstate energy. But at this point tunneling does not imply motion, just a violation of a very important classical concept - kinetic energy cannot be negative!
We also see that the ground and first-excited states are very nearly degenerate. In the presence of a small perturbing electromagnetic field, the system would be forced into a superposition of these states which would exhibit oscillatory behavior. This effect is now demonstrated for the two lowest energy states. In the presence of this perturbation we see movement from one well to the other because the system is no longer in a stationary state.
This effect is demonstrated using Mathcad's animation capabilities.
Select quantum numbers of superposition states:
$\begin{matrix} vi = 1 & vf = 2 & t = \frac{ \text{FRAME}}{0.005}\end{matrix} \nonumber$
Timed-dependent superposition of these states:
$Phi_i = \frac{ \Psi(vi)_i exp(-i E_{vi}t) + \Psi (vf)_i exp (-i E_{vf}t)}{ \sqrt{2}} \nonumber$
To animate click on Tools, Animation, Record and follow the instructions in the dialog box. Recommended setting: From 0 To 40 in 5 Frames/Sec.
4.06: Analyses of the Pure Rotational Spectrum of HCl
This exercise deals with simple algebraic analyses of the microwave spectrum of HCl. This problem is dealt with in a number of current physical chemistry texts. Here the rigid rotor and non‐rigid rotor models will be used to analyze the data that follows.
Rotational absorption lines from H35Cl were found at the following wavenumbers (cm‐1): 83.32, 104.13, 124.73, 145.37, 165.89, 186.23, 206.60, and 226.86. Given this data our goal is to calculate the rotational constant, the bond length and the centrifugal distortion constant. We also want to assign J quantum numbers to each of the transitions.
Rigid Rotor Model for HCl
For the rigid rotor model the rotational energy levels (in cm‐1) are given by the following equation.
$E_J = BJ(J+1) \nonumber$
where
$B = \dfrac{h}{8 \pi^2 c \mu r^2} \nonumber$
For absorption spectroscopy, the rotational selection rule is $ΔJ = \pm 1$. Therefore, the energies of the allowed rotational transitions are,
$\begin{matrix} \Delta E(J) = B(J+1)(J+2) - BJ(J+1) \text{ simplify } \rightarrow \Delta E(J) = 2BJ + 2B \ \Delta E(J) = 2 \textcolor{red}{B} (J + 1) \end{matrix} \nonumber$
We do not know if the first frequency listed above (83.32 cm‐1) is the J = 0 to J = 1 transition. However, we do know that the frequencies represent adjacent transitions. So we can use the first two transitions to calculate both J and B.
$\begin{matrix} h = 6.62608 (10)^{-34} \text{joule sec} & c = 2.99792458 (10)^8 \dfrac{m}{sec} & u = 1.66054 (10)^{-27} kg \ m_H = 1.0078 u & m_{Cl} = 34.9688 u & pm = 10^{-12} m \end{matrix} \nonumber$
Calculate the reduced mass:
$\begin{matrix} \mu = \dfrac{m_H m_{Cl}}{m_H + m_{Cl}} & \mu = 1.62661 \times 10^{-27} kg \end{matrix} \nonumber$
Calculate the bond length:
$\begin{array}{c|c} B = \dfrac{h}{8 \pi^2 c \mu r^2} & _{float,~4}^{solve,~r} \rightarrow \begin{bmatrix} \dfrac{.1286e-8}{ \text{m kg}} \text{(m kg joule cm)}^{ \dfrac{1}{2}} \text{sec} \ \dfrac{-.1286e-8}{ \text{m kg}} \text{(m kg joule cm)}^{ \dfrac{1}{2}} \text{sec} \end{bmatrix} = \begin{pmatrix} 128.6 \ -128.6 \end{pmatrix} pm \end{array} \nonumber$
Clear memory of J and B to facilitate subsequent calculations:
$\begin{matrix} J = J & B = B \end{matrix} \nonumber$
Non‐Rigid Rotor Model for HCl
The non‐rigid rotor model adds a centrifugal distortion term to accommodate the classical idea that the H‐Cl bond will stretch as the rotational energy increases causing the rotational states to become closer together at higher J values.
The rotational energy levels (in cm‐1) for the non‐rigid rotor are given by,
$E(J) = BJ (J+1) - \textcolor{red}{D} J^2 (J+1)^2 \nonumber$
The rotational transitions, therefore, appear at the following energies
$\begin{matrix} \Delta E (J) = E(J+1) - E(J) \text{ simplify } \rightarrow 2B(J+1) = 2BJ + 2B - 4DJ^3 - 12 DJ^2 - 12DJ - 4D \ \Delta E(J) = 2BJ + 2B - 4 \textcolor{red}{D} J^3 - 12 DJ^2 - 12DJ - 4D \end{matrix} \nonumber$
Because J values have now been assigned to the rotational transitions, we can use two of them to calculate B and D.
$\begin{array}{c|c} (B ~D) = \begin{pmatrix} \Delta E(3) = 83.2 cm^{-1} \ \Delta E (4) = 104.13 cm^{-1} \end{pmatrix} & _{float,~4}^{solve,~ \begin{pmatrix} B \ D \end{pmatrix}} \rightarrow \begin{pmatrix} \dfrac{10.42}{cm} \ \dfrac{.1111e-3}{cm} \end{pmatrix} \end{array} \nonumber$
Next we calculate the HCl bond length under the non-rigid rotor approximation.
$\begin{array}{c|c} B = \dfrac{h}{8 \pi^2 c \mu r^2} & _{float,~4}^{solve,~r} \rightarrow \begin{bmatrix} \dfrac{.1285e-8}{ \text{m kg}} \text{(m kg joule cm)}^{ \dfrac{1}{2}} \text{sec} \ \dfrac{-.1285e-8}{ \text{m kg}} \text{(m kg joule cm)}^{ \dfrac{1}{2}} \text{sec} \end{bmatrix} = \begin{pmatrix} 128.5 \ -128.5 \end{pmatrix} pm \end{array} \nonumber$
Finally we summarize the calculated results and compare them to the literature values. Given the simplicity of the models and the rudimentary method of analysis, the comparisons are satisfactory. The large error in D is not surprising given its small magnitude.
$\begin{pmatrix} \text{Molecular Parameter} & \text{Rigid Rotor} & \text{Nonrigid Rotor} & \text{Literature} \ \dfrac{ \text{Rotational Constant}}{cm^{-1}} & 10.41 & 10.42 & 10.59 \ \dfrac{ \text{Bond Length}}{pm} & 128.6 & 128.5 & 127.5 \ \dfrac{ \text{Centrifugal Distortion Constant}}{cm^{-1}} & . & 1.11 (10)^{-4} & 5.32 (10)^{-4} \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.05%3A_A_Symmetric_Double_Well_Potential_Illustrating_Tunneling.txt
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This analysis assumes an harmonic‐oscillator, non‐rigid rotor model for the vibrational and rotational degrees of freedom of gas‐phase HCl. In other words, the magnitude of the rotational constant depends on the vibrational state of the molecule. Using the portion of the H35Cl vibrational‐rotational spectrum provided below, this model will be used to calculate the following molecular parameters: ν0, B0, B1, Be, αe, r0, r1, re, and k.
A simple algebraic method, rather than a sophisticated statistical analysis, will be used to extract HClʹs molecular parameters from the spectroscopic data. We will see that although the model and the method of analysis are rudimentary, the results compare rather well with literature values for the molecular parameters. This exercise might serve as an introduction to a more rigorous and thorough statistical analysis. The equations for the R‐branch and P‐branch transitions appropriate for this model are given below.
$\begin{matrix} \nu_R(J) = \textcolor{red}{ \nu_0} B_1 (J+1)(J+2) - B_0 J(J+1) & \nu_p (J) = \textcolor{red}{ \nu_0} + B_1 (J-1)J-B_0 J(J+1) \end{matrix} \nonumber$
where
$\begin{matrix} B_{ \nu} = \frac{h}{8 \pi^2 c \mu r_{ \nu}^2} & B_{ \nu} = B_e - \alpha_e \left( \nu + \frac{1}{2} \right) \end{matrix} \nonumber$
Fundamental constants, conversion factors and atomic masses:
$\begin{matrix} h = 6.6260755 (10)^{-34} \text{joule sec} & c = 2.99792458 (10)^8 \frac{m}{sec} & u = 1.66054 10^{-27} kg & pm = 10^{-12} m \ m_H = 1.0078 u & m_{Cl} = 34.9688 u \end{matrix} \nonumber$
Calculate the reduced mass of HCl:
$\begin{matrix} \mu = \frac{m_H + m_{Cl}}{m_H + m_{Cl}} & \mu = 1.627 \times 10^{-27} kg \end{matrix} \nonumber$
Obtain several P- and R-branch transitions from the spectrum:
$\begin{pmatrix} \text{Transition} & P(2) & P(1) & R(0) & R(1) \ \frac{ \text{Frequency}}{cm^{-1}} & 2841.80 & 2864.35 & 2905.63 & 2923.87 \end{pmatrix} \nonumber$
Set up and solve a system of equations to calculate ν0, B0 and B1 by selecting data from the table above.
$\begin{array}{c|c} \begin{pmatrix} \nu_0 & B_0 & B_1 \end{pmatrix} = \begin{pmatrix} \nu_p (2) = 2841.80 cm^{-1} \ \nu_p (1) = 2864.35 cm^{-1} \ \nu_R (0) = 2905.63 cm^{-1} \end{pmatrix} & _{float,~5}^{solve,~ \begin{pmatrix} \nu_0 \ B_0 \ B_1 \end{pmatrix}} \rightarrow \begin{pmatrix} \frac{2885.6}{cm} \ \frac{10.638}{cm} \ \frac{10.002}{cm} \end{pmatrix} \end{array} \nonumber$
Use the values of B0 and B1 to calculate Be and αe:
$\begin{array}{c|c} \begin{pmatrix} B_e & \alpha_e \end{pmatrix} = \begin{pmatrix} B_0 = B_e - \alpha_e \frac{1}{2} \ B_1 = B_e - \alpha_e \frac{3}{2} \ \nu_R (0) = 2905.63 cm^{-1} \end{pmatrix} & _{float,~5}^{solve,~ \begin{pmatrix} B_e \ \alpha_e \end{pmatrix}} \rightarrow \begin{pmatrix} \frac{10.956}{cm} \ \frac{.63600}{cm} \end{pmatrix} \end{array} \nonumber$
Now calcuate r0, r1, and re using the values of B0, B1 and Be.
$\begin{matrix} r_0 = \sqrt{ \frac{h}{8 \pi^2 c \mu B_0}} & r_0 = 127.189 pm & r_1 = \sqrt{ \frac{h}{8 \pi^2 c \mu B_1}} & r_1 = 131.171 pm \ r_e = \sqrt{ \frac{h}{8 \pi^2 c \mu B_e}} & r_e = 125.33 pm \end{matrix} \nonumber$
Calculate the force constant using the value of ν0.
$\begin{matrix} \begin{array}{c|c} k = \nu_0 = \frac{1}{2 \pi c} \sqrt{ \frac{k}{ \mu}} & _{float,~4}^{solve,~k} \rightarrow .4806e-1 \frac{m^2}{sec^2 cm^2} kg \end{array} & k = 480.6 \frac{newton}{m} \end{matrix} \nonumber$
Compare the calculated parameter values with the literature values.
$\begin{pmatrix} \frac{ \text{Molecular}}{ \text{Parameter}} & \frac{ \text{Calcuated}}{ \text{Value}} & \frac{ \text{Literature}}{ \text{Value}} & \% \text{Error} \ \frac{ \nu_0}{cm^{-1}} & 2885.6 & 2886 & 0.014 \ \frac{B_0}{cm^{-1}} & 10.638 & 10.440 & 1.90 \ \frac{B_1}{cm^{-1}} & 10.002 & 10.136 & 1.32 \ \frac{B_e}{cm^{-1}} & 10.956 & 10.593 & 3.43 \ \frac{ \alpha_e}{ cm_{-1}} & 0.6360 & 0.307 & 107 \ \frac{r_0}{pm} & 127.2 & 128.3 & 0.86 \ \frac{r_1}{pm} & 131.2 & 130.2 & 0.77 \ \frac{r_e}{pm} & 125.3 & 127.4 & 1.65 \ \frac{k}{newton~m^{-1}} & 480.6 & 516.3 & 6.92 \end{pmatrix} \nonumber$
Summary: Given the simplicity of the model (harmonic‐oscillator, non‐rigid rotor) and the rudimentary algebraic (as opposed to rigorous statistical) method of analysis, the results are quite respectable. Naturally, results will vary depending on the P‐ and R‐branch transitions used to calculate the molecular parameters.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.07%3A_A_Rudimentary_Analysis_of_the_Vibrational-Rotational_HCl_Spectrum.txt
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Using the harmonic oscillator model, the selection rule for molcecular vibrations is ∆v = +/- 1. However, overtone transitions (∆v = 2, 3, ..) are observed experimentally. In what follows it will be shown that under the Morse oscillator (anharmonic) model, overtone vibrational transitions are weakly allowed.
Schroedinger's equation for the Morse oscillator model for $\ce{HCl}}$ is integrated numerically for the first five energy states. (Integration algorithm taken from J. C. Hansen, J. Chem. Educ. Software, 8C2, 1996.)
Set integration parameters (in atomic units):
$\begin{matrix} n = 150& xmin = 1.4 & xmax = 8 & \Delta = \frac{xmax - xmin}{n-1} \end{matrix} \nonumber$
Enter the Morse parameters for HCl (in atomic units):
$\begin{matrix} \mu = 1785.64 & D = .1655 & \beta = 1.0051 & x_e = 2.4086 \end{matrix} \nonumber$
Calculate position vector, the potential energy matrix, and the kinetic energy matrix. Then combine them into a total energy matrix.
$\begin{matrix} i = 1 .. n & j = 1 .. n & x_i = xmin + (i-1) \Delta \end{matrix} \nonumber$
$\begin{matrix} V_{i,~j} = \text{if} \left[ i = j,~D \left[ 1 - exp \left[ - \beta (x_i - x_e ) \right] \right]^2,~0 \right] & T_{i,~j} = \text{if} \left[ i=j,~ \frac{ \pi^2}{6 \mu \Delta^2},~ \frac{(-1)^{i-j}}{(i-j)^2 \mu \Delta} \right] \end{matrix} \nonumber$
Hamiltonian matrix: H = T + V
Calculate eigenvalues: E = sort(eigenvals(H))
Display selected eigenvalues: m = 1 .. 6
$E_m = \begin{array}{|c|} \hline 0.00677149 \ \hline 0.01989016 \ \hline 0.03244307 \ \hline 0.04443024 \ \hline 0.05585166 \ \hline \end{array} \nonumber$
Calculate selected eigenvectors: Ψ(k) = eigenvec(H, Ek)
Plot diagonal elements of potential energy matrix and selected eigenfunctions: Vpoti = Vi, i
Because of the way the integration algorithm is implemented the initial index is m = 1. This corresponds to the v = 0 vibrational ground state. In other words, subtract 1 from the m index to get the v quantum number.*
Demonstrate numerically that the solutions form an orthonormal set:
$\begin{matrix} \sum_i ( \Psi (1)_i )^2 = 1.00 & \sum_i ( \Psi (2)_i )^2 = 1.00 & \sum_i ( \Psi (3)_i )^2 = 1.00 \ \sum_i ( \Psi (1)_i \Psi (2)_i ) = 0.00 & \sum_i ( \Psi (1)_i \Psi (3)_i ) = 0.00 & \sum_i ( \Psi (2)_i \Psi (3)_i ) = 0.00 \end{matrix} \nonumber$
Calculate the transition dipole moment to demonstrate that the v = 0 to v = 2 transition, which is forbidden for the harmonic oscillator, is (slightly) allowed. Compare this result to v = 0 to v = 1, v = 1 to v=2 transitions which are allowed. (See * above.)
$\begin{matrix} \sum_i ( \Psi (1)_i x_i \Psi (3)_i ) = -0.01 & \sum_i ( \Psi (1)_i x_i \Psi (2)_i ) = 0.14 & \sum_i ( \Psi (2)_i x_i \Psi (3)_i ) = 0.21 \end{matrix} \nonumber$
Animate these transitions to confirm that they lead to oscillating dipole character.
Select quantum numbers of initial and final states:
$\begin{matrix} v_i = 1 & v_f = 3 \end{matrix} \nonumber$
Energy of transition:
$\begin{matrix} \Delta E = E_{vf} - E_{vi} & \Delta E = 0.02567158 \end{matrix} \nonumber$
Define animation parameter:
$t = \frac{ \text{FRAME}}{10^{-8}} \nonumber$
Time-dependent superposition of initial and final states:
$\Phi_i = \frac{ \Psi (vi)_i exp (-i E_{vi} t) + \Psi (vf)_i exp(-i E_{vf} t)}{ \sqrt{2}} \nonumber$
To animate click on Tools, Animation, Record and follow the instructions in the dialog box. Recommended setting: From 0 To 40 in 5 Frames/Sec.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.08%3A_Visualizing_the_Formally_Forbidden_Overtone_Vibrational_Transitions_in_HCl.txt
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Quantum Jumps for an Electron in a One-dimensional Box
The phrases "quantum jump" and "quantum leap" are used in everyday discourse. This disguises the fact that scientists have always been somewhat troubled by the nature of the process by which a quantum system passes from one allowed energy state to another. McMillin [J. Chem. Ed. 55, 7 (1978)] has described an appealing model for "quantum jumps" that is referred to as the fluctuating dipole mechanism. This mechanism will be illustrated by considering spectroscopic transitions for an electron in a one-dimensional box of length ao. Since only a brief outline of the mechanism can be provided here, please consult this reference for a thorough presentation of the theoretical background.
In order for an electron in a one-dimensional box to undergo a transition from one allowed energy state to another under the influence of electromagnetic radiation two criteria must be met according to the mechanism described by McMillin. First, the photons of the electromagnetic field must satisfy the Bohr frequency condition and have an energy which equals the difference in energy between the two states under consideration. Second, there must be a dipolar coupling between the electromagnetic field and the oscillating electron density in the box. This later criterion is, of course, the selection rule for the transition.
According to this model, when an electron in a box is subjected to a perturbation such as electromagnetic radiation, the electron responds by going into a state which is a linear superposition of the unperturbed states.
$\Psi (x,~t) = \sum_i \psi_i \text{exp} \left( - \frac{iEt}{ \hbar} \right) \nonumber$
If the square modulus, Ψ(x,t)*Ψ(x,t), of the time-dependent wavefunction associated with this linear superposition exhibits an asymmetric fluctuating charge density (an oscillating dipole moment) in the box, a coupling with the dipolar character of the electromagnetic field exists and a transition can occur. If the linear combination of states leads to a symmetric fluctuating charge density there is no coupling between the field and the electron density in the box and a transition is not possible.
The model is illustrated below first for the n = 1 to n = 2 allowed transition for the electron in a one- dimensional box. Note that for the 1 ---> 2 transition, shown immediately below, the electron density does oscillates from one side of the box to the other satisfying the mechanism's criterion for coupling with the external electromagnetic field.
The n = 1 to n = 2 Transition for the Particle in a Box is Allowed
In the space immediately below the wavefunction for the linear superposition of states is calculated and Ψ*Ψ is plotted to demonstrate that the 1 ---> 2 transition is allowed.
Initial and final energy states for the transition under study
$\begin{matrix} n_i = 1 & n_f = 2 & E_i = \frac{n_i^2 \pi^2}{2} & E_f = \frac{n_f^2 \pi^2}{2} \end{matrix} \nonumber$
Plot the wavefunction:
$\begin{matrix} j = 0 .. 40 & x_j = \frac{j}{40} & k = 0 .. 40 & t_k = \frac{k}{40} \end{matrix} \nonumber$
Linear combination of ground state and excited states:
$\Psi (x,~t) = \sin (n_i \pi x) exp (-i E_i t) + \sin (n_f \pi x) exp (-i E_f t) \nonumber$
Calculate and plot Ψ*Ψ:
$\Psi \Psi_{(j,~k)} = \overline{ \Psi (x_j,~t_k} \Psi (x_j,~t_k ) \nonumber$
In this contour plot the horizontal axis is the spatial axis and time is graphed on the vertical axis.
The n = 1 to n = 3 Transition for the Particle in a Box is Forbidden
However, for the 1 ---> 3 transition the electron density fluctuates symmetrically about the center of the box, and, therefore does not provide a mechanism for coupling with the external electromagnetic field.
Initial and final energy states for the transition under study
$\begin{matrix} n_i = 1 & n_f = 3 & E_i = \frac{n_i^2 \pi^2}{2} & E_f = \frac{n_f^2 \pi^2}{2} \end{matrix} \nonumber$
Plot the wavefunction:
$\begin{matrix} j = 0 .. 40 & x_j = \frac{j}{40} & k = 0 .. 40 & t_k = 4 \frac{k}{40} \end{matrix} \nonumber$
Linear combination of ground state and excited states:
$\Psi (x,~t) = \sin (n_i \pi x) exp (-i E_i t) + \sin (n_f \pi x) exp (-i E_f t) \nonumber$
Calculate and plot Ψ*Ψ:
$\Psi \Psi_{(j,~k)} = \overline{ \Psi (x_j,~t_k} \Psi (x_j,~t_k ) \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.09%3A_The_Quantum_Jump.txt
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Consider an electron in a one‐bohr, one‐dimensional box. This tutorial will explore, in an elementary way, the selection rule for the transition of the electron from one allowed energy level to another. The condition for an allowed transition is two‐fold: the photon exciting the transition must satisfy the Bohr frequency condition [hν = Ef ‐ Ei ] and the expectation value for the position of the electron must exhibit oscillatory dipole character as a function of time. This latter requirement provides a coupling mechanism between the oscillating electromagnetic field and the oscillating charge density of the electron in the box.
Assuming that the Bohr frequency condition is met we will look at the second at the second criterion for a number of possible electronic transitions or ʺquantum jumps.ʺ Consider first the transition from the ground state to the first excited state.
$\begin{matrix} \text{Initial state:} & n_i = 1 & E_i = \frac{n_i^2 \pi^2}{2} & \text{Final state:} & n_f = 2 & E_f = \frac{n_f^2 \pi^2}{2} \end{matrix} \nonumber$
Time‐dependent superposition of initial and final states:
$\Psi (x,~t) = \sin (n_i \pi x) exp ( -i E_i t) + \sin (n_f \pi x) exp (-i E_f t) \nonumber$
Time-dependent expectation value for position:
$\begin{matrix} \text{position(t)} = \int_0^1 x (| \Psi \text{(x, t)} |)^2 dx & t = 0, .001 .. 1 \end{matrix} \nonumber$
Since the expectation value for position fluctuates in time (has time dependence) the transition between the n =1 and n = 2 states is allowed.
Now consider the n = 1 to n = 3 electronic transition.
$\begin{matrix} \text{Initial state:} & n_i = 1 & E_i = \frac{n_i^2 \pi^2}{2} & \text{Final state:} & n_f = 3 & E_f = \frac{n_f^2 \pi^2}{2} \end{matrix} \nonumber$
Time-dependent superposition of initial and final states:
$\Psi (x,~t) = \sin (n_i \pi x) exp (-i E_i t) + \sin (n_f \pi x) exp (-i E_f t) \nonumber$
$\begin{matrix} \text{position}(t) = \int_0^1 x (| \Psi (x,~t) |)^2 dx & t = 0, .002 .. 1 \end{matrix} \nonumber$
For this case the expectation value for position does not fluctuate with time, providing no mechanism for coupling with the oscillating dipole character of the electromagnetic field. Therefore, the n = 1 to n = 3 electronic transition is not allowed. The selection rule that emerges after study of more cases is that Δn = an odd integer.
Simple Harmonic Oscillator
The same method is now used to look at allowed and forbidden transitions for the simple harmonic oscillator.
$\begin{matrix} \text{Initial state:} & E_i = \nu_i + \frac{1}{2} & \text{Final state:} & E_f = \nu_f + \frac{1}{2} & nu = 0,~1, \text{etc.} \end{matrix} \nonumber$
Time‐dependent superposition of initial and final harmonic oscillator states:
$\Psi (x,~t,~\nu_i,~\nu_f ) = \frac{1}{ \sqrt{2}} \text{exp} \left( - \frac{x^2}{2} \right) \left[ \frac{ \text{Her} ( \nu_i,~ x) \text{exp} \left[ -i \left( \nu_i + \frac{1}{2} \right) t \right] }{ \sqrt{ \nu_i! 2^{\nu_i} \sqrt{ \pi} }} + \frac{ \text{Her} ( \nu_f,~ x) \text{exp} \left[ -i \left( \nu_f + \frac{1}{2} \right) t \right] }{ \sqrt{ \nu_f! 2^{\nu_f} \sqrt{ \pi} }} \right] \nonumber$
Time-dependent expectation value for position:
$\begin{matrix} \text{position} \left( t,~ \nu_i,~ \nu_f \right) = \int_{- \infty}^{ \infty} x \left( \left| \Psi (x,~t,~ \nu_i,~ \nu_f ) \right| \right)^2 dx & t = 0,~.05 .. 20 \end{matrix} \nonumber$
Plot the time‐dependent position expectation value for three transitions: 0‐1; 0‐2; 1‐2. The 0‐1 and 1‐2 transitions are allowed and the 0‐2 transition is forbidden.
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Quantum beats are the oscillatory behavior in the intensity of radiation emitted by atomic or molecular systems that are in a superposition of excited states created by off‐resonance excitation. The quantum mechanical analysis of quantum beats is essentially the same as it is for the double‐slit experiment. In the double‐slit experiment there are two paths to each position on the detector and the probability amplitudes for these paths interfere. As shown in the figure below, in this example the ground‐state can be reached from two excited states having different energies. The probability amplitudes for these paths to the ground state (| a > ‐‐‐> | g > and | b > ‐‐‐> | g >) also interfere constructively and destructively causing oscillations in the intensity of the emitted radiation.
The superposition of transitions to the ground state in the time domain is as follows,
$\langle g | t \rangle \langle t | a \rangle + \langle g | t \rangle \langle t | b \rangle \nonumber$
where, for example, the Dirac brackets in atomic units are of the form
$\langle t | a \rangle = \langle a | t \rangle* = \text{exp}(-i E_a t) \nonumber$
Using relations of the type given in equation (2) we can write equation (1) as
$\text{exp}(iE_g t) \text{exp}(-iE_a t) + \text{exp}(iE_g t) \text{exp}(-i E_b t) \nonumber$
This equation can be simplified by using the ground state as a reference by setting Eg = 0.
$\text{exp}(-iE_a t) + \text{exp}(-iE_b t) \nonumber$
The excited states decay exponentially, but not necessarily with the same decay constant. The fraction of the atoms or molecules not decayed at time t is given by first‐order kinetics.
$\frac{A_t}{A_0} = \text{exp} (-k_a t) \nonumber$
Thus, each term in equation (4) is weighted by the probability that state exists at time t.
$\text{exp}(-i E_a t) \text{exp}(-k_a t) + \text{exp} (-iE_b t) \text{exp}(-k_b t) \nonumber$
The time‐dependence of the intensity of the emitted radiation is proportional to the square of the absolute magnitude of equation (6)
$\left| \text{exp} (-iE_a t) \text{exp} (-k_a t) + \text{exp}(-i E_b t) \text{exp} (-k_bt) \right|^2 \nonumber$
The time‐dependence as calculated below for arbitrary values for the parameters clearly shows the oscillatory character of the intensity of the emitted radiation.
$\begin{matrix} E_a = 1 & k_a = .1 & E_b = 2 & k_b = .05 \end{matrix} \nonumber$
$\text{Beat(t)} = \text{exp} (-i E_b t) \text{exp} (-k_b t) + \text{exp}(-i E_a t) \text{exp}(-k_a t) \nonumber$
4.12: The 1s-2s Electronic Transition in the 1D Hydrogen Atom
This tutorial examines the 1s-2s electronic transition in the hydrogen atom. These L = 0 states of the hydrogen atom can be written as one-dimensional, time-dependent wave functions as shown below.
The condition for an allowed electronic transition is two-fold:
1. the photon exciting the transition must satisfy the Bohr frequency condition [hν = E_f - E_i ] and
2. the expectation value for the position of the electron must exhibit oscillatory dipole character as a function of time.
This latter requirement provides a coupling mechanism between the oscillating electromagnetic field and the oscillating charge density of the electron. For further details see McMillin J. Chem. Ed. 55, 7 (1978).
The subsequent calculations are carried out using atomic units: $h = 2 π$; $m_e = 1$.
$\begin{matrix} \text{Initial state:} & n_i = 1 & E_i = \frac{-0.5}{n_i^2} & \text{Final state:} & n_f = 2 & E_f = \frac{-0.5}{n_f^2} \end{matrix} \nonumber$
Time-dependent 1s and 2s electron wave functions:
$\begin{matrix} \Psi_1 (x,~t) = 2 x \text{exp}(-x) \text{exp}(-i E_i t) & \Psi_2 (x,~t) = \frac{1}{ \sqrt{8}} x(2-x) \text{exp} \left( - \frac{x}{2} \right) \text{exp} (-i E_f t) \end{matrix} \nonumber$
The following graphic shows the separate 1s and 2s electron densities at $t =0$.
In the presence of a perturbing electromagnetic field the hydrogen atom electron is no longer in a stationary state and is represented by the following time-dependent superposition of the 1s and 2s electronic states.
$\Psi (x,~t) = \frac{1}{ \sqrt{2}} \left( \Psi_1 (x,~t) + \Psi_2 (x,~t) \right) \nonumber$
The square of the absolute magnitude for this time-dependent superposition is shown below for three time values. It clearly shows the oscillatory character of the perturbed electron density, and it appears that the oscillations have dipole character.
That the oscillations of the electron density do indeed have dipole character can also be seen by plotting the average position of the electron from the nucleus as a function of time.
$\begin{matrix} \text{Average Position(t)} = \int_0^{ \infty} x \left( \left| \Psi (x,~t) \right| \right)^2 dx & t = 0 .. 100 \end{matrix} \nonumber$
Since the expectation value for position fluctuates in time creating an oscillating dipole the transition between the $n =1$ and $n = 2$ states is allowed.
It is also possible to animate the time-dependent superposition. Click on Tools, Animation, Record and follow the instructions in the dialog box. Choose From: 0, To: 100, At: 10 Frames/Sec .
t = FRAME
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.11%3A_Quantum_Beats.txt
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This tutorial is a companion to "The Quantum Jump" which deals with the quantum jump from the perspective of the coordinate-space wave function. This tutorial accomplishes the same thing in momentum space.
The time-dependent momentum wave function for a particle in a one-dimensional box of width 1a0 is shown below.
$\psi (n,~p,~t) = n \sqrt{ \pi} \left[ \frac{1-(-1)^n \text{exp} (-i p)}{n^2 - \pi^2 - p^2} \right] \text{exp}(-i E_i t) \nonumber$
The n = 1 to n = 2 Transition for the Particle in a Box is Allowed
This transition is allowed because it yields a momentum distribution that is asymmetric in time as is shown in the figure below. Consequently it allows for coupling with the perturbing electromagnetic field.
$\begin{matrix} \text{Momentum increment} & P = 100 & \text{Time Increment} & T = 100 & \text{Initial} & n_i = 1 & \text{Final state} & n_f = 2 \end{matrix} \nonumber$
Initial and final energy states for the transition under study:
$\begin{matrix} E_i = \frac{n_i^2 \pi^2}{2} & E_f = \frac{n_f^2 \pi^2}{2} \end{matrix} \nonumber$
Plot the wavefunction:
$\begin{matrix} j = 0 .. P & p_j = -10 + \frac{20j}{P} & k = 0 .. T & t_k = \frac{k}{T} \end{matrix} \nonumber$
In the presence of electromagnetic radiation the particle in the box goes into a linear superposition of the stationary states. The linear superposition for the n = 1 and n = 2 states is given below.
$\Psi (p,~t) = n_i \sqrt{ \pi} \left[ \frac{1-(-1)^{n_i} \text{exp}(-ip)}{n_i^2 \pi^2 - p^2} \right] \text{exp} (-i E_i t) + n_f \sqrt{ \pi} \left[ \frac{1-(-1)^{n_f} \text{exp}(-ip)}{n_f^2 \pi^2 - p^2}\right] \text{exp} (-i E_f t) \nonumber$
Calculate and plot the momentum distribution: Ψ*Ψ:
$\Psi \Psi_{(j,~k)} = \left( \left| \Psi (p_j,~t_k ) \right| \right)^2 \nonumber$
The n = 1 to n = 3 Transition for the Particle in a Box is Not Allowed
This transition is allowed because it yields a momentum distribution that is symmetric in time as is shown in the figure below. Consequently it does not allow for coupling with the perturbing electromagnetic field.
$\begin{matrix} \text{Momentum increment} & P = 100 & \text{Time Increment} & T = 100 & \text{Initial} & n_i = 1 & \text{Final state} & n_f = 3 \end{matrix} \nonumber$
Initial and final energy states for the transition under study:
$\begin{matrix} E_i = \frac{n_i^2 \pi^2}{2} & E_f = \frac{n_f^2 \pi^2}{2} \end{matrix} \nonumber$
Plot the wavefunction:
$\begin{matrix} j = 0 .. P & p_j = -10 + \frac{20j}{P} & k = 0 .. T & t_k = \frac{k}{T} \end{matrix} \nonumber$
In the presence of electromagnetic radiation the particle in the box goes into a linear superposition of the stationary states. The linear superposition for the n = 1 and n = 3 states is given below.
$\Psi (p,~t) = n_i \sqrt{ \pi} \left[ \frac{1-(-1)^{n_i} \text{exp}(-ip)}{n_i^2 \pi^2 - p^2} \right] \text{exp} (-i E_i t) + n_f \sqrt{ \pi} \left[ \frac{1-(-1)^{n_f} \text{exp}(-ip)}{n_f^2 \pi^2 - p^2}\right] \text{exp} (-i E_f t) \nonumber$
Calculate and plot the momentum distribution: Ψ*Ψ:
$\Psi \Psi_{(j,~k)} = \left( \left| \Psi (p_j,~t_k ) \right| \right)^2 \nonumber$
4.14: The Harmonic Oscillator Quantum Jump
This Mathcad worksheet determines whether an SHO spectroscopic transition is allowed assuming that the Bohr frequency condition is satisfied. It requires only the quantum numbers of the initial and final states.
The $v = 0$ to $v = 1$ transition is allowed because the position distribution function, $Y^{*}Y$, exhibits oscillating dipole character.
$\begin{matrix} \text{Initial State:} & v_i = 0 & E_i = v_i + \frac{1}{2} & \text{Final state:} & v_f = 1 & E_f = v_f + \frac{1}{2} \end{matrix} \nonumber$
$\begin{matrix} \text{Set plot parameters:} & \text{Space} = 60 & \text{Time} = 10 & \text{Xmin} = 3 \ j = 0 .. \text{Space} & x_j = - \text{Xmin} + \frac{2 \text{Xmin}j}{ \text{Space}} & k = 0 .. \text{Time} & t_k = \frac{k20}{ \text{Time}} \end{matrix} \nonumber$
Construct time-dependent superposition of the initial and final states:
$\Psi (x,~t) = \text{exp} \left( - \frac{x^2}{2} \right) ( \text{Her} (v_i,~x) \text{exp} (-i E_i t) + \text{Her} (v_f,~x) \text{exp} (-i E_f t)) \nonumber$
Calculate and plot Y*Y:
$\Psi \Psi_{(j,~k)} = \overline{ \Psi (x_j,~t_k)} \Psi (x_j,~t_k ) \nonumber$
In this contour plot the horizontal axis is the spatial axis. Time is graphed on the veritcal axis. If Y*Y asymmetric in time the transition is allowed.
The v = 0 to v = 2 transition is allowed because the position distribution function, Y*Y, does not exhibit oscillating dipole character.
$\begin{matrix} \text{Initial State:} & v_i = 0 & E_i = v_i + \frac{1}{2} & \text{Final state:} & v_f = 1 & E_f = v_f + \frac{1}{2} \end{matrix} \nonumber$
$\begin{matrix} \text{Set plot parameters:} & \text{Space} = 60 & \text{Time} = 150 & \text{Xmin} = 3 \ j = 0 .. \text{Space} & x_j = - \text{Xmin} + \frac{2 \text{Xmin}j}{ \text{Space}} & k = 0 .. \text{Time} & t_k = \frac{k20}{ \text{Time}} \end{matrix} \nonumber$
Construct time-dependent superposition of the initial and final states:
$\Psi (x,~t) = \text{exp} \left( - \frac{x^2}{2} \right) ( \text{Her} (v_i,~x) \text{exp} (-i E_i t) + \text{Her} (v_f,~x) \text{exp} (-i E_f t)) \nonumber$
Calculate and plot Y*Y:
$\Psi \Psi_{(j,~k)} = \overline{ \Psi (x_j,~t_k)} \Psi (x_j,~t_k ) \nonumber$
In this contour plot the horizontal axis is the spatial axis. Time is graphed on the vertical axis. If Y*Y asymmetric in time the transition is allowed.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.13%3A_The_Quantum_Jump_in_Momentum_Space.txt
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The quantum mechanical harmonic oscillator eigenstates are stationary states and therefore cannot be used individually to represent classical oscillatory motion. (Atomic units are used in this tutorial.)
$\Psi (v,~x,~t) = \frac{1}{ \sqrt{ 2^v v~ \sqrt{ \pi}}} \text{Her}(v,~x) \text{exp} \left( \frac{-x^2}{2} \right) \text{exp} \left[ -i \left( v + \frac{1}{2} \right) t \right] \nonumber$
For example, suppose we choose to represent the ground vibrational state of a homonuclear diatomic molecule as a simple harmonic oscillator with vibrational quantum number v = 0. We see that probability distribution |Ψ(0,x,t)|2, is independent of time. There is no oscillatory motion; the molecule is in a stationary state which weighted superposition of all possible internuclear separations.
However,simple superpositions of the vibrational eigenstates do show oscillatory behavior. This is due to exponential term involving the vibrational energy, exp[-iE(v)t]. This term oscillates with a dependence on vibrational quantum number. Thus, the different eigenstates oscillate with different frequencies giving rise constructive and destructive interference. The figures below show the time dependence of the v = 0/v =1 the v = 0/v = 2 superpositions. Both show oscillatory behavior, but the first is asymmetric and the second i symmetric. This has significance for harmonic oscillator selection rules, as will be discussed below.
$\text{Sup(x, t)} = \frac{ \Psi (0,~x,~t) + \Psi(1,~x,~t)}{ \sqrt{2}} \nonumber$
The asymmetry of this time-dependent probability distribution gives it electric oscillating dipole character mechanism for coupling with the oscillating dipole of the electromagnetic field. Thus we could argue this i basis for the fact that the v = 0 to v = 1 vibrational transition is allowed.
$\text{Sup(x, t)} = \frac{ \Psi (0,~x,~t) + \Psi (2,~x,~t)}{ \sqrt{2}} \nonumber$
By comparison the symmetry of this time-dependent probability distribution does not have oscillatory dipole character, so there is no coupling with the external electromagnetic field. Therefore, the v = 0 to v = 2 vibrational transition is formally forbidden. Further detail on this interpretation of the "quantum jump" can found in the Spectroscopy section of Quantum Potpourri.
By comparison, coherent states (also called Glauber states) of the harmonic oscillator are more elaborate superpositions that maintain the well-defined shape of the ground state distribution while exhibiting the kind of classical oscillatory motion that is absent in the previous examples. The time-dependence and "classical" oscillatory behavior of a coherent superposition of 25 vibrational eigenstates is illustrated below. See any contemporary text on quantum optics for further information on coherent states of the harmonic oscillator.
$\begin{matrix} n = 25 & x = -8, -7.98 .. 8 & \alpha = 3.5 \end{matrix} \nonumber$
$\Psi (x,~t) = \frac{1}{ \sqrt{n}} \text{exp} \left( \frac{-x^2}{2} \right) \text{exp} \left( \frac{- \alpha^2}{2} \right) \sum_{v=0}^n \left[ \text{Her(v, x) exp} \left[ -i \left( v + \frac{1}{2} \right) t \right] \frac{ \alpha^v}{v! \sqrt{2^v \sqrt{2}}} \right] \nonumber$
Time-dependent superpositions of coherent states have been used to model Schrödinger cat states. Below show the interaction of two coherent states moving in opposite directions from opposite sides of a harmonic potential well. The interference observed when they meet in the middle has been observed experimentally Bose-Einstein condensates.
$\begin{matrix} \alpha = 3.5 & \beta = -3.5 \end{matrix} \nonumber$
$\begin{matrix} \Psi (x,~t) = & \frac{1}{ \sqrt{n}} \text{exp} \left( \frac{-x^2}{2} \right) \text{exp} \left( \frac{- \alpha^2}{2} \right) \sum_{v=0}^n \left[ \text{Her(v, x) exp} \left[ -i \left( v + \frac{1}{2} \right) t \right] \frac{ \alpha^v}{v~ \sqrt{ 2^v \sqrt{2}}} \right] ... \ & + \frac{1}{ \sqrt{n}} \text{exp} \left( \frac{-x^2}{2} \right) \text{exp} \left( \frac{- \beta^2}{2} \right) \sum_{v=0}^n \left[ \text{Her(v, x) exp} \left[ -i \left( v + \frac{1}{2} \right) t \right] \frac{ \beta^v}{v~ \sqrt{ 2^v \sqrt{2}}} \right] \end{matrix} \nonumber$
We finish with a calculation of the Wigner phase-space distribution for a Schrödinger cat state at t = 0. In t interest of computational expediency a superposition of only 10 harmonic eigenstates is calculated. The Wigner function is itself a superposition of all phase-space trajectories and is called a quasi probability distribution because it can take on negative values as is shown in the figure below. The interference fringe the center are closely related to those that appear in the figures above.
$\begin{matrix} \Psi (x) = & \frac{1}{ \sqrt{n}} \text{exp} \left( \frac{-x^2}{2} \right) \text{exp} \left( \frac{- \alpha^2}{2} \right) \sum_{v=0}^{10} \left( \text{Her(v, x)} \frac{ \alpha^v}{v~ \sqrt{ 2^v \sqrt{2}}} \right) ... \ ~ & + \left[ \frac{1}{ \sqrt{n}} \text{exp} \left( \frac{-x^2}{2} \right) \text{exp} \left( \frac{- \beta^2}{2} \right) \sum_{v=0}^{10} \left( \text{Her(v, x)} \frac{ \beta^v}{v~ \sqrt{ 2^v \sqrt{2}}} \right) \right] \end{matrix} \nonumber$
Wigner distribution:
$\text{W(x, p)} = \frac{1}{ \pi^{ \frac{3}{2}}} \int_{- \infty}^{ \infty} \Psi \left( x + \frac{s}{2} \right) \text{exp(i s p)} \Psi \left( x - \frac{s}{2} \right) ds \nonumber$
$\begin{matrix} N = 80 & i = 0 .. N & x_i = -5 + \frac{10i}{N} & j = 0 .. N & p_j = -5 + \frac{10j}{N} & \text{Wigner}_{i,~j} = W(x_i,~p_j) \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.15%3A_Coherent_States_of_the_Harmonic_Oscillator.txt
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This spectroscopy exercise deals with the interpretation of the visible spectrum of $\ce{Ti(H2O)6^{3+}}$ shown below.
The analysis begins with the assumption that $\ce{Ti(H2O)6^{3+}}$ has octahedral, $O_h$, symmetry. This assumption accounts for the gross features of the spectrum, but does not explain the shoulder that appears on the main absorption peak. The hexaaquatitanium(III) complex is orbitally degenerate and is therefore subject to a Jahn-Teller distortion which reduces the symmetry from octahedral to square planar. It will be shown that D4h symmetry is fully consistent with the experimental spectrum. The d-orbital energy level diagrams for Oh and D4h symmetry are shown below and will be referred to later in the analysis.
On the basis of these energy level diagrams the following predictions would be plausible. For octahedral symmetry there is one electronic transition in the visible region. For square planar symmetry there are two electronic transitions in the visible region and one in the infrared. In the analysis that follows it will be shown that none of the predicted electronic transitions is orbitally allowed, but they are observed through the agency of vibronic coupling. Furthermore the square planar geometry is in better agreement with the experimental spectrum than octahedral symmetry.
Because of the importance of vibronic coupling to electronic transitions in this example the determination of the symmetry of the vibrational degrees of freedom is essential. In the vibrational analysis the water ligands will be treated as composite entities. We will only consider the titanium water molecule vibrational modes and ignore the internal vibrations of the water molecule.
Ti(H2O)63+ - Octahedral Symmetry
$\begin{matrix} \begin{array} E & & E & C_3 & C_2 & C_4 & C_2" & i & S_{4} & S_{6} & \sigma_h & \sigma_d \end{array} & ~ \ \text{C}_{Oh} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & 1 & 1 \ 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 \ 3 & 0 & -1 & 1 & -1 & 3 & 1 & 0 & -1 & 0 \ 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 \ 2 & -1 & 0 & 0 & 2 & -2 & 0 & 1 & -2 & 0 \ 3 & 0 & -1 & 1 & -1 & -3 & -1 & 0 & 1 & 1 \ 3 & 0 & 1 & -1 & -1 & -3 & 1 & 0 & 1 & -1 \end{pmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2,~x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g: }xy,~xz,~yz \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} & \text{Oh} = \begin{pmatrix} 1 \ 8 \ 6 \ 6 \ 3 \ 1 \ 6 \ 8 \ 3 \ 6 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 7 \ 1 \ 1 \ 3 \ 3 \ 1 \ 1 \ 1 \ 5 \ 3 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{A}_{1g} = ( \text{C}_{Oh}^T )^{<1>} & \text{A}_{2g} = ( \text{C}_{Oh}^T )^{<2>} & \text{E}_{g} = ( \text{C}_{Oh}^T )^{<3>} & \text{T}_{1g} = ( \text{C}_{Oh}^T )^{<4>} & \text{T}_{2g} = (C_{Oh}^T)^{<5>} \ \text{A}_{1u} = ( \text{C}_{Oh}^T )^{<6>} & \text{A}_{2u} = ( \text{C}_{Oh}^T )^{<7>} & \text{E}_{u} = ( \text{C}_{Oh}^T )^{<8>} & \text{T}_{1u} = ( \text{C}_{Oh}^T )^{<9>} & \text{T}_{2u} = (C_{Oh}^T)^{<10>} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{tot}^T = \begin{pmatrix} 21 & 0 & -1 & 3 & -3 & -3 & -1 & 0 & 5 & 3 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - T_{1u} - T_{1g} \end{matrix} \nonumber$
Determine which irreducible representations contribute to Γvib.
$\begin{matrix} i = 1 .. 10 & \text{Vib}_i = \frac{ \sum \overrightarrow{ \left[ \text{Oh} (C_{Oh}^T )^{<i>} \Gamma_{vib} \right]}}{h} & \text{Vib} = \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \ 0 \ 0 \ 0 \ 2 \ 1 \end{pmatrix} \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2,~x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g: }xy,~xz,~yz \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} \end{matrix} \nonumber$
Thus we see that the vibrational modes have the following symmetry:
$\Gamma_{vib} = A_{1g} + E_g + T_{2g} + 2T_{1u} + T_{2u} \nonumber$
Inspection of the character table shows that in octahedral symmetry the degeneracy of the d-orbitals is split into a three-fold degenerate T2g (dxy, dxz, dyz) level and a two-fold degenerate Eg (dz2, dx2 - y2) level. Electrostatic arguments (see your general chemistry text) predict that the T2g level is lower in energy. Thus, according to this model Ti3+ has one d-electron in the T2g level. As the spectrum above shows, the complex absorbs in the visible region at 20,000 cm-1 (500 nm). However, the T2g ---> Eg transition is orbitally forbidden as is shown below.
$\begin{matrix} \int \Psi_{ex} \mu_e \Psi_{eg} d \tau_e = 0 & \frac{ \sum \overrightarrow{(Oh E_g T_{1u} T_{2g})}}{h} = 0 \end{matrix} \nonumber$
On the left we have the transition moment integral expressed in the language of integral calculus (see any quantum chemistry text) and on the right we have the same integral in the group-theoretical vector-matrix representation.
In calculating the transition moment for T2g ---> Eg electronic transition it has been assumed that there was no change in the vibrational state of the molecule. However, it is possible for formally forbidden electronic transitions to become allowed through coupling to changes in the vibrational state of the molecule. In other words pure electronic transitions do not actually occur, because the vibrational (and rotational) states of the molecule change at the same time. These are called vibronic transitions and they are allowed if the integral shown below is nonzero.
$\int \int \Psi_{ex} \Psi_{vx} \mu_e \Psi_{eg} \Psi_{vg} d \tau_e d \tau_v \nonumber$
The calculations below show that vibronic transitions involving the T1u and T2u vibrational modes are allowed because the transition moment is not zero. It is important to note that the ground states of all vibrational modes of all point groups have A1g symmetry. (See F. A. Cotton, Chemical Applications of Group Theory, J. Wiley, 1963, p. 262.)
$\begin{matrix} \frac{ \sum \overrightarrow{(Oh T_{1u} E_g T_{1u} T_{2g}A_{1g})}}{h} = 2 & \frac{ \sum \overrightarrow{(Oh T_{2u} E_g T_{1u} T_{2g} A_{1g})}}{h} = 2 \end{matrix} \nonumber$
The A1g, Eg, and T2g modes do not provide vibrational assistance as is shown below.
$\begin{matrix} \frac{ \sum \overrightarrow{(Oh A_{1g} E_g T_{1u} T_{2g} A_{1g})}}{h} = 0 & \frac{ \sum \overrightarrow{(Oh E_g E_g T_{1u} T_{2g} A_{1g})}}{h} = 0 & \frac{ \sum \overrightarrow{(Oh T_{2g} E_g T_{1u} T_{2g} A_{1g})}}{h} = 0 \end{matrix} \nonumber$
At this point we have shown that the vibrationally assisted T2g ---> Eg electronic transition is allowed. However, the shoulder on the experimental spectrum suggest that more than one electronic transition is occurring. In the next section we will see that a reduction to square planar symmetry under the Jahn-Teller effect leads to a d-orbital energy level diagram that is consistent with the experimental spectrum.
Ti(H2O)63+ - Square Planar Symmetry
The Jahn-Teller effect predicts a tetragonal distortion of the octahedral complex to the lower D4h square planar symmetry. The energy level diagram is shown above - essentially the ligands on the z-axis move in toward the titanium ion.
$\begin{matrix} ~ & \begin{array} E & C_4 & C_2 & C_2' & C_2" & i & S_4 & \sigma_h & \sigma_v & \sigma_d \end{array} \ CD4h = & \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 \ 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 \ 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 \ 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 \ 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 \ 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \ 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 \end{bmatrix} & \begin{array} A_{1g}:~x^2 + y^2,~z^2 \ A_{2g}:~Rz \ B_{1g}:~ x^2-y^2 \ B_{2g}:~ xy \ E_{g}:~ \text{(Rx, Ry), (xz, yz)} \ A_{1u}: \ A_{2u}:~z \ B_{1u} \ B_{2u} \ E_{u}:~ \text{x, y)} \end{array} & \text{D4h} = \begin{bmatrix} 1 \ 2 \ 1 \ 2 \ 2 \ 1 \ 2 \ 1 \ 2 \ 2 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 7 \ 3 \ 3 \ 3 \ 1 \ 1 \ 1 \ 5 \ 5 \ 3 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{A}_{1g} = ( \text{CD4h}^T )^{<1>} & \text{A}_{2g} = ( \text{CD4h}^T )^{<2>} & \text{B}_{1g} = ( \text{CD4h}^T )^{<3>} & \text{B}_{2g} = ( \text{CD4h}^T )^{<4>} & \text{E}_{g} = (\text{CD4h}^T)^{<5>} \ \text{A}_{1u} = ( \text{CD4h}^T )^{<6>} & \text{A}_{2u} = ( \text{CD4h}^T )^{<7>} & \text{B}_{1u} = ( \text{CD4h}^T )^{<8>} \text{B}_{2u} = ( \text{CD4h}^T )^{<9>} & \text{E}_{u} = ( \text{CD4h}^T)^{<10>} \end{matrix} \nonumber$
$\begin{matrix} h = \sum \text{D4h} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} (A2u + Eu))} & \Gamma_{tot}^T = \begin{pmatrix} 21 & 3 & -3 & -3 & -1 & -3 & -1 & 5 & 5 & 3 \end{pmatrix} \end{matrix} \nonumber$
Symmetry of the vibrational modes.
$\Gamma_{vib} = \Gamma_{tot} - A_{2g} - E_g - A_{2u} - E_u \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{[D4h (CD4h^T )^{<i>} \Gamma_{vib}]}}{h} \ \text{Vib} = \begin{pmatrix} 2 \ 0 \ 1 \ 1 \ 1 \ 0 \ 2 \ 0 \ 1 \ 3 \end{pmatrix} & \begin{array} A_{1g}:~x^2 + y^2,~z^2 \ A_{2g}:~Rz \ B_{1g}:~ x^2-y^2 \ B_{2g}:~ xy \ E_{g}:~ \text{(Rx, Ry), (xz, yz)} \ A_{1u}: \ A_{2u}:~z \ B_{1u} \ B_{2u} \ E_{u}:~ \text{x, y)} \end{array} \end{matrix} \nonumber$
$\Gamma_{vib} = 2A_{1g} + B_{1g} + B_{2g} + E_g + 2A_{2u} B_{2u} + 3E_u \nonumber$
The D4h energy level diagram shows three possible electronic transitions, one IR transition and two transitions in visible region of the spectrum. The following calculations show that all three are formally forbidden. Again, vibronic coupling is invoked to explain the appearance of the two electronic transitions in the visible region.
$\begin{matrix} \frac{ \sum \overrightarrow{(D4h~E_g(A_{2u} + E_u)B_{2g}))}}{h} = 0 & \frac{ \sum \overrightarrow{(D4h~B_{1g}(A_{2u} + E_u)B_{2g}))}}{h} = 0 & \frac{ \sum \overrightarrow{(D4h~A_{1g}(A_{2u} + E_u)B_{2g}))}}{h} = 0 \end{matrix} \nonumber$
An A1u or Eu vibrational mode can provide vibronic coupling for the B2g --> B1g transition. Again it is important to recall that the ground states of all vibrational modes have A1g symmetry.
$\begin{matrix} \frac{ \sum \overrightarrow{(D4h~A_{1u}~B_{1g}(A_{2u} + E_u)B_{2g}~A_{1g}))}}{h} = 1 & \frac{ \sum \overrightarrow{(D4h~E_u~B_{1g}(A_{2u} + E_u)B_{2g}~A_{1g}))}}{h} = 1 \end{matrix} \nonumber$
A B1u or Eu vibrational mode can provide vibronic coupling for the B2g --> A1g transition.
$\begin{matrix} \frac{ \sum \overrightarrow{(D4h~B_{1u}~A_{1g}(A_{2u} + E_u)B_{2g}~A_{1g}))}}{h} = 1 & \frac{ \sum \overrightarrow{(D4h~E_u~A_{1g}(A_{2u} + E_u)B_{2g}~A_{1g}))}}{h} = 1 \end{matrix} \nonumber$
A close examination of the experimental spectrum indicates the presence of two electronic transitions of similar energy (shoulder). So the energy level diagram and the vibronic analysis are consistent with the actual spectrum.
Primary reference: Daniel C. Harris and Michael D. Bertolucci, "Symmetry and Spectroscopy: An Introduction to Vibrational and Electronic Spectroscopy." Dover Publications, Inc., New York, 1989.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.16%3A_Analysis_of_the_Electronic_Spectrum_of_Ti%28H2O%293.txt
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In spite of Max Born's warning, we are going to present a simple model for a quantum jump that is based on visualization.
This tutorial deals with the quantum mechanical particle (quon to use Nick Herbert's term) on a ring. Previously the behavior of the π-electrons of benzene were analyzed using this model. Links to the tutorials are provided below.
In the absence of perturbations, such as oscillating electromagnetic fields, quantum systems are in the stationary states obtained from the solution of the time-independent Schrödinger equation. When a quantum system is subject to a perturbation such as an oscillating electromagnetic field it is forced into a time-dependent superposition of its stationary states.
$\Psi (q,~t) = \sum_n \left( \psi_n (q) \text{exp} \left( -i \frac{2 \pi E_n t}{h} \right) \right) \nonumber$
For the system to absorb energy from the electromagnetic field and undergo a transition from one allowed energy state to another two conditions must be met. First, the photons of the electromagnetic field must satisfy the Bohr frequency condition by having an energy which equals the energy difference between the two states under consideration. Second, there must be a dipolar coupling between the electromagnetic field and the oscillating electron density of the system. This later criterion is, of course, the selection rule for the transition.
According to this model, when an electron on a ring is subjected to a electromagnetic perturbation, the electron responds by going into a state which is a linear superposition of the unperturbed stationary states. Assuming the Bohr frequency condition is satisfied for two states, if the square modulus, |Ψ(x,t)|2, of the time-dependent wavefunction associated with this linear superposition exhibits an asymmetric fluctuating charge density (an oscillating dipole moment) on the ring, a coupling with the dipolar character of the electromagnetic field exists and a transition can occur. If the linear combination of states leads to a symmetric fluctuating charge density there is no coupling between the field and the electron density in the box and a transition is not possible.
The model is illustrated below for several possible transitions for an electron on a ring and the selection rule determined. Previously this model was used for the particle in a box and the harmonic oscillator. Links to these tutorials are provided below.
http://www.users.csbsju.edu/~frioux/...mp/jumpsho.htm
http://www.users.csbsju.edu/~frioux/q-jump/p-jump.htm
Set up for graphics:
$\begin{matrix} \text{numpts} = 100 & i = 0 .. \text{numpts} & j = 0 .. \text{numpts} & \phi_i = \frac{2 \pi i}{ \text{numpts}} \ x_{i,~j} = \cos ( \phi_i ) & y_{i,~j} = \sin ( \phi_i ) & z_{i,~j} = 0 & t = \frac{ \text{FRAME}}{3} \end{matrix} \nonumber$
Form the time-dependent superposition of initial and final states:
$\begin{matrix} \text{Initial state:} & n_i = 0 & E_i = 2 n_i^2 \pi^2 & \text{Final state:} & n_f = 1 & E_f = 2 n_f^2 \pi^2 \end{matrix} \nonumber$
$\Psi_{i,~j} = \frac{1}{2 \sqrt{ \pi}} \left( \left| \text{exp} (i n_1 \phi_i ) \text{exp} (-i E_i t) + \text{exp} ( i n_f \phi_i ) \text{exp} (-i E_ft ) \right| \right)^2 \nonumber$
Select Animation and then Record from the Tools menu and follow instructions. Choose from 0 to 30 at 3 frames/sec for a reasonable animated display.
Before running the animation note that this superposition shows dipole character in that the electron density is a minimum at the opposite side of the ring at which it is a maximum. This dipole rotates around the ring during the animation, providing a mechanism for coupling with the electromagnetic field illuminating the particle on a ring, if the Bohr frequency condition is satisfied. Thus the transition from the n = 0 state to the n = 1 state is allowed. It is easy to show in this way that any transition for which ∆n = +/- 1 is allowed.
Now we look at a transition for which ∆n = 2. It is clear from the symmetry of the superposition wave function that the electron density distribution does not have dipole character, and, therefore, there is no mechanism for coupling with the electromagnetic field. It is easy to show that the same is true for ∆n = +/- 3, 4, etc.
$\begin{matrix} \text{Initial state:} & n_i = 1 & E_i = 2n_i^2 \pi^2 & \text{Final state:} & n_f = 3 & E_f = 2n_f^2 \pi^2 \end{matrix} \nonumber$
$\Psi_{i,~j} = \frac{1}{2 \sqrt{ \pi}} \left( \left| \text{exp} (in_i \phi_i ) \text{exp} (-i E_i t) + \text{exp}(i n_f \phi_i ) \text{exp} (-i E_f t) \right| \right)^2 \nonumber$
It is easy to show that while n = 1 to 2 and n = -1 to -2 transitions are allowed, the n = 1 to -2 and n = -1 to 2 transitions are forbidden, even though all of these transitions involve the same energy change.
Acknowledgement
The preparation of this tutorial and others on my page dealing with quantum jumps was stimulated by reading "Fluctuating Electric Dipoles and the Absorption of Light," by David R. McMillin [J. Chem. Ed. 55, 7 (1978)] .
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This tutorial deals with the interpretation of the vibrational and electronic spectra of benzene using group theory.
Analysis of the Vibrational Spectrum
The vibrational analysis uses the method of unmoved atoms (uma). The number of unmoved atoms for each symmetry operation is stored in a vector, Γuma. This is converted into a reducible representation (Γvib) for the vibrational degrees of freedom in several steps as is shown below. Next the irreducible representations that contribute to Γvib is determined.
$\begin{matrix} D6h = \begin{pmatrix} 1 & 2 & 2 & 1 & 3 & 3 & 1 & 2 & 2 & 1 & 3 & 3 \end{pmatrix} & D6h = D6h^T & h = \sum D6h & h =24 \end{matrix} \nonumber$
$\begin{matrix} \begin{array} E & & E & C_6 & C_3 & C_2 & C_2' & C_2" & i & S_{3} & S_{6} & \sigma_h & \sigma_d & \sigma_v \end{array} & ~ \ \text{CD6h} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & 1 & -1 & -1 \ 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \ 1 & -1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & -1 & -1 & 1 \ 2 & 1 & -1 & -2 & 0 & 0 & 2 & 1 & -1 & -2 & 0 & 0 \ 2 & -1 & -1 & 2 & 0 & 0 & 2 & -1 & -1 & 2 & 0 & 0 \ 1 & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 \ 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 \ 1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & 1 \ 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 \ 2 & 1 & -1 & -2 & 0 & -0 & -2 & -1 & 1 & 2 & 0 & 0 \ 2 & -1 & -1 & 2 & 0 & 0 & -2 & 1 & 1 & -2 & 0 & 0 \end{pmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g: Rz} \ \text{B1g} \ \text{B2g} \ \text{E1g: (Rx, Ry), (xz, yz)} \ \text{E2g: }(x^2 - y^2,~xy) \ \text{A1u:} \ \text{A2u: z} \ \text{B1u:} \ \text{B2u:} \ \text{E1u: (x, y)} \ \text{E2u:} \end{array} & \Gamma_{uma} = \begin{pmatrix} 12 \ 0 \ 0 \ 0 \ 4 \ 0 \ 0 \ 0 \ 0 \ 12 \ 0 \ 4 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{trans} = (CD6h^T)^{<g>} + (CD6h^T)^{<11>} & \Gamma_{rot} = (CD6h^T)^{<2>} + (CD6h^T)^{<5>} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} \Gamma_{trans})} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{vib} = \Gamma_{rot} - \Gamma_{trans} - \Gamma_{rot} & i = 1 .. 12 & \text{Vib}_i = \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T )^{ <i>} \Gamma_{vib} \right] }}{h} \end{matrix} \nonumber$
$\text{Vib}^T = \begin{pmatrix} 2 & 1 & 0 & 2 & 1 & 4 & 0 & 1 & 2 & 2 & 3 & 2 \end{pmatrix} \nonumber$
$\Gamma_{vib} = 2A_{1g} + A_{2g} + 2B_{2g} + E_{1g} + 4E_{2g} + A_{2u} + 2B_{1u} + 1B_{2u} + 3E_{1u} + 2E_{2u} \nonumber$
The symmetry of the vibrational modes and their IR and Raman activity are given below:
IR active: A2u and 3E1u
Raman active: 2A1g, E1g, and 4E2g
IR active modes are observed at 675, 1035, 1479, and 3036 cm-1, which is consistent with the above analysis. The Raman spectrum is not as clearly resolved.
Analysis of the Electronic Spectrum
The electronic spectrum to be analyzed (see below) is due to transitions involving benzene's π electrons. The symmetry of the relevant π−electron molecular orbitals is determined by examining how the π orbitals transform under the symmetry operations of the D6h group.
$\begin{matrix} \Gamma_{ \pi} = \begin{pmatrix} 6 & 0 & 0 & -2 & 0 & 0 & 0 & -6 & 0 & 2 \end{pmatrix} & \Pi_i = \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{<i>} \tau_{ \pi}^T \right]}}{h} \end{matrix} \nonumber$
$\begin{matrix} \Pi = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g: Rz} \ \text{B1g} \ \text{B2g} \ \text{E1g: (Rx, Ry), (xz, yz)} \ \text{E2g: }(x^2 - y^2,~xy) \ \text{A1u:} \ \text{A2u: z} \ \text{B1u:} \ \text{B2u:} \ \text{E1u: (x, y)} \ \text{E2u:} \end{array} \end{matrix} \nonumber$
The symmetry of the π-molecular orbitals is Γπ = B2g + E1g + A2u + E2u. The order of the levels from a Huckel calculation is as shown above: A2u, E1g, E2u, and B2g. The ground electronic state is A2u(2), E1g(4) and has A1g symmetry because the A2u and E1g orbitals are full. The first electronic excited state is A2u(2), E1g(3), E2u(1). This has the symmetry properties of E1g(1)E2u(1) which gives rise to the manifold of states: B1u, B2u, and E1u as is shown below.
$\begin{matrix} X_i = \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{<i>} (CD6h^T)^{<5>} (CD6h^T)^{<12>} \right]}}{h} & X^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
The electronic energy level diagram consistent with this analysis is shown below. Note that the excited state splits into a set of singlet and triplet exited states. Transitions from the singlet ground state to the triplet excited states are formally forbidden.
For an electronic transition to be allowed the transition moment integral must be greater than zero.
$\int \Psi_{ex} \mu_e \Psi_{eg} d \tau_e > 0 \nonumber$
Only the A1g ---> E1u transition is orbitally allowed as is shown below.
$\begin{matrix} A_{1g} \rightarrow B_{1u} & \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{<9>} \left[ (CD6h^T)^{<g>} + (CD6h^T)^{<11>} \right] (CD6h^T )^{<1>} \right]}}{h} = 0 \ A_{1g} \rightarrow B_{2u} & \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{<10>} \left[ (CD6h^T)^{<g>} + (CD6h^T)^{<11>} \right] (CD6h^T )^{<1>} \right]}}{h} = 0 \ A_{1g} \rightarrow E_{1u} & \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{<11>} \left[ (CD6h^T)^{<g>} + (CD6h^T)^{<11>} \right] (CD6h^T )^{<1>} \right]}}{h} = 0 \end{matrix} \nonumber$
Electronic transitions that are orbitally forbidden can occur if they are properly coupled to vibrational transitions. This occurs when the following integral is non-zero. These are called vibronic or vibrationally assisted electronic transitions.
$\int \int \Psi_{ex} \Psi_{vx} \mu_e \Psi_{eg} \Psi_{vg} d \tau_e d \tau_v \nonumber$
The orbitally forbidden A1g --> B1u is vibronically assisted by B2g or E2g vibrations.
$X_i = \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{9} (CD6h^T)^{i} \left[ (CD6h^T)^{g} + (CD6h^T)^{11} \right] (CD6h^T)^{1} (CD6h^T)^{1} \right]}}{h} \ X^T = \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
The orbitally forbidden A1g --> B2u is vibronically assisted by B1g or E2g vibrations. However there is no B1g vibration.
$X_i = \frac{ \sum \overrightarrow{ \left[ D6h (CD6h^T)^{10} (CD6h^T)^{i} \left[ (CD6h^T)^{g} + (CD6h^T)^{11} \right] (CD6h^T)^{1} (CD6h^T)^{1} \right]}}{h} \ X^T = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \nonumber$
The fully allowed A1g ---> E1u transition is assigned to the most intense transition which occurs at 180 nm. The vibronically assisted A1g --> B1u and A1g --> B2u transitions are assigned to the less intense bands at 200 and 260 nm, respectively. The spin-forbidden 1 A1g --> 3 B1u is assigned to the lowest energy and lowest intensity transition at 340 nm.
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In this computer/spectroscopy lab the variational method is used to determine the nuclear spin states and allowed transitions in a three proton spin system. The theoretical procedure is essentially the same as that used in the other applications of the variational method that we have studied (LCAO-MO for H2+, Roothaan SCF, and various examples presented in class).
To prepare for this lab study the section on NMR in your textbook and review your class notes on the quantum mechanical treatment of the NMR spectroscopy of one- and two-proton systems. This lab will compare the low field NMR spectra of vinyl acetate and acrylonitrile. The high field (300 MHz) NMR spectrum of acrylonitrile will be measured and analyzed, and compared to the 60 MHz spectra of vinyl acetate and acrylonitrile.
The major steps in this exercise are listed below. Further detail is provided following the list. As you examine this sequence of steps you will see the similarities with previous variational calculations that we have looked at in class and lab.
1. Choose a trial nuclear spin wavefunction.
2. Write the energy operator for the three spin system which includes all the pertinent interactions.
3. Input parameters (in this case chemical shifts and coupling constants) and evaluate the matrix elements of the secular determinant for the system.
4. Solve the secular determinant for the energy eigenvalues and eigenfunctions.
5. Using the appropriate selection rule, DIz = 1, construct a transition probability matrix.
6. Using the results from parts 4 and 5 above, calculate the allowed transitions and their intensities and then generate a model spectrum.
7. Compare the theoretical generated spectrum with the experimentally determined NMR spectrum.
Fortunately for us, the computer does most of the work and certainly all the difficult math in this exercise.
1. A molecule with three protons can be in any of eight possible spin states. To determine these states we choose as our trial wavefunction the following linear combination. $| \Psi_{spin} \rangle = c_1 | \alpha \alpha \alpha \rangle + c_2 | \alpha \alpha \beta \rangle + c_3 | \alpha \beta \alpha \rangle + c_4 | \beta \alpha \alpha \rangle + c_5 | \alpha \beta \beta \rangle + c_6 | \beta \alpha \beta \rangle + c_7 | \beta \beta \alpha \rangle + c_8 | \beta \beta \beta \rangle \nonumber$ In this particular notation, |aaa> stands for all three nuclear spins in the spin-up (Iz = +�) spin state, while |aba> indicates that nuclei 1 and 3 are in the spin-up state and nucleus 2 is in the spin-down state (Iz = -�). In the absence of an external magnetic field and any interaction between the protons, all eight states have the same energy.
2. The complete energy operator is the sum of equations (3) and (4). There is a term like (3) for each proton and there is a term like (4) for each distinct pair of protons. For a three proton system, ABC, we have, $\hat{H}_{mag} = -g_n \beta_n \hat{B}_z \left[ (1- \sigma_a ) \hat{I}_z (a) + (1- \sigma_b ) \hat{I})_z (b) + (1 - \sigma_c ) \hat{I}_z (c) \right] + \hat{J}_{ab} \hat{I}_a \hat{I}_b + J_{ac} \hat{I}_a \hat{I}_c + J_{bc} \hat{I}_b \hat{I}_c \nonumber$
3. We are now ready to apply the variational method as outlined earlier in your text. The variational integral for this application is $E = \langle \Psi_{spin} | \hat{H}_{mag} | \Psi_{spin} \rangle \nonumber$
Treatment of this integral by the usual techniques yields an 8 x 8 secular determinant of the form:
$\begin{vmatrix} H_{11} - E & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & H_{22} - E & \frac{J_{bc}}{2} & \frac{J_{ac}}{2} & 0 & 0 & 0 & 0 \ 0 & \frac{J_{bc}}{2} & H_{33} - E & \frac{J_{ab}}{b} & 0 & 0 & 0 & 0 \ 0 & \frac{J_{ac}}{2} & \frac{J_{ab}}{2} & H_{44} - E & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & H_{55} - E & \frac{J_{ab}}{2} & \frac{J_{ac}}{2} & 0 \ 0 & 0 & 0 & 0 & \frac{J_{ab}}{2} & H_{66} - E & \frac{J_{bc}}{2} & 0 \ 0 & 0 & 0 & 0 & \frac{J_{ac}}{2} & \frac{J_{bc}}{2} & H_{ \eta} - E & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & H_{88} - E \end{vmatrix} = 0 \nonumber$
where the diagonal matrix elements are, for example, of the form
$\begin{matrix} H_{11} = \frac{1}{2} \left[ v_a + v_b + v_c \right] + \frac{1}{4} \left[ J_{ab} + J_{ac} + J_{bc} \right] \ H_{22} = \frac{1}{2} \left[ v_a + v_b - v_c \right] + \frac{1}{4} \left[ J_{ab} - J_{ac} - J_{bc} \right] \ \vdots \ H_{88} = \frac{1}{2} \left[ -v_a - v_b - v_c \right] + \frac{1}{4} \left[ J_{ab} + J_{ac} + J_{bc} \right] \end{matrix} \nonumber$
Here the ni represent the terms gn bn (1 - si) Bz which are the resonant frequencies of the protons in their respective chemical environments. The various matrix elements can be evaluated once the chemical shifts and coupling constants are supplied.
4. The Mathcad document which is appended solves the secular determinant (7) after chemical shifts and coupling constants are provided. It yields the nuclear spin energy levels, their associated wavefunctions, the allowed transitions between these levels, the frequencies at which the transitions occur and their intensities. It also generates a model spectrum. Study the Mathcad document to make certain you understand each of the steps in this quantum mechanical calculation.
5. The calculation of the allowed transitions deserves further comment. Because the three proton system has eight spin states, there are 28 potential transitions amongst these energy levels. How many transitions are actually observed and at what intensities is determined by the selection rule for NMR spectroscopy. This selection rule states that only one spin can be flipped in any given transition. In other words D Iz = 1.
A transition is allowed if the product of the three matrices shown below is non-zero. The intensity of the transition is equal to the square of the product of these matrices.
$\begin{pmatrix} c_1 & c_2 & c_3 & c_4 & c_5 & c_6 & c_7 & c_8 \end{pmatrix} \begin{pmatrix} 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \ 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \ 1& 0 & 0 & 0 & 0 & 1 & 1 & 0 \ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \end{pmatrix} \begin{pmatrix} c_1 \ c_2 \ c'_3 \ c_4 \ c_5 \ c_6 \ c_7 \ c`_8 \end{pmatrix} \nonumber$
For example this matrix determines whether the transition from state 1 to state 2 is allowed. The matrix on the left contains the coefficients of the wavefunction of state 1 written in row-matrix form, while the matrix on the right contains the coefficients of the wavefunction of state 2 written in column-matrix form. The center matrix is the transition matrix and is an 8x8 square matrix which summarizes the allowed transitions. For example, the (1,1) matrix element represents the |aaa> ---> |aaa> transition and is 0 because it violates the selection rule (no spins are flipped). However, the (1,2) matrix element is 1 because the transition |aaa> ---> |aab> is allowed (only one spin is flipped). Following the rules of matrix multiplication, Mathcad examines each possible transition. If its intensity is non-zero it makes a record of it and later displays the transition in tabular and graphical form. For example, the matrix product shown below indicates that the |aaa> ---> |bbb> transition is forbidden.
Experimental:
The 300 MHz NMR spectrum of acrylonitrile will be measured and compared to its 60 MHz NMR and also to the 60 MHz spectrum of vinyl acetate.
The methyl protons are not coupled to the vinyl protons so it is possible to treat the vinyl protons as an ABC system. The low field (60 MHz) NMR spectra of acrylonitrile and vinyl acetate are shown below.
Interpretation of the Spectrum
Acrylonitrile is a true ABC system while the vinyl protons of vinyl acetate behave like an ABC system because they are not coupled to the methyl protons. Vinyl acetate presents a classic ABC spectrum. Each proton has a distinct chemical shift and each proton's resonance is split by the two other protons. Thus what is observed is a spectrum which consists of three separate resonances each of which is a quartet of peaks (a doublet of doublets). The chemical shifts are obtained by locating the center of the doublet of doublets. The determination of the coupling constants is described below.
In the absence of any coupling the resonance of an individual proton, A, would appear as a singlet.
The presence of two non-equivalent protons, X and Y , splits A's singlet into a quartet of peaks as shown in the diagram below. This occurs because there are four possible orientations of the non-equivalent protons (up-up, up-down, down-up, down-down) and, therefore, both the ground state and excited state of proton A are split into four levels.
Remember that it is proton A that is undergoing the resonance and that the spins of X and Y don't flip during A's resonance. That is why there are only four transitions amongst these eight states. Note that the difference between transition 1 and 2 is the spin state of proton Y. Thus the frequency difference between these peaks is the coupling constant Jay expressed in hertz. Similarly, the difference between 3 and 4 is Jay, the difference between 1 and 3 is Jax, the difference between 2 and 4 is Jax, the difference between 1 and 4 is Jax + Jay, and the difference between 2 and 3 is Jax - Jay. Similar arguments would be used to discuss the resonances of protons B and C.
To obtain the three coupling constants for vinyl acetate use the following procedure. Label the doublet of doublets farthest down field "A", the middle one "B" and the last one "C". The analysis described above provides you with two direct measurements of two coupling constants. For example, Jab and Jac can be obtained from the analysis of the "A" resonance but you don't know which proton is "B" and which one is "C" at this point. However, the analysis of the "B" resonance provides determinations of Jab and Jbc. Comparison of the results for "A" and "B" enable you to assign values to Jab, Jac, and Jbc. Use the average of the four values for Jab, but hold off on Jac and Jbc until you have analyzed the resonance of proton "C". This proceedure provides you with four determinations of Jac and Jbc. Use the average values for each of these parameters.
Assignment
1. First compare the 60 MHz NMR spectra of acrylonitrile and vinyl acetate shown below. Notice the difference in appearance of the spectra in spite of the fact that both spectra show the resonances of three vinyl protons.
2. Use the methods outlined above to obtain chemical shifts and coupling constants from the vinyl acetate NMR spectrum.
3. With assistance, obtain the 300 MHz spectrum of acrylonitile. Note its basic similarity to the low field vinyl acetate spectrum. Explain why increasing the magnetic field simplifies the acrylonitrile spectrum. Analyze the spectrum to obtain the chemical shifts and coupling constants.
4. Run the Mathcad program supplied for the low field NMR spectrum of vinyl acetate.
1. Use the eigenvalues and eigenvectors generated by Mathcad to construct an energy level diagram for the eight allowed spin states. Use the format indicated below for each energy level.
E8 ................................................................ Y8
E7 ................................................................ Y7
E6 ................................................................ Y6
E5 ................................................................ Y5
E4 ................................................................ Y4
E3 ................................................................ Y3
E2 ................................................................ Y2
E1 ................................................................ Y1
2. Now use the intensity matrix generated by Mathcad to sketch in the allowed transitions on the diagram above.
3. Also prepare a table which compares the experimental transition frequencies with the computer generated frequencies given in the frequency matrix. If you do not find a favorable comparison it means either that the theoretical model used in this exercise does not adequately represent the vinyl acetate nuclear spin system or that you have misread your spectrum.
4. Qualitatively compare the appearance of the experimental spectrum with the model spectrum generated by Mathcad. In other words do you find satisfactory agreement for the relative values of the frequencies and intensities for each resonance. Explain.
5. Run the Mathcad program supplied for the high field NMR spectrum of acrylonitrile.
1. Use the eigenvalues and eigenvectors generated by Mathcad to construct an energy level diagram for the eight allowed spin states. Use the format given in 4.1 for each energy level.
2. Now use the intensity matrix generated by Mathcad to sketch in the allowed transitions on the diagram from part 4.1 above.
3. Also prepare a table which compares the experimental transition frequencies with the computer generated frequencies given in the frequency matrix. If you do not find a favorable comparison it means either that the theoretical model used in this exercise does not adequately represent the acrylonitrile nuclear spin system or that you have misread your spectrum.
4. Qualitatively compare the appearance of the experimental spectrum with the model spectrum generated by Mathcad. In other words do you find satisfactory agreement for the relative values of the frequencies and intensities for each resonance. Explain.
6. Run the Mathcad program to simulate the low field (60 MHz) spectrum of acrylonitrile. Do this by dividing the high field chemical shifts by 5 (60 = 300/5). Compare the Mathcad model spectrum with the experimental 60 MHz spectrum provided.
1. Use the eigenvalues and eigenvectors generated by Mathcad to construct an energy level diagram for the eight allowed spin states. Use the format given in 4.1 for each energy level.
2. Now use the intensity matrix generated by Mathcad to sketch in the allowed transitions on the diagram from part 4.1 above.
3. Qualitatively compare the appearance of the experimental spectrum with the model spectrum generated by Mathcad. In other words do you find satisfactory agreement for the relative values of the frequencies and intensities for each resonance. Explain.
4. Before the existence of high field nmr an analysis of the 60 MHz spectrum of acrylonitrile yielded the following chemical shifts and coupling constants in Hz: sa = 372.2; sb = 364.4; sc= 342.0; Jab = 0.91; Jac = 17.90; Jbc = 11.75. Compare your values for the chemical shifts and coupling constants with these values.
7. The low-field acrylonitrile spectrom shows that 14 resonances are observed [a 15th resonance (see intensity matrix) is too weak to be seen experimentally]. The high-field acrylonitrile and low-field vinyl acetate spectra show 12 resonances. Examine the nuclear spin wave functions for all three cases and use the superposition principle to explain these differences.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.19%3A_NMR_-_Quantum_Mechanics_of_a_Three_Proton_System.txt
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The purpose of this tutorial is to deviate from the traditional matrix mechanics approach to the AB proton nmr system in order to illustrate a related method of analysis which uses tensor algebra.
The nuclear magnetic energy operator for the AB system is given below.
$\begin{matrix} \hat{H}_{mag} = - \nu_A \hat{I}_z^A - \nu_B \hat{I}_z^B + J_{AB} \hat{I}^A \hat{I}^B & \text{where} & \nu_A = g_n \beta_n B_n (1- \sigma_A ) \hat{I}_z^A & \nu_B = g_n \beta_n B_z (1- \sigma_B ) \hat{I}_z^B \ ~ & \text{and} & \hat{I}^A \hat{I}^B = \hat{I}_x^A \hat{I}_x^B + \hat{I}_y^A \hat{I}_y^B + \hat{I}_z^A \hat{I}_z^B \end{matrix} \nonumber$
The customary matrix mechanics analysis requires the following mathematical structures.
• Nuclear spin wave function:
$| \Psi_{mag} \rangle = c_1 | \alpha_A \alpha_B \rangle + c_2 | \alpha_A \beta_B \rangle + c_3 | \beta_A \alpha_B \rangle + c_4 | \beta_A \beta_B \rangle = c_1 | \alpha_A \rangle | \alpha_B \rangle + c_2 | \alpha_A \rangle | \beta_B \rangle + c_3 | \beta_A \rangle | \alpha_B \rangle + c_4 | \beta_A \rangle | \beta_B \rangle \nonumber$
• Spin operators (using atomic units, h = 2 π) in the x-, y- and z-directions, plus the identity operator (needed later):
$\begin{matrix} I_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & I_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
• Spin eigenfunctions in the x-, y- and z-directions:
$\begin{matrix} \alpha_a = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \beta_z = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \alpha_x = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ 1 \end{pmatrix} & \beta_x = \frac{1}{ \sqrt{2}} \begin{pmatrix} -1 \ 1 \end{pmatrix} & \alpha_y = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ i \end{pmatrix} & \beta_y = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 \ -i \end{pmatrix} \end{matrix} \nonumber$
Note that while the spin wave function is a linear combination of four spin states, the operators are 2x2 matrices. This requires that special care is taken to insure that the operator is operating on the correct spin wave function component. This can be avoided by writing both the spin states and magnetic operator in tensor format.
First tensor vector multiplication is used to construct the nuclear spin states.
$\begin{matrix} | \alpha \alpha \rangle = | \alpha \rangle | \alpha \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \begin{pmatrix} 1 \ 0 \end{pmatrix} \ 0 \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & | \alpha \beta \rangle = | \alpha \rangle | \beta \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 1 \begin{pmatrix} 1 \ 0 \end{pmatrix} \ 0 \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} \ | \beta \alpha \rangle = | \beta \rangle | \alpha \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \begin{pmatrix} 1 \ 0 \end{pmatrix} \ 1 \begin{pmatrix} 1 \ 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & | \beta \beta \rangle = | \beta \rangle | \beta \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \begin{pmatrix} 0 \ 1 \end{pmatrix} \ 1 \begin{pmatrix} 0 \ 1 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
We now write these spin states in the Mathcad code.
$\begin{matrix} \alpha \alpha = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \end{pmatrix} & \alpha \beta = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix} & \beta \alpha = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix} & \beta \beta = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
Next, tensor matrix multiplication is used to construct the magnetic energy operator.
$\begin{matrix} - \nu_A \hat{I}_z^A \rightarrow - \nu_A \hat{I}_z^A \otimes \hat{I} = \frac{- \nu_A}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \frac{- \nu_A}{2} \begin{pmatrix} 1 \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & 0 \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \ 0 \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} & -1 \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{pmatrix} = \frac{- \nu_A}{2} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix} \ - \nu_B \hat{I}_z^B \rightarrow - \nu_B \hat{I}_z^B \otimes \hat{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \otimes \frac{- \nu_B}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} = \frac{- \nu_B}{2} \begin{pmatrix} 1 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & 0 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \ 0 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & 1 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{pmatrix} = \frac{- \nu_B}{2} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix} \ J_{AB} \hat{I}_x^A \otimes \hat{I}_x^B = \frac{J_{AB}}{4} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} = \frac{J_{AB}}{4} \begin{pmatrix} 0 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & 1 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \ 1 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & 0 \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \end{pmatrix} = \frac{J_{AB}}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & -1 \end{pmatrix} \ J_{AB} \hat{I}_y^A \otimes \hat{I}_y^B = \frac{J_{AB}}{4} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} = \frac{J_{AB}}{4} \begin{pmatrix} 0 \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & -i \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \ i \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & 0 \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} \end{pmatrix} = \frac{J_{AB}}{4} \begin{pmatrix} 0 & 0 & 0 & -1 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ -1 & 0 & 0 & 0 \end{pmatrix} \ J_{AB} \hat{I}_z^A \otimes \hat{I}_z^B = \frac{J_{AB}}{4} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} = \frac{J_{AB}}{4} \begin{pmatrix} 1 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & 0 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \ 0 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & -1 \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} \end{pmatrix} = \frac{J_{AB}}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \ J_{AB} \left( \right) = J_{AB} \hat{I}^A \otimes \hat{I}^B = \frac{J_{AB}}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 2 & 0 \ 0 & 2 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The magnetic Hamiltonian can now be written in Mathcad code.
$\begin{matrix} \text{IzAIzB} = \frac{-1}{2} \begin{pmatrix} \textcolor{red}{ \nu_A} + \nu_B & 0 & 0 & 0 \ 0 & \nu_A - \nu_B & 0 & 0 \ 0 & 0 & - \nu_A + \nu_B & 0 \ 0 & 0 & 0 - \nu_A - \nu_B \end{pmatrix} & \text{IAIB} = \frac{ \textcolor{red}{J_{AB}}}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & -1 & 2 & 0 \ 0 & 2 & -1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} & H_{mag} = \textcolor{red}{IzAIzIzB} + \text{IAIB} \end{matrix} \nonumber$
$H_{mag} \rightarrow \begin{pmatrix} \frac{-1}{2} \nu_A - \frac{1}{2} \nu_B + \frac{1}{4} J_{AB} & 0 & 0 & 0 \ 0 & \frac{-1}{2} \nu_A + \frac{1}{2} \nu_B - \frac{1}{4} J_{AB} & \frac{1}{2} J_{AB} & 0 \ 0 & \frac{1}{2} J_{AB} & \frac{1}{2} \nu_A - \frac{1}{2} \nu_B - \frac{1}{4} J_{AB} & 0 \ 0 & 0 & 0 & \frac{1}{2} \nu_A + \frac{1}{2} \nu_B + \frac{1}{4} J_{AB} \end{pmatrix} \nonumber$
The eigenvaues of the magnetic energy matrix can now be calculated.
$\text{eigenvals}(H_{mag}) \rightarrow \begin{bmatrix} \frac{-1}{2} \nu_A - \frac{1}{2} \nu_B + \frac{1}{4} J_{AB} \ \frac{1}{2} \nu_A + \frac{1}{2} \nu_B + \frac{1}{4} J_{AB} \ \frac{-1}{4} J_{AB} + \frac{1}{2} \left( J_{AB}^2 + \nu_A^2 - 2 \nu_A \nu_B + \nu_B^2 \right)^{ \frac{1}{2}} \ \frac{-1}{4} J_{AB} - \frac{1}{2} \left( J_{AB}^2 + \nu_A^2 - 2 \nu_A \nu_B + \nu_B^2 \right)^{ \frac{1}{2}} \end{bmatrix} \nonumber$
We can also calculate the magnetic energy matrix elements individually, as follows.
$\begin{matrix} \alpha \alpha^T H_{mag} \alpha \alpha \rightarrow \frac{-1}{2} \nu_A - \frac{1}{2} \nu_B + \frac{1}{4} J_{AB} & \alpha \beta H_{mag} \alpha \alpha \rightarrow 0 & \beta \alpha^T H_{mag} \alpha \alpha \rightarrow 0 & \beta \beta^T H_{mag} \alpha \alpha \rightarrow 0 \ \alpha \beta^T H_{mag} \alpha \beta \rightarrow \frac{-1}{2} \nu_A + \frac{1}{2} \nu_B - \frac{1}{4} J_{AB} & \beta \alpha H_{mag} \alpha \beta \rightarrow \frac{1}{2} J_{AB} & \beta \beta^T H_{mag} \alpha \beta \rightarrow 0 \ \beta \alpha^T H_{mag} \beta \alpha \rightarrow \frac{1}{2} \nu_A - \frac{1}{2} \nu_B - \frac{1}{4} J_{AB} & \alpha \beta H_{mag} \beta \alpha \rightarrow \frac{1}{2} J_{AB} & \beta \beta^T H_{mag} \beta \alpha \rightarrow 0 \ \beta \beta^T H_{mag} \beta \beta \rightarrow \frac{1}{2} \nu_A + \frac{1}{2} \nu_B + \frac{1}{4} J_{AB} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.20%3A_AB_Proton_NMR_Using_Tensor_Algebra.txt
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The purpose of this tutorial is to deviate from the usual matrix mechanics approach to the ABC proton nmr system in order to illustrate a related method of analysis which uses tensor algebra. For a discussion of the traditional approach for the ABC system visit http://www.users.csbsju.edu/~frioux/nmr/Speclab4.htm. This site also provides general information on the quantum mechanics of nmr spectroscopy.
$\begin{matrix} \text{Nuclear spin and identity operators:} & I_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & I_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \ \text{Chemical shifts:} & \nu_A = 250 & \nu_B = 300 & \text{Coupling constant:} & \text{Jab} = 10 \end{matrix} \nonumber$
Hamiltonian representing the interaction of nuclear spins with the external magnetic field in tensor format:
$\begin{matrix} \hat{H}_{mag} = \nu_A \hat{I}_z^A - \nu_B \hat{I}_z^B = - \nu_A \hat{I}_z^A \otimes \hat{I} + \hat{I} \otimes \left( - \nu_B \hat{I}_z^B \right) & \text{where, for example,} & \nu_A = g_n \beta_n B_z(1- \sigma_A) \end{matrix} \nonumber$
Implementing the operator using Mathcad's command for the tensor product, kronecker, is as follows.
$H_{mag} = - \nu_A \text{kronecker}(I_z,~I) - \nu_B \text{kronecker}(I,~I_z) \nonumber$
Hamiltonian representing the interaction of nuclear spins with each other in tensor format:
$\hat{H}_{spin} = Jab \left( \hat{I}_x^A \otimes \hat{I}_x^B + \hat{I}_y^B \otimes \hat{I}_y^B + \hat{I}_z^A \otimes \hat{I}_z^B \right) \nonumber$
Implementation of the operator in the Mathcad programming environment:
$H_{spin} = Jab \left( \text{kronecker}(I_x,~I_x) + \text{kronecker}(I_y,~I_y) + \text{kronecker}(I_z,~I_z) \right) \nonumber$
The total Hamiltonian spin operator is now calculated and displayed.
$H = H_{mag} + H_{spin} \nonumber$
$\begin{matrix} \begin{array} \alpha \alpha \alpha & & \alpha \beta & & \beta \alpha & & \beta \beta \end{array} \ \begin{pmatrix} -272.5 & 0 & 0 & 0 \ 0 & 22.5 & 5 & 0 \ 0 & 5 & -27.5 & 0 \ 0 & 0 & 0 & 277.5 \end{pmatrix} \begin{array} \alpha \alpha \alpha \ \alpha \beta \ \beta \alpha \ \beta \beta \end{array} \end{matrix} \nonumber$
Calculate and display the energy eigenvalues and associated eigenvectors of the Hamiltonian.
$\begin{matrix} i = 1 .. 4 & E = \text{sort(eigenvals(H))} & C^{<i>} = \text{eigenvec}(H,~E_i) \end{matrix} \nonumber$
$\begin{matrix} \text{augment}(E,~C^T )^T = \begin{pmatrix} -272.5 & -27.995 & 22.995 & 277.5 \ 1 & 0 & 0 & 0 \ 0 & -0.099 & 0.995 & 0 \ 0 & 0.005 & 0.099 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \begin{array} \alpha \alpha \alpha \ \alpha \beta \ \beta \alpha \ \beta \beta \end{array} \end{matrix} \nonumber$
The nmr selection rule is that only one nuclear spin can flip during a transition. Therefore, the transition probability matrix for the AB spin system is:
$\begin{matrix} T = \begin{pmatrix} ' & \alpha \alpha & \alpha \beta & \beta \alpha & \beta \beta \ \alpha \alpha & 0 & 1 & 1 & 0 \ \alpha \beta & 1 & 0 & 0 & 1 \ \beta \alpha & 1 & 0 & 0 & 1 \ \beta \beta & 0 & 1 & 1 & 0 \end{pmatrix} & T = \begin{pmatrix} 0 & 1 & 1 & 0 \ 1 & 0 & 0 & 1 \ 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Calculate the intensities and frequencies of the allowed transitions.
$\begin{matrix} i = 1 .. 4 & j = 1 .. 4 & I_{i,~j} = \left[ C^{<i>} \left( TC^{<j>} \right) \right]^2 & V_{i,~j} = \text{if} \left( I_{i,~j} > .001,~ \left| E_i - E_j \right|,~ 0 \right) \end{matrix} \nonumber$
$\begin{matrix} \text{Intensity matrix:} & I = \begin{pmatrix} 0 & 0.8 & 1.2 & 0 \ 0.8 & 0 & 0 & 0.8 & \ 1.2 & 0 & 0 & 1.2 \ 0 & 0.8 & 1.2 & 0 \end{pmatrix} & \text{Frequency matrix:} & V = \begin{pmatrix} 0 & 244.5 & 295.5 & 0 \ 244.5 & 0 & 0 & 244.5 \ 295.5 & 0 & 0 & 254.5 \ 0 & 305.5 & 254.5 & 0 \end{pmatrix} \end{matrix} \nonumber$
Display the calculated AB nmr spectrum:
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.21%3A_Calculating_the_AB_Proton_NMR_Using_Tensor_Algebra.txt
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The chemical shifts and the coupling constant for the two protons in 2,3-dibromothiophene are obtained from its nmr spectrum. These paramaters serve as inputs to a quantum mechanical calculation of the nmr spectrum. The theoretically generated spectrum is in good agreement with experiment, justifying the Hamiltonian operator chosen to represent the interaction of the protons with the external magnetic field and with each other.
Theory
Nuclear spin and identity operators:
$\begin{matrix} I_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & I_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{Chemical shifts:} & \nu_A = 427.57 & \nu_B = 408.42 & \text{Coupling constant:} & J_{AB} = 5.70 \end{matrix} \nonumber$
Hamiltonian representing the interaction of nuclear spins with the external magnetic field in tensor format:
$\begin{matrix} \hat{H}_{mag} = - \nu_A \hat{I}_z^A - \nu_B \hat{I}_z^B = - \nu_A \hat{I}_z^A \otimes \hat{I} + \hat{I} \otimes \left( - \nu_B \hat{I}_z^B \right) & \text{where for example,} & \nu_A = g_n \beta_n B_z (1- \sigma_A ) \end{matrix} \nonumber$
Implementing the operator using Mathcad's command for the tensor product, kronecker, is as follows.
$H_{mag} = \nu_A \text{kronecker} \left(I_z,~I \right) - \nu_B \text{kronecker} \left(I,~I_z \right) \nonumber$
Hamiltonian representing the interaction of nuclear spins with each other in tensor format:
$\widehat{H}_{spin} = J_{AB} \left( \hat{I}_x^A \otimes \hat{I}_x^B + \hat{I}_y^A \otimes \hat{I}_y^B + \hat{I}_z^A \otimes \hat{I}_z^B \right) \nonumber$
Implementation of the operator in the Mathcad programming environment:
$H_{spin} = J_{AB} \left( \text{kronecker} \left( I_x,~I_x \right) + \text{kronecker} \left( I_y,~I_y \right) + \text{kronecker} \left( I_z,~I_z \right) \right) \nonumber$
The total Hamiltonian spin operator is now calculated and displayed.
$H = H_{mag} + H_{spin} \nonumber$
$H = \begin{matrix} \begin{array} \alpha \alpha \alpha & \alpha \beta & \beta \alpha & \beta \beta \end{array} \ \begin{pmatrix} -416.57 & 0 & 0 & 0 \ 0 & -11 & 2.85 & 0 \ 0 & 2.85 & 8.15 & 0 \ 0 & 0 & 0 & 419.42 \end{pmatrix} \begin{array} \alpha \alpha \alpha \ \alpha \beta \ \beta \alpha \ \beta \beta \end{array} \end{matrix} \nonumber$
Calculate and display the energy eigenvalues and associated eigenvectors of the Hamiltonian.
$\begin{matrix} i = 1 .. 4 & E = \text{sort(eigenvals(H))} & C^{<i>} = \text{eigenvec}(H,~E_i ) \end{matrix} \nonumber$
$\text{augment}(E,~C^T )^T = \begin{pmatrix} -416.57 & -11.415 & 8.565 & 419.42 \ 1 & 0 & 0 & 0 \ 0 & 0.99 & 0.144 & 0 \ 0 & -0.144 & 0.99 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix} \begin{array} \ \alpha \alpha \ \alpha \beta \ \beta \alpha \ \beta \beta \end{array} \nonumber$
The nmr selection rule is that only one nuclear spin can flip during a transition. Therefore, the transition probability matrix for the two spin system is:
$\begin{matrix} T = \begin{pmatrix} ' & \alpha \alpha & \alpha \beta & \beta \alpha & \beta \beta \ \alpha \alpha & 0 & 1 & 1 & 0 \ \alpha \beta & 1 & 0 & 0 & 1 \ \beta \alpha & 1 & 0 & 0 & 1 \ \beta \beta & 0 & 1 & 1 & 0 \end{pmatrix} & T = \begin{pmatrix} 0 & 1 & 1 & 0 \ 1 & 0 & 0 & 1 \ 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \end{pmatrix} \end{matrix} \nonumber$
Calculate the transition intensities and frequencies.
$\begin{matrix} i = 1 .. 4 & j = 1 .. 4 & I_{i,~j} = \left[ C^{<i>} \left(T C^{<j>} \right) \right]^2 & V_{i,~j} = \left| E_i - E_j \right| \end{matrix} \nonumber$
Display the calculated 2,3-dibromothiophene nmr spectrum:
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.22%3A_AB_Proton_NMR_Analysis_for_23-dibromothiophene.txt
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The purpose of this tutorial is to deviate from the usual matrix mechanics approach to the ABC proton nmr system in order to illustrate a related method of analysis which uses tensor algebra. For a discussion of the traditional approach visit http://www.users.csbsju.edu/~frioux/nmr/Speclab4.htm. This site also provides general information on the quantum mechanics of nmr spectroscopy.
Nuclear spin operators and identity:
$\begin{matrix} I_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & I_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The following experimentally determined chemical shifts and coupling constants (both in Hz) are for the vinyl protons of vinyl acetate at 60 MHz.
$\begin{matrix} \text{Chemical shifts:} & \nu_A = 345.6 & \nu_B = 237.6 & Jab = 6.1 \ \text{Coupling constants:} & Jab = 7.00 & Jbc = 1.50 & Jac = 15.00 \end{matrix} \nonumber$
Hamiltonian representing the interaction of nuclear spins with the external magnetic field in tensor format:
$\widehat{H}_{mag} = - \nu_A \hat{I}_z^A - \nu_B \hat{I}_z^B - \nu_C \hat{I}_z^C = - \nu_A \hat{I}_z^A \otimes \hat{I} \otimes \hat{I} + \hat{I} \otimes \left( - \nu_B \hat{I}_z^B \right) \otimes \hat{I} + \hat{I} \otimes \hat{I} \otimes \left( - \nu_B \hat{I}_z^B \right) \nonumber$
where, for example,
$\nu_A = g_n \beta_n B_z (1 - \sigma_A) \nonumber$
Implementing the operator using Mathcad's command for the tensor product, kronecker, is as follows.
$H_{mag} = - \nu_A \text{kronecker} \left( I_z,~ \text{kronecker(I, I)} \right) - \nu_B \text{kronecker} (I,~ \text{kronecker} (I_z,~I)) - \nu_B \text{kronecker}(I,~ \text{kronecker}(I,~I_z)) \nonumber$
Hamiltonian representing the interaction of nuclear spins with each other in tensor format:
$\begin{matrix} \widehat{H}_{spin} = J_{AB} \left( \hat{I}_x^A \otimes \hat{I}_x^B \otimes \hat{I} + \hat{I}_y^A \otimes \hat{I}_y^B \otimes \hat{I} + \hat{I}_z^A \otimes \hat{I}_z^B \otimes \hat{I} \right) \ + J_{AB} \left( \hat{I}_x^A \otimes \hat{I} \otimes \hat{I}_x^B + \hat{I}_y^A \otimes \hat{I} \otimes \hat{I}_y^B + \hat{I}_z^A \otimes \hat{I} \otimes \hat{I}_z^B \right) \ +J_{BC} \left( \hat{I} \otimes \hat{I}_x^B \otimes \hat{I}_x^C + \hat{I} \otimes \hat{I}_y^ \otimes \hat{I}_y^C + \hat{I} \otimes \hat{I}_z^B \otimes \hat{I}_z^C \right) \end{matrix} \nonumber$
Implementation of the operator in the Mathcad programming environment:
$H_{spin} = \begin{matrix} Jab \left( \text{kronecker} (I_x,~ \text{kronecker} (I_x,~I)) \text{kronecker}(I_y,~ \text{kronecker}(I_y,~I)) + \text{kronecker}(I_z,~ \text{kronecker} (I_z,~I)) \right) ... \ + \begin{bmatrix} Jab \left( \text{kronecker} (I_x,~ \text{kronecker}(I,~I_x)) + \text{kronecker} (I_y,~ \text{kronecker} (I,~I_y)) + \text{kronecker} (I_z,~ \text{kronecker}(I,~I_z)) \right) ... \ + Jbc \left( \text{kronecker} (I,~ \text{kronecker}(I_x,~I_x)) + \text{kronecker} (I,~ \text{kronecker} (I_y,~I_y)) + \text{kronecker} (I,~ \text{kronecker}(I_z,~I_z)) \right) \end{bmatrix} \end{matrix} \nonumber$
The total Hamiltonian spin operator is now calculated and displayed.
$H = H_{mag} + H_{spin} \nonumber$
The indexing of the matrix elements of the Hamiltonial spin operator is discussed in the Appendix.
i = 1 .. 8
$H = \begin{matrix} \begin{array} \alpha \alpha \alpha \alpha & \alpha \alpha \beta & \alpha \beta \alpha & \alpha \beta \beta & \beta \alpha \alpha & \beta \alpha \beta & \beta \beta \alpha & \beta \beta \beta \end{array} \begin{pmatrix} -491.63 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & -199.88 & 0.75 & 0 & 7.5 & 0 & 0 & 0 \ 0 & 0.75 & -230.88 & 0 & 3.5 & 0 & 0 & 0 \ 0 & 0 & 0 & 62.38 & 0 & 3.5 & 7.5 & 0 \ 0 & 7.5 & 3.5 & 0 & -72.63 & 0 & 0 & 0 \ 0 & 0 & 0 & 3.5 & 0 & 234.13 & 0.75 & 0 \ 0 & 0 & 0 & 7.5 & 0 & 0.75 & 195.13 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 503.38 \end{pmatrix} \end{matrix} \nonumber$
Calculate and display the energy eigenvalues and associated eigenvectors of the Hamiltonian.
$\begin{matrix} E = \text{sort(eigenvals(H))} & C^{<i>} = \text{eigenvec}(H,~E_i ) \end{matrix} \nonumber$
$\text{augment} \left( E,~C^T \right)^T = \begin{matrix} \begin{pmatrix} -491.625 & -230.963 & -200.306 & -72.106 & 61.883 & 195.524 & 234.217 & 503.375 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & -0.019 & 0.998 & 0.059 & 0 & 0 & 0 & 0 \ 0 & 1 & 0.018 & 0.022 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0.998 & 0.056 & 0.021 & 0 \ 0 & -0.021 & -0.059 & 0.998 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & -0.02 & -0.024 & 1 & 0 \ 0 & 0 & 0 & 0 & -0.056 & 0.998 & 0.023 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{array} \ \alpha \alpha \alpha \alpha \ \alpha \alpha \beta \ \alpha \beta \alpha \ \alpha \beta \beta \ \beta \alpha \alpha \ \beta \alpha \beta \ \beta \beta \alpha \ \beta \beta \beta \end{array} \end{matrix} \nonumber$
Notice that the ground state |ααα> and the highest excited state |βββ> are pure states. The other six states are strictly speaking superpositions.
The nmr selection rule is that only one nuclear spin can flip during a transition. Therefore, the transition probability matrix for the ABC spin system is:
$T = \begin{matrix} \begin{array} \alpha \alpha \alpha \alpha & \alpha \alpha \beta & \alpha \beta \alpha & \alpha \beta \beta & \beta \alpha \alpha & \beta \alpha \beta & \beta \beta \alpha & \beta \beta \beta \end{array} \ \begin{pmatrix} 0 & 1.00 & 1.00 & 0 & 1.00 & 0 & 0 & 0 \ 1.00 & 0 & 0 & 1.00 & 0 & 1.00 & 0 & 0 \ 1.00 & 0 & 0 & 1.00 & 0 & 0 & 1.00 & 0 \ 0 & 1.00 & 1.00 & 0 & 0 & 0 & 0 & 1.00 \ 1.00 & 0 & 0 & 0 & 0 & 1.00 & 1.00 & 0 \ 0 & 1.00 & 0 & 0 & 1.00 & 0 & 0 & 1.00 \ 0 & 0 & 1.00 & 0 & 1.00 & 0 & 0 & 1.00 \ 0 & 0 & 0 & 1.00 & 0 & 1.00 & 1.00 & 0 \end{pmatrix} \begin{array} \alpha \alpha \alpha \alpha \ \alpha \alpha \beta \ \alpha \beta \alpha \ \alpha \beta \beta \ \beta \alpha \alpha \ \beta \alpha \beta \ \beta \beta \alpha \ \beta \beta \beta \end{array} \end{matrix} \nonumber$
Calculate the intensities and frequencies of the allowed transitions.
$\begin{matrix} i = 1 .. 8 & j = 1 .. 8 & I_{i,~j} = \left[ C^{<i>} \left( TC^{<j>} \right) \right]^2 & V_{i,~j} = \text{ if}(I_{i,~j} . .001,~ \left| E_i - E_j \right|,~0) \end{matrix} \nonumber$
Intensity matrix:
$i = \begin{pmatrix} 0 & 0.92 & 0.92 & 1.16 & 0 & 0 & 0 & 0 \ 0.92 & 0 & 0 & 0 & 0.86 & 1.07 & 0 & 0 \ 0.92 & 0 & 0 & 0 & 1 & 0 & 0.92 & 0 \ 1.16 & 0 & 0 & 0 & 0 & 0.99 & 1.17 & 0 \ 0 & 0.86 & 1 & 0 & 0 & 0 & 0 & 0.85 \ 0 & 1.07 & 0 & 0.99 & 0 & 0 & 0 & 1.06 \ 0 & 0 & 0.92 & 1.17 & 0 & 0 & 0 & 1.09 \ 0 & 0 & 0 & 0 & 0.85 & 1.06 & 1.09 & 0 \end{pmatrix} \nonumber$
Frequency matrix:
$\text{V} = \begin{pmatrix} 0 & 260.66 & 291.32 & 419.52 & 0 & 0 & 0 & 0 \ 260.66 & 0 & 0 & 0 & 292.85 & 426.49 & 0 & 0 \ 291.32 & 0 & 0 & 0 & 262.19 & 0 & 434.52 & 0 \ 419.52 & 0 & 0 & 0 & 0 & 267.63 & 306.32 & 0 \ 0 & 292.85 & 262.19 & 0 & 0 & 0 & 0 & 441.49 \ 0 & 426.49 & 0 & 267.63 & 0 & 0 & 0 & 269.16 \ 0 & 0 & 434.52 & 306.32 & 0 & 0 & 0 & 269.16 \ 0 & 0 & 0 & 0 & 441.49 & 307.85 & 269.16 & 0 \end{pmatrix} \nonumber$
Display the calculated vinyl acetate nmr spectrum:
The calculated spectrum compares favorably with experimental spectrum, indicating that the spin Hamiltonian used adequately represents the magnetic interaction of the vinyl protons in vinyl acetate. By comparison, acrylonitrile, a pure ABC system, has a very complicated 60 MHz spectrum. See the link provide above for a quantum mechanical explanation for the greater complexity of the acrylonitrile spectrum.
Appendix
The tensor product of three spinors 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 this type of 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.
$\begin{matrix} \text{Spin-up in the z-direction:} & \alpha = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{Spin-down in the z-direction:} & \beta = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \text{Null vector:} & N = \begin{pmatrix} 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The eight possible spin states of a three-proton system are calculated as shown below.
$\Psi \text{(a, b, c) = submatrix(kronecker(augment(a, N), kronecker(augment(b, N), augment(c, N))), 1, 8, 1, 1)} \nonumber$
$\begin{matrix} \Psi ( \alpha, \alpha, \alpha )^T = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \Psi ( \alpha, \alpha, \beta )^T = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \Psi ( \alpha, \beta, \alpha )^T = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \Psi ( \alpha, \beta, \beta )^T = \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix} \ \Psi ( \beta, \alpha, \alpha )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} & \Psi ( \beta, \alpha, \beta )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} \ \Psi ( \beta, \beta, \alpha )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} & \Psi ( \beta, \beta, \beta )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Thus the indexing in Hamiltonian matrix is: |ααα>, |ααβ>, |αβα>, |αββ>, |βαα>, |βαβ>, |ββα>, |βββ>.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.23%3A_ABC_Proton_NMR_Using_Tensor_Algebra.txt
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The purpose of this tutorial is to deviate from the usual matrix mechanics approach to the ABC proton nmr system in order to illustrate a related method of analysis which uses tensor algebra. For a discussion of the traditional approach visit http://www.users.csbsju.edu/~frioux/nmr/Speclab4.htm. This site also provides general information on the quantum mechanics of nmr spectroscopy.
Nuclear spin operators and identity:
$\begin{matrix} I_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & I_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The following experimentally determined chemical shifts and coupling constant (all in Hz) are for the AB2 proton system 1,1,2-trichloroethane at 60 MHz.
$\begin{matrix} \text{Chemical shifts:} & \nu_A = 345.6 & \nu_B = 237.6 & Jab = 6.1 \end{matrix} \nonumber$
Hamiltonian representing the interaction of nuclear spins with the external magnetic field in tensor format:
$\widehat{H}_{mag} = - \nu_A \hat{I}_z^A - \nu_B \hat{I}_z^B - \nu_C \hat{I}_z^C = \nu_A \hat{I}_z^A \otimes \hat{I} \otimes \hat{I} + \hat{I} \otimes \left( - \nu_B \hat{I}_z^B \right) \otimes \hat{I} + \hat{I} \otimes \hat{I} \otimes \left( - \nu_B \hat{I}_z^B \right) \nonumber$
where, for example,
$\nu_A = g_n \beta_n B_z (1 - \sigma_A) \nonumber$
Implementing the operator using Mathcad's command for the tensor product, kronecker, is as follows.
$H_{mag} = - \nu_A \text{kronecker} \left( I_z,~ \text{kronecker(I, I)} \right) - \nu_B \text{kronecker} (I,~ \text{kronecker} (I_z,~I)) - \nu_B \text{kronecker}(I,~ \text{kronecker}(I,~I_z)) \nonumber$
Hamiltonian representing the interaction of nuclear spins with each other in tensor format:
$\begin{matrix} \widehat{H}_{spin} = J_{AB} \left( \hat{I}_x^A \otimes \hat{I}_x^B \otimes \hat{I} + \hat{I}_y^A \otimes \hat{I}_y^B \otimes \hat{I} + \hat{I}_z^A \otimes \hat{I}_z^B \otimes \hat{I} \right) \ + J_{AB} \left( \hat{I}_x^A \otimes \hat{I} \otimes \hat{I}_x^B + \hat{I}_y^A \otimes \hat{I} \otimes \hat{I}_y^B + \hat{I}_z^A \otimes \hat{I} \otimes \hat{I}_z^B \right) \end{matrix} \nonumber$
Implementation of the operator in the Mathcad programming environment:
$H_{spin} = \begin{matrix} Jab \left( \text{kronecker} (I_x,~ \text{kronecker} (I_x,~I)) \text{kronecker}(I_y,~ \text{kronecker}(I_y,~I)) + \text{kronecker}(I_z,~ \text{kronecker} (I_z,~I)) \right) ... \ + \begin{bmatrix} Jab \left( \text{kronecker} (I_x,~ \text{kronecker}(I,~I_x)) + \text{kronecker} (I_y,~ \text{kronecker} (I,~I_y)) + \text{kronecker} (I_z,~ \text{kronecker}(I,~I_z)) \right) \end{bmatrix} \end{matrix} \nonumber$
The total Hamiltonian spin operator is now calculated and displayed.
$H = H_{mag} + H_{spin} \nonumber$
The indexing of the matrix elements of the Hamiltonial spin operator is discussed in the Appendix.
i = 1 .. 8
$H = \begin{pmatrix} -407.35 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & -172.8 & 0 & 0 & 3.05 & 0 & 0 & 0 \ 0 & 0 & -172.8 & 0 & 3.05 & 0 & 0 & 0 \ 0 & 3.05 & 3.05 & 0 & -67.85 & 0 & 0 & 0 \ 0 & 0 & 0 & 3.05 & 0 & 172.8 & 0 & 0 \ 0 & 0 & 0 & 3.5 & 0 & 234.13 & 0.75 & 0 \ 0 & 0 & 0 & 3.05 & 0 & 0 & 172.8 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 413.45 \end{pmatrix} \nonumber$
Calculate and display the energy eigenvalues and associated eigenvectors of the Hamiltonian.
$\begin{matrix} E = \text{sort(eigenvals(H))} & C^{<i>} = \text{eigenvec}(H,~E_i ) \end{matrix} \nonumber$
$\text{augment} \left( E,~C^T \right)^T = \begin{matrix} \begin{pmatrix} -407.35 & -172.977 & -172.8 & -67.673 & 61.583 & 172.8 & 172.967 & 413.45 \ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0.707 & -0.707 & 0.029 & 0 & 0 & 0 & 0 \ 0 & 0.707 & 0.707 & 0.029 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0.999 & 0 & 0.039 & 0 \ 0 & -0.041 & 0 & 0.999 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & -0.027 & -0.707 & 0.707 & 0 \ 0 & 0 & 0 & 0 & -0.027 & 0.707 & 0.707 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{array} \ \alpha \alpha \alpha \alpha \ \alpha \alpha \beta \ \alpha \beta \alpha \ \alpha \beta \beta \ \beta \alpha \alpha \ \beta \alpha \beta \ \beta \beta \alpha \ \beta \beta \beta \end{array} \end{matrix} \nonumber$
Notice that the ground state |ααα> and the highest excited state |βββ> are pure states. The other six states are strictly speaking superpositions.
The nmr selection rule is that only one nuclear spin can flip during a transition. Therefore, the transition probability matrix for the ABC spin system is:
$T = \begin{matrix} \begin{array} \alpha \alpha \alpha \alpha & \alpha \alpha \beta & \alpha \beta \alpha & \alpha \beta \beta & \beta \alpha \alpha & \beta \alpha \beta & \beta \beta \alpha & \beta \beta \beta \end{array} \ \begin{pmatrix} 0 & 1.00 & 1.00 & 0 & 1.00 & 0 & 0 & 0 \ 1.00 & 0 & 0 & 1.00 & 0 & 1.00 & 0 & 0 \ 1.00 & 0 & 0 & 1.00 & 0 & 0 & 1.00 & 0 \ 0 & 1.00 & 1.00 & 0 & 0 & 0 & 0 & 1.00 \ 1.00 & 0 & 0 & 0 & 0 & 1.00 & 1.00 & 0 \ 0 & 1.00 & 0 & 0 & 1.00 & 0 & 0 & 1.00 \ 0 & 0 & 1.00 & 0 & 1.00 & 0 & 0 & 1.00 \ 0 & 0 & 0 & 1.00 & 0 & 1.00 & 1.00 & 0 \end{pmatrix} \begin{array} \alpha \alpha \alpha \alpha \ \alpha \alpha \beta \ \alpha \beta \alpha \ \alpha \beta \beta \ \beta \alpha \alpha \ \beta \alpha \beta \ \beta \beta \alpha \ \beta \beta \beta \end{array} \end{matrix} \nonumber$
Calculate the intensities and frequencies of the allowed transitions.
$\begin{matrix} i = 1 .. 8 & j = 1 .. 8 & I_{i,~j} = \left[ C^{<i>} \left( TC^{<j>} \right) \right]^2 & V_{i,~j} = \text{ if}(I_{i,~j} . .001,~ \left| E_i - E_j \right|,~0) \end{matrix} \nonumber$
Intensity matrix:
$i = \begin{pmatrix} 0 & 1.88 & 0 & 1.12 & 0 & 0 & 0 & 0 \ 1.88 & 0 & 0 & 0 & 1.89 & 0 & 0.99 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 1.12 & 0 & 0 & 0 & 0 & 0 & 2.12 & 0 \ 0 & 1.89 & 0 & 0 & 0 & 0 & 0 & 0.89 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0.99 & 0 & 2.12 & 0 & 0 & 0 & 2.11 \ 0 & 0 & 0 & 0 & 0.89 & 0 & 2.11 & 0 \end{pmatrix} \nonumber$
Frequency matrix:
$\text{V} = \begin{pmatrix} 0 & 234.37 & 234.55 & 339.68 & 468.93 & 580.15 & 580.32 & 820.8 \ 234.37 & 0 & 0.18 & 105.3 & 234.56 & 345.78 & 345.94 & 586.43 \ 234.55 & 0.18 & 0 & 105.13 & 234.38 & 345.6 & 345.77 & 586.25 \ 339.68 & 105.3 & 105.13 & 0 & 129.26 & 240.47 & 240.64 & 481.12 \ 468.93 & 234.56 & 234.38 & 129.26 & 0 & 111.22 & 111.38 & 351.87 \ 580.15 & 345.78 & 345.6 & 240.47 & 111.22 & 0 & 0.17 & 240.65 \ 580.32 & 345.94 & 345.77 & 240.64 & 111.38 & 0.17 & 0 & 240.48 \ 820.8 & 586.43 & 586.25 & 481.12 & 351.87 & 240.65 & 240.48 & 0 \end{pmatrix} \nonumber$
Display the calculated vinyl acetate nmr spectrum:
The calculated spectrum compares favorably with experimental spectrum, indicating that the spin Hamiltonian used adequately represents the magnetic interaction of the protons in 1,1,2-trichloroethane at 60 MHz.
Appendix
The tensor product of three spinors 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 this type of 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.
$\begin{matrix} \text{Spin-up in the z-direction:} & \alpha = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{Spin-down in the z-direction:} & \beta = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \text{Null vector:} & N = \begin{pmatrix} 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The eight possible spin states of a three-proton system are calculated as shown below.
$\Psi \text{(a, b, c) = submatrix(kronecker(augment(a, N), kronecker(augment(b, N), augment(c, N))), 1, 8, 1, 1)} \nonumber$
$\begin{matrix} \Psi ( \alpha, \alpha, \alpha )^T = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \Psi ( \alpha, \alpha, \beta )^T = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \Psi ( \alpha, \beta, \alpha )^T = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} & \Psi ( \alpha, \beta, \beta )^T = \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix} \ \Psi ( \beta, \alpha, \alpha )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} & \Psi ( \beta, \alpha, \beta )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} \ \Psi ( \beta, \beta, \alpha )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} & \Psi ( \beta, \beta, \beta )^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Thus the indexing in Hamiltonian matrix is: |ααα>, |ααβ>, |αβα>, |αββ>, |βαα>, |βαβ>, |ββα>, |βββ>.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.24%3A_AB2_Proton_NMR_Using_Tensor_Algebra.txt
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The purpose of this tutorial is to calculate the NMR spectrum of a four proton AB3 system.
$\begin{matrix} \text{Nuclear spin operators and identity:} & I_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & I_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & I_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
The following experimentally determined chemical shifts and coupling constant (all in Hz) are for the AB3 proton system 1,1‐dichloroethane at 60 MHz.
$\begin{matrix} ]nu_A = 350.0 & \nu_B = 120.0 & Jab = 10.00 \end{matrix} \nonumber$
Hamiltonian representing the interaction of nuclear spins with the external magnetic field in tensor format:
$\begin{matrix} H_{mag} & = - \nu_A \text{kronecker} \left( I_z,~ \text{kronecker(I, kronecker(I, I))} \right) - \nu_B \text{kronecker} \left( \text{I, kronecker} \left( I_z,~ \text{kronecker(I, I)} \right) \right) ... \ ~ & + - \nu_B \text{kronecker} \left( \text{I, kronecker} \left( \text{I, kronecker} \left( I_z,~ I \right) \right) \right) - \nu_B \text{kronecker} \left( \text{I, kronecker} \left( \text{I, kronecker} \left( I,~I_z \right) \right) \right) \end{matrix} \nonumber$
Hamiltonian representing the nuclear spin‐spin interaction in tensor format:
$\begin{matrix} H_{spin} = Jab \begin{pmatrix} \text{kronecker} \left( I_x,~ \text{kronecker} \left(I_x, ~ \text{kronecker(I, I)} \right) \right) + \text{kronecker} \left(I_y,~ \text{kronecker} \left( I_y,~ \text{kronecker(I, I)} \right) \right) ... \ \end{pmatrix} ... \ + Jab \begin{pmatrix} \text{kronecker} \left( I_x,~ \text{kronecker} \left(I, ~ \text{kronecker}(I_x,~ I) \right) \right) + \text{kronecker} \left(I_y,~ \text{kronecker} \left( I,~ \text{kronecker} \left(I_y, ~I \right) \right) \right) ... \ \end{pmatrix} ... \ + Jab \begin{pmatrix} \text{kronecker} \left( I_x,~ \text{kronecker} \left(I, ~ \text{kronecker}(I,~ I_x) \right) \right) + \text{kronecker} \left(I_y,~ \text{kronecker} \left( I,~ \text{kronecker} \left(I, ~I_y \right) \right) \right) ... \ \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{Total Hamiltonian operator:} & H = H_{mag} + H_{spin} & i = 1 .. 16 & E = \text{sort(eigenvals(H))} & C^{<i>} = \text{eigenvec} \left( H,~ E_i \right) \end{matrix} \nonumber$
The first half of the results matrix:
$\text{augment} \left( E,~C^T \right) ^T = \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline -347.5 & 232.84 & -232.5 & -232.5 & -117.93 & -117.61 & -117.61 & -12.16 & -2.81 \ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ \hline 0 & 0.58 & -0.76 & -0.76 & 0 & 0 & 0 & 0.02 & 0 \ \hline 0 & 0.58 & 0.11 & 0.11 & 0 & 0 & 0 & 0.02 & 0 \ \hline 0 & 0 & 0 & 0 & 0.58 & -0.18 & -0.18 & 0 & 0 \ \hline 0 & 0.58 & 0.64 & 0.64 & 0 & 0 & 0 & 0.02 & 0 \ \hline 0 & 0 & 0 & 0 & 0.58 & -0.6 & -0.6 & 0 & 0 \ \hline 0 & 0 & 0 & 0 & 0.58 & 0.78 & 0.78 & 0 & 0 \ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ \hline 0 & -0.04 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ \hline 0 & 0 & 0 & 0 & -0.03 & 0.02 & 0.02 & 0 & 0 \ \hline 0 & 0 & 0 & 0 & -0.03 & -0.01 & -0.01 & 0 & 0 \ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.02 \ \hline 0 & 0 & 0 & 0 & -0.03 & -0 & -0 & 0 & 0 \ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.02 \ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -0.02 \ \hline \end{array} \nonumber$
The nmr selection rule is that only one nuclear spin can flip during a transition. Therefore, the transition probability matrix for the four spin system is shown below. See the Appendix for detail on how this matrix is constructed.
$T = \begin{pmatrix} 0&1&1&0&1&0&0&0&1&0&0&0&0&0&0&0 \ 1&0&0&1&0&1&0&0&0&1&0&0&0&0&0&0 \ 1&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0 \ 0&1&1&0&0&0&0&1&0&0&0&1&0&0&0&0 \ 1&0&0&0&0&1&1&0&0&0&0&0&1&0&0&0 \ 0&1&0&0&1&0&0&1&0&0&0&0&0&1&0&0 \ 0&0&1&0&1&0&0&1&0&0&0&0&0&0&1&0 \ 0&0&0&1&0&1&1&0&0&0&0&0&0&0&0&1 \ 1&0&0&0&0&0&0&0&0&1&1&0&1&0&0&0 \ 0&1&0&0&0&0&0&0&1&0&0&1&0&1&0&0 \ 0&0&1&0&0&0&0&0&1&0&0&1&0&0&1&0 \ 0&0&0&1&0&0&0&0&0&1&1&0&0&0&0&1 \ 0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0 \ 0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&1 \ 0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&1 \ 0&0&0&0&0&0&0&1&0&0&0&1&0&1&1&0 \end{pmatrix} \nonumber$
Calculate the intensities and frequencies of the allowed transitions, and display the spectrum.
$\begin{matrix} i = 1 .. 16 & j = 1 .. 16 & I_{i,~j} = \left[ C^{<i>} \left( TC^{<i>} \right) \right] & V_{i,~j} = \left| E_i - E_j \right| \end{matrix} \nonumber$
Appendix
The construction of the transition probability matrix requires the proper indexing of the 16 spin states of the four proton system. In tensor format the states are represented in this manner.
$\begin{pmatrix} a \ b \end{pmatrix} \otimes \begin{pmatrix} c \ d \end{pmatrix} \otimes \begin{pmatrix} e \ f \end{pmatrix} \otimes \begin{pmatrix} g \ h \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix} \otimes \begin{pmatrix} c \ d \end{pmatrix} \otimes \begin{pmatrix} eg \ eh \ fg \ fh \end{pmatrix} = \begin{pmatrix} aceg \ aceh \ acfg \ acfh \ adeg \ adeh \ adfg \ adfh \ bceg \ bceh \ bcfg \ bcfh \ bdeg \ bedh \ bdfg \ bdfh \end{pmatrix} \nonumber$
Mathcad does not have a command for this type of 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.
$\begin{matrix} \text{Spin-up in the z-direction:} & \alpha = \begin{pmatrix} 1 \ 0 \end{pmatrix} & \text{Spin-down in the z-direction:} & \beta = \begin{pmatrix} 0 \ 1 \end{pmatrix} & \text{Null vector:} & N = \begin{pmatrix} 0 \ 0 \end{pmatrix} \end{matrix} \nonumber$
The 16 possible spin states of a three‐proton system are calculated as shown below.
$\Psi \text{(a, b, c, d) = submatrix(kronecker (augment(b, N), kronecker(augment(c, N), augment(d, N)))), 1, 16, 1, 1)} \nonumber$
Representing the spin states in tensor format facillitates their proper indexing and the formation of the transition probability matrix.
$\begin{pmatrix} \Psi ( \alpha, ~ \alpha,~ \alpha,~ \alpha )^T \ \Psi ( \alpha, ~ \alpha,~ \alpha,~ \beta )^T \ \Psi ( \alpha, ~ \alpha,~ \beta,~ \alpha )^T \ \Psi ( \alpha, ~ \alpha,~ \beta,~ \beta )^T \ \Psi ( \alpha, ~ \beta,~ \alpha,~ \alpha )^T \ \Psi ( \alpha, ~ \beta,~ \alpha,~ \beta )^T \ \Psi ( \alpha, ~ \beta,~ \beta,~ \alpha )^T \ \Psi ( \beta, ~ \alpha,~ \alpha,~ \alpha )^T \ \Psi ( \beta, ~ \alpha,~ \alpha,~ \beta )^T \ \Psi ( \beta, ~ \alpha,~ \beta,~ \alpha )^T \ \Psi ( \beta, ~ \alpha,~ \beta,~ \beta )^T \ \Psi ( \beta, ~ \beta,~ \alpha,~ \alpha )^T \ \Psi ( \beta, ~ \beta,~ \alpha,~ \beta )^T \ \Psi ( \beta, ~ \beta,~ \beta,~ \alpha )^T \ \Psi ( \beta, ~ \beta,~ \beta,~ \beta )^T \ \end{pmatrix} = \begin{bmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix} \ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{bmatrix} \nonumber$
$\begin{pmatrix} ' & \alpha \alpha \alpha \alpha & \alpha \alpha \alpha \beta & \alpha \alpha \beta \alpha & \alpha \alpha \beta \beta & \alpha \beta \alpha \alpha & \alpha \beta \alpha \beta & \alpha \beta \beta \alpha & \alpha \beta \beta \beta & \beta \alpha \alpha \alpha & \beta \alpha \alpha \beta & \beta \alpha \beta \alpha & \beta \alpha \beta \beta & \beta \beta \alpha \alpha & \beta \beta \alpha \beta & \beta \beta \beta \alpha & \beta \beta \beta \beta \ \alpha \alpha \alpha \alpha & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ \alpha \alpha \alpha \beta & 1& 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ \alpha \alpha \beta \alpha & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ \alpha \alpha \beta \beta & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ \alpha \beta \alpha \beta & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ \alpha \beta \beta \alpha & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ \alpha \beta \beta \beta & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ \beta \alpha \alpha \alpha & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \ \beta \alpha \alpha \beta & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \ \beta \alpha \beta \alpha & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \ \beta \alpha \beta \beta & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \ \beta \beta \alpha \alpha & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \ \beta \beta \alpha \beta & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \ \beta \beta \beta \alpha & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \ \beta \beta \beta \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.25%3A_AB3_Proton_NMR_Using_Tensor_Algebra.txt
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In section 7.12.3 of Atoms and Molecules, Karplus and Porter consider the nuclear magnetic resonance of HD. In this tutorial it is treated as analogous to the traditional AB system, except that HD consists of a spin 1/2 nucleus interacting with a spin 1 nucleus. Using arbitrary values for the chemical shifts and coupling constant, a tensor algebra calculation yields a model nmr spectrum. Related tutorials on the AB and ABC systems are also available in this tutorial section.
$\begin{matrix} \text{Chemical shifts:} & \nu_H = 400 & \nu_D = 200 & \text{Coupling constant:} & J_{HD} = 30 \end{matrix} \nonumber$
Nuclear spin and identity operators for the proton (the proton is a spin 1/2 particle):
$\begin{matrix} IH_x = \frac{1}{2} \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} & IH_y = \frac{1}{2} \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix} & IH_z = \frac{1}{2} \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix} & IH = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Nuclear spin and identity operators for the deuteron (the deuteron is a spin 1 particle):
$\begin{matrix} ID_x = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \ 1 & 0 & 1 \ 0 & 1 & 0 \end{pmatrix} & ID_y = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0 & -i & 0 \ i & 0 & -i \ 0 & i & 0 \end{pmatrix} & ID_z = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & -1 \end{pmatrix} & ID = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \end{matrix} \nonumber$
Hamiltonian representing the interaction of nuclear spins with the external magnetic field in tensor format:
$\widehat{h}_{MagField} = - \nu_H \hat{I}_z^H - \nu_D \hat{I}_z^H = - \nu_H \hat{I}_z^H \otimes \hat{I}_D - \nu_D \hat{I}_H \otimes \hat{I}_z^D \nonumber$
Implementing the operator using Mathcadʹs command for the tensor product, kronecker, is as follows.
$H_{MagField} = - \nu_H \text{kronecker} \left( IH_z,~ID \right) - \nu_D \text{kronecker} \left(IH,~ID_z \right) \nonumber$
Hamiltonian representing the interaction of nuclear spins with each other in tensor format:
$\hat{H}_{SpinSpin} = J_{HD} \left( \hat{I}_x^H \otimes \hat{I}_x^D + \hat{I}_y^H \otimes \hat{I}_y^D + \hat{I}_z^H \otimes \hat{I}_z^D \right) \nonumber$
Implementation of the operator in the Mathcad programming environment:
$H_{SpinSpin} = J_{HD} \left( \text{kronecker} \left( IH_x,~ID_x \right) + \text{kronecker} \left( IH_y,~ID_y \right) + \text{kronecker} \left( IH_z,~ID_z \right) \right) \nonumber$
The total Hamiltonian spin operator is now calculated and displayed.
$H = H_{MagField} + H_{SpinSpin} \nonumber$
$H = \begin{pmatrix} -385 & 0 & 0 & 0 & 0 & 0 \ 0 & -200 & 0 & 21.213 & 0 & 0 \ 0 & 0 & -15 & 0 & 21.213 & 0 \ 0 & 21.213 & 0 & -15 & 0 & 0 \ 0 & 0 & 21.213 & 0 & 200 & 0 \ 0 & 0 & 0 & 0 & 0 & 415 \end{pmatrix} \nonumber$
Calculate and display the energy eigenvalues and associated eigenvectors of the Hamiltonian.
$\begin{matrix} i = 1 .. 6 & E = \text{sort(eigenvals(H))} & C^{<i>} = \text{eigenvec} \left( H,~E_i \right) \end{matrix} \nonumber$
$\text{augment} \left( E,~C^T \right)^T = \begin{pmatrix} -385 & -202.401 & -17.073 & -12.599 & 202.073 & 415 \ 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0.994 & 0 & 0.112 & 0 & 0 \ 0 & 0 & 0.995 & 0 & 0.097 & 0 \ 0 & -0.112 & 0 & 0.994 & 0 & 0 \ 0 & 0 & -0.097 & 0 & 0.995 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \nonumber$
The HD composite nuclear spin states expressed in tensor format are as follows:
$\begin{matrix} | \frac{1}{2},~ 1 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & | \frac{1}{2},~ 0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} & | \frac{1}{2},~ -1 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \ 0 \end{pmatrix} \ | - \frac{1}{2},~ 1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} & | - \frac{1}{2},~ 0 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 1 \ 0 \end{pmatrix} & | - \frac{1}{2},~ -1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \otimes \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
The nmr selection rule is that only one nuclear spin can flip during a transition and that ΔI = +/‐ 1. Therefore, the transition probability matrix given the above nuclear spin assignments for the HD system is:
$T = \begin{pmatrix} 0 & 1 & 0 & 1 & 0 & 0 \ 1 & 0 & 1 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 0 & 1 & 0 \end{pmatrix} \nonumber$
Calculate the transition intensities and frequencies.
$\begin{matrix} i = 1 .. 6 & j = 1 .. 6 & I_{i,~j} = \left[ C^{<i>T} \left( TC^{<j>} \right) \right] & V_{i,~j} = \left| E_i - E_j \right| \end{matrix} \nonumber$
Display the calculated HD nmr spectrum:
As expected, we see that the single proton resonance on the left is split into three transitions due to the three deuteron spin orientations, and that the two deuteron resonances are each split into two transitions due to the two proton spin orientations.
The frequencies of all possible transitions are presented in the following matrix.
$V = \begin{pmatrix} 0 & 182.599 & 367.927 & 372.401 & 587.073 & 800 \ 182.599 & 0 & 185.328 & 189.803 & 404.474 & 617.401 \ 367.927 & 185.328 & 0 & 4.474 & 219.146 & 432.073 \ 372.401 & 189.803 & 4.474 & 0 & 214.672 & 427.599 \ 587.073 & 404.474 & 219.146 & 214.672 & 0 & 212.927 \ 800 & 617.401 & 432.073 & 427.599 & 212.927 & 0 \end{pmatrix} \nonumber$
Some of these transitions are forbidden by the selection rule as can be seen in the following intensity matrix.
$I = \begin{pmatrix} 0 & 0.776 & 0 & 1.224 & 0 & 0 \ 0.776 & 0 & 0.816 & 0 & 0.948 & 0 \ 0 & 0.816 & 0 & 1.903 \times 10^{-5} & 0 & 0.806 \ 1.224 & 0 & 1.903 \times 10^{-5} & 0 & 1.236 & 0 \ 0 & 0.948 & 0 & 1.236 & 0 & 1.194 \ 0 & 0 & 0.806 & 0 & 1.194 & 0 \end{pmatrix} \nonumber$
On the basis of the frequency and intensity tables we make the following assignments for the various allowed transition for the proton and deuteron.
$\begin{matrix} \left| \frac{1}{2},~-1 \right\rangle \xrightarrow{432.1} \left| \frac{-1}{2},~ -1 \right\rangle & \left| \frac{1}{2},~ 0 \right\rangle \xrightarrow{404.5} \left| \frac{-1}{2},~0 \right\rangle & \left| \frac{1}{2},~1 \right\rangle \xrightarrow{372.4} \left| \frac{-1}{2},~1 \right\rangle \ \left| \frac{-1}{2},~1 \right\rangle \xrightarrow{214.7} \left| \frac{-1}{2},~ 0 \right\rangle & \left| \frac{1}{2},~ 0 \right\rangle \xrightarrow{212.9} \left| \frac{-1}{2},~-1 \right\rangle & \left| \frac{1}{2},~0 \right\rangle \xrightarrow{185.3} \left| \frac{1}{2},~-1 \right\rangle & \left| \frac{1}{2},~1 \right\rangle \xrightarrow{182.6} \left| \frac{-1}{2},~0 \right\rangle \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.26%3A_HD-Like_NMR_Spectrum_Calculated_Using_Tensor_Algebra.txt
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A Michelson interferometer with a movable mirror in one arm is an essential element in Fourier Transform (FT) spectroscopy.
A source illuminates a 50‐50 beam splitter (BS). After the beam splitter the photons are in a linear superposition of being both transmitted and reflected. By convention a 90o phase shift is assigned to the reflected beam.
$| S \rangle \rightarrow \frac{1}{ \sqrt{2}} \left[ |T \rangle + i | R \rangle \right] \nonumber$
Mirrors return the reflected and transmitted photons to the beam splitter where they evolve into the following superpositions.
$| R \rangle \rightarrow \frac{1}{ \sqrt{2}} \left[ |D \rangle + i | S \rangle \right] \nonumber$
$| T \rangle \rightarrow \frac{ \text{exp} \left( i \frac{2 \pi \delta}{ \lambda} \right)}{ \sqrt{2}} \left[ S \rangle + i|D \rangle \right] \nonumber$
The reflected beam is in a superposition of being transmitted to the detector and reflected back to the source, while the transmitted beam is in a superposition of being transmitted to the source and reflected to the detector. The exponential term in equation (3) is the phase shift acquired by the transmitted beam due to any path difference between the two arms created by the movable mirror.
The history of the source photons is obtained by substitution of equations (2) and (3) into equation (1).
$|S \rangle \rightarrow \frac{1}{2} \left[ \left( \text{exp} \left( i \frac{2 \pi \delta}{ \lambda} \right) - 1 \right) |S \rangle + i \left( \text{exp} \left( i \frac{2 \pi \delta}{ \lambda} \right) +1 \right) |D \rangle \right] \nonumber$
The intensity of the radiation arriving at the detector (D) is the absolute magnitude squared of its coefficient in equation(4).
$I_D ( \delta) = \left| i \left( \text{exp} \left( i \frac{2 \pi \delta}{ \lambda} \right) + 1 \right) \right|^2 = \frac{1}{2} \left[ \cos \frac{2 \pi \delta}{ \lambda} +1 \right] \nonumber$
Thus for a single‐frequency source the intensity at the detector as a function of δ traces a cosine function.
In FT‐IR spectroscopy the source is a ʺglowerʺ that emits a broad range of infrared frequencies. In this case the signal at the detector is the sum of of intensities of the individual photons. This is called an interferogram and has the following mathematical form.
$I ( \delta) = \sum_j \left| i \left( \text{exp} \left( i \frac{2 \pi \delta}{ \lambda_j} \right) +1 \right) \right|^2 = \sum_j \frac{1}{2} \left[ \cos \left( \frac{2 \pi \delta}{ \lambda_j} \right) + 1 \right] \nonumber$
The interferogram is obtained by measuring the intensity at the detector as a function of the path difference (δ) also called the retardation. In the next step the intensity of the frequencies present in the source beam are recovered by the following Fourier Transform.
$B( \nu) = \int I ( \delta) \text{exp} (i 2 \pi \delta \nu) d \delta \nonumber$
To illustrate the FT method we will assume that the source beam only contains four wavelengths or frequencies.
$\begin{matrix} \text{Wavelength:} & \lambda_1 = 20 & \lambda_2 = 10 & \lambda_3 = 5 & \lambda_4 = 2.5 \ \text{Intensities:} & B_1 = 2 & B_2 = 5 & B_3 = 3 & B_4 = 6 & \text{Frequencies:} & \frac{1}{ \lambda_1} = 0.05 & \frac{1}{ \lambda_2} = 0.1 & \frac{1}{ \lambda_3} = 0.2 & \frac{1}{ \lambda_4} = 0.4 \end{matrix} \nonumber$
Generate interferogram:
$\begin{matrix} I ( \delta) = & \frac{1}{2} \left( \cos \left( 2 \pi \frac{ \delta}{ \lambda_1} \right) +1 \right) B_1 + \frac{1}{2} \left( \cos \left( 2\pi \frac{ \delta}{ \lambda_2} \right) + 1 \right) B_2 ... \ & + \frac{1}{2} \left( \cos \left( 2 \pi \frac{ \delta}{ \lambda_3} \right) +1 \right) B_3 + \frac{1}{2} \left( \cos \left( 2 \pi \frac{ \delta}{ \lambda_4} \right) +1 \right) B_4 \end{matrix} \nonumber$
Display interferogram:
$\delta = 0, .1 ... 200 \nonumber$
Fourier transform interferogram to recover frequencies and intensities:
$\begin{matrix} B( \nu) = \int_{0}^{200} I ( \delta) \text{exp}(i 2 \pi \delta \nu) d \delta & \nu = 0, 0.005 .. .5 \end{matrix} \nonumber$
This calculation reveals how a Michelson interferometer is used to measure the wavelengths or frequencies present in the source beam. In a FT‐IR instrument the source ideally contains all the IR frequencies. The sample is placed between the beam splitter and the detector and an absorption spectrum is measured.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.27%3A_The_Michelson_Interferometer_and_Fourier_Transform_Spectroscopy.txt
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A Michelson interferometer with a movable mirror in one arm is an essential element in a Fourier transform infrared (FTIR) spectrophotometer. Feynman's "sum over histories" approach to quantum mechanics will be used to analyze its operation. The analysis is similar to that used for the Mach-Zehnder interferometer in Chapter 6.
There are two photon paths (histories) to the detector from the source. The photon can be transmitted by the beam splitter into the arm with the movable mirror, returned to the beam splitter by the mirror, and then reflected at the beam splitter to the detector. And it can be reflected at the beam splitter into the arm with the fixed mirror, returned to the beam splitter, and then transmitted to the detector. The probability amplitudes for transmission and reflection at the beam splitter are given below, along with the phase shift that occurs due to any path difference in the arms of the interferometer.
Probability amplitude for photon transmission at a 50-50 beam splitter: $\langle T|S \rangle = \frac{1}{ \sqrt{2}}$
Probability amplitude for photon reflection at a 50-50 beam splitter: $\langle R|S \rangle = \frac{i}{ \sqrt{2}}$
Phase shift due to the path difference between the beam splitter arms: $\text{exp} \left( i 2 \pi \frac{ \delta}{ \lambda} \right)$
The probability amplitude for photon detection is given below the figure. Using the information above yields the following expression.
$\frac{1}{ \sqrt{2}} \text{exp} \left( i 2 \pi \frac{ \delta}{ \lambda} \right) \frac{i}{ \sqrt{2}} + \frac{i}{ \sqrt{2}} \frac{1}{ \sqrt{2}} \nonumber$
The probability the photon will arrive at the detector is the magnitude squared of the probability amplitude.
$\left( \left| \frac{1}{ \sqrt{2}} \text{exp} \left( i 2 \pi \frac{ \delta}{ \lambda} \right) \frac{i}{ \sqrt{2}} + \frac{i}{ \sqrt{2}} \frac{1}{ \sqrt{2}} \right| \right)^2 \text{ simplify } \rightarrow \frac{1}{2} \cos \left( 2 \pi \frac{ \delta}{ \lambda} \right) + \frac{1}{2} \nonumber$
For a single wavelength source the intensity at the detector as a function of δ traces a cosine.
In FTIR spectroscopy the source is a "glower" that emits a broad range of infrared frequencies. In this case the signal at the detector is the sum of of intensities of the individual photons. This is called an interferogram and has the following mathematical form.
$I ( \delta) = \sum_j \left( \frac{1}{2} \cos \left( 2 \pi \frac{ \delta}{ \textcolor{red}{ \lambda_j }} \right) + \frac{1}{2} \right) \nonumber$
The interferogram is obtained by measuring the intensity at the detector as a function of the path difference (δ) also called the retardation. In the next step the intensity of the frequencies present in the source beam are recovered by the following Fourier transform. The mirror moves at a constant speed, so δ is proportional to time. Therefore, this is a time-frequency Fourier transform.
$B( \nu) = \int I ( \delta) \text{exp} (i 2 \pi \delta \nu ) d \delta \nonumber$
To illustrate the FT method we will assume that the source beam only contains four wavelengths or frequencies.
$\begin{matrix} \text{Wavelengths:} & \lambda_1 = 20 & \lambda_2 = 10 & \lambda_3 = 5 & \lambda_4 = 2.5 \ \text{Intensities:} & B_1 = 3 & B_2 = 5 & B_3 = 4 & B_4 = 6 \ \text{Frequencies:} & \frac{1}{ \lambda_1} = 0.05 & \frac{1}{ \lambda_2} = 0.1 & \frac{1}{ \lambda_3} = 0.2 & \frac{1}{ \lambda_4} = 0.4 \end{matrix} \nonumber$
Generate interferogram:
$\begin{matrix} I ( \delta) = & \frac{1}{2} \left( \cos \left( 2 \pi \frac{ \delta}{ \lambda_1} \right) +1 \right) B_1 + \frac{1}{2} \left( \cos \left( 2\pi \frac{ \delta}{ \lambda_2} \right) + 1 \right) B_2 ... \ & + \frac{1}{2} \left( \cos \left( 2 \pi \frac{ \delta}{ \lambda_3} \right) +1 \right) B_3 + \frac{1}{2} \left( \cos \left( 2 \pi \frac{ \delta}{ \lambda_4} \right) +1 \right) B_4 \end{matrix} \nonumber$
Display interferogram:
$\delta = 0, .1 .. 200 \nonumber$
Fourier transform interferogram to recover frequencies and intensities:
$\begin{matrix} B ( \nu ) = \int_0^{200} I ( \delta) \text{exp} (i 2 \pi \delta \nu) d \delta & \nu = 0, .005 .. 5 \end{matrix} \nonumber$
This calculation reveals how a Michelson interferometer is used to measure the wavelengths or frequencies present in the source beam. In a FT‐IR instrument the source ideally contains all the IR frequencies. The sample is placed between the beam splitter and the detector and an absorption spectrum is measured.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.28%3A_A_Sum_Over_Histories_Approach_to_Fourier_Transform_Infrared_Spectroscopy.txt
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From previous work we know that the momentum eigenfunction in coordinate space is given by
$\langle x | p \rangle = \frac{1}{ \sqrt{2 \pi}} \text{exp} \left( \frac{ipx}{ \hbar} \right) \nonumber$
Quantum mechanical wave functions must be single-valued, which in this application (electron on a ring) requires a cyclical boundary condition: the wave function must match at points separated by one circumference, C = 2πR.
$\begin{matrix} \langle x + C|p \rangle = \langle x | p \rangle \ \text{exp} \left( \frac{pi(x+C)}{ \hbar} \right) = \text{exp} \left( \frac{ipx}{ \hbar} \right) \ \text{exp} \left( \frac{}{} \right) \text{exp} \left( \frac{pix}{ \hbar} \right) = \text{exp} \left( \frac{ipC}{ \hbar} \right) \ \text{exp} \left( \frac{ipC}{ \hbar} \right) = 1 \end{matrix} \nonumber$
This equation is satisfied if $\frac{pC}{ \hbar} = 2 \pi m$ where $m = 0, \pm 1, \pm 2...$ You can easily verify this with Mathcad. Thus the quantum number is required in order to satisfy the cyclic boundary condition. Next we calculate the kinetic energy and express it in terms of the circumference of the ring.
$T_m = \frac{p^2}{2m_e} = \frac{m^2h^2}{2m_eC^2} ~ \text{ where } m = 0, \pm 1, \pm 2 ... \nonumber$
Substitution of $\frac{2 \pi m \hbar}{C}$ for p and recognizing that $\phi = \frac{2 \pi x}{C}$ allow one to transform the first equation to,
$\langle \phi | m \rangle = \frac{1}{ \sqrt{2 \pi}} \text{exp} (i m \hi) \nonumber$
1. Sketch an energy level diagram for a particle on a ring.
2. Use the aufbau principle and the Pauli exclusion principle to place the π-electrons of benzene in the allowed energy levels.
3. Calculate the wavelength of the lowest allowed electronic transition (HOMO ÷ LUMO). The average c - c bond length in benzene is 140 pm.
4.30: Modeling the Pi-electrons of Benzene as Particles on a Ring - Version 2
Wave-particle duality for massive objects as expressed by the de Broglie equation (λ = h/mv = h/p) is a foundational concept of quantum theory. Classical potential energy carries over to quantum mechanics unchanged, but classical kinetic energy is, as shown below, transformed into a quantum mechanical confinement energy by the de Broglie relation. Confined objects with wave properties are subject to interference which restricts the allowed values of λ, which in turn leads to energy quantization.
$T = \frac{p^2}{2m} \xrightarrow{p= \frac{h}{ \lambda}} \frac{h^2}{2m \lambda^2} \nonumber$
A general one-dimensional function, $\Psi (x) = \frac{1}{ \sqrt{2 \pi}} \text{exp} \left( i 2 \pi \frac{x}{ \lambda} \right)$, is used to represent the wave character of a particle on a ring (POR) where in order to avoid self-interference the head and tail of the wave function must match after one revolution.
$\begin{matrix} \Psi (x) = \Psi (x + C) & C = 2 \pi R \end{matrix} \nonumber$
This requirement restricts the allowed values of the wavelength and leads to a manifold of quantized energies as is now demonstrated. The structure of the manifold of allowed energies depends on the nature of the confinement, each problem gives a characteristic energy manifold. For the particle on a ring we find,
$\frac{1}{ \sqrt{2 \pi}} \text{exp} \left( i 2 \pi \frac{x}{ \lambda} \right) = \frac{1}{ \sqrt{2 \pi }} \text{exp} \left( i 2 \pi \frac{x+C}{ \lambda} \right) = \frac{1}{ \sqrt{2 \pi}} \text{exp} \left( i 2 \pi \frac{x}{ \lambda} \right) \frac{1}{ \sqrt{2 \pi}} \text{exp} \left( i 2 \pi \frac{C}{ \lambda} \right) \nonumber$
$\begin{matrix} \text{It follows that} & \frac{1}{ \sqrt{2 \pi}} \text{exp} \left( i 2 \pi \frac{C}{ \lambda} \right) = 1 & \text{which requires} & \lambda = \frac{C}{n} & \text{where n} = 0,~ \pm1,~ \pm 2 ...\end{matrix} \nonumber$
Substitution of this restriction for λ into the quantum expression for kinetic energy yields an equation for the allowed energy states in terms of Planck's constant, the particle mass, the ring circumference and the quantum number, n.
$T = \frac{h^2}{2m \lambda^2} \text{substitute,}~ \lambda = \frac{C}{n} \rightarrow T = \frac{h^2 n^2}{2 C^2 m} \nonumber$
An obvious POR application is to treat the π-electrons of benzene as particles on a ring. The energy level diagram is constructed using the above energy expression and the allowed values for the quantum number, n. Then the energy level diagram is populated with six π-electrons using the aufbau principle and the Pauli exclusion principle.
The validity of the model is tested by calculating the wavelength of the photon required for the HOMO-LUMO transition. The average c-c bond length in benzene is 140 pm, so the ring circumference is approximated as 6x140 pm. As shown below the photon wavelength required for the HOMO-LUMO transition is 194 nm, a value that might be described as "in the ball park."
Energy conservation for the HOMO-LUMO transition requires:
$\frac{n_i^2 h^2}{2m_eC^2} + \frac{hc}{ \lambda} = \frac{n_f^2 h^2}{2m_eC^2} \nonumber$
Fundamental constants and conversion factors:
$\begin{matrix} pm = 10^{-12}m & aJ = 10^{-18} J \end{matrix} \nonumber$
$\begin{matrix} h = 6.626-755 (10^{-34} \text{joule sec} & c = 2.99792458 (10^8) \frac{m}{sec} & m_e = 9.1093897 (10^{-31}) kg \end{matrix} \nonumber$
Calculate the photon wavelength for the HOMO-LUMO electronic transition.
$\begin{matrix} \text{HOMO:} & n_i = 1 & \text{LUMO:} & n_f = 2 & \text{Benzene circumference:} & C = 6(140) pm \end{matrix} \nonumber$
$\begin{array}{c|c} \lambda = \frac{n_i^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{n_f^2 h^2}{2 m_e C^2} & _{ \text{solve, } \lambda}^{ \text{float, 3}} \rightarrow \frac{1.94e-7 \text{kg m}^3}{ \text{joule sec}^2 } ~ \lambda = 194 nm \end{array} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.29%3A_Modeling_the_Pi-electrons_of_Benzene_as_Particles_on_a_Ring.txt
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Calculate the wavelength of the photon required for the first allowed (HOMO-LUMO) electronic transition involving the π−electrons of benzene.
Energy conservation requirements:
$\frac{n_i^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{n_f^2 h^2}{2 m_e C^2} \nonumber$
Fundamental constants and conversion factors:
$\begin{matrix} pm = 10^{-12} m & aJ = 10^{-18} J \end{matrix} \nonumber$
$\begin{matrix} h = 6.6260755 (10^{-34}) \text{joule sec} & c = 2.99792458 (10^8) \frac{m}{sec} & m_e = 9.1093897 (10^{-31}) kg \end{matrix} \nonumber$
Calculate the photon wavelength for the HOMO-LUMO electronic transition.
$\begin{matrix} \text{HOMO:} & n_i = 1 & \text{LUMO:} & n_f = 2 & \text{Benzene circumference:} & C = 6(140) pm \end{matrix} \nonumber$
$\begin{array}{c|c} \lambda = \frac{n_i^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{n_f^2 h^2}{2 m_e C^2} & _{ \text{solve, } \lambda} ^{ \text{float, 3}} \rightarrow .194e-6m^3 \frac{kg}{ joule~ sec^2} ~ \lambda = 194 nm \end{array} \nonumber$
Calculate the photon energy and frequency.
$\begin{matrix} \text{energy} & \frac{c}{ \lambda} = 1.024 aJ & \text{frequency} & \frac{c}{ \lambda} = 1.545 \times 10^{15} Hz \end{matrix} \nonumber$
Plot Wave Functions
See Figure 7.6 (page 111) in Quantum Chemistry and Spectroscopy, by Engel.
The real part of the wave function is plotted below.
Quantum number: n = 5
$\begin{matrix} \text{numpts} = 100 & i = 0 .. \text{numpts} & j = 0 .. \text{numpts} & \phi_i = \frac{2 \pi i}{ \text{numpts}} \ x_{i,~j} = \cos \left( \phi_i \right) & y_{i,~j} = \sin \left( \phi_i \right) & z_{i,~j} = \frac{1}{ \sqrt{2 \pi}} \text{exp} \left(i n \phi_i \right) & zz_{i,~j} = 0 \end{matrix} \nonumber$
The square of the absolute magnitude for all the wave functions (for all values of the quantum number n) is 1/2π, as shown below.
$\begin{matrix} \left( \left| \frac{1}{ \sqrt{2 \pi}} \text{exp} (i n \phi ) \right| \right)^2 & \text{simplifies to} & \frac{1}{2 \pi} \end{matrix} \nonumber$
The wave functions for the electron on a ring are eigenstates of the momentum operator. In other words the momentum is precisely known: p = nh/C, where n is the quantum number and C is the ring circumference. According to the uncertainty principle, the elctron position must be uncertain. The result above confirms this; the electron density is distributed uniformly over the entire ring.
4.32: Modeling the Pi-electrons of Benzene as Particles in a Ring
In this exercise benzene's six π electrons will be modeled as particles in a ring or circular corral. Schrödinger's equation in plane polar coordinates and its energy eigenvalues are given below. R is the ring radius and C the ring circumference.
$\begin{matrix} \frac{-h^2}{8 \pi^2 m_e} \left( \frac{d^2}{dr^2} \Psi (r) + \frac{1}{r} \frac{d}{dr} \Psi (r) - \frac{L^2}{r^2} \Psi (r) \right) = E \Psi (r) & E_{n,~L} = \frac{ \left( Z_{n,~L} \right)^2 h^2}{8 \pi^2 m_e R^2} = \frac{ \left( Z_{n,~L} \right)^2 n^2}{2 m_e C^2} \end{matrix} \nonumber$
JL is the Lth order Bessel function, L is the angular momentum quantum number, n is the principle quantum number, Zn,L is the nth root of JL. Dirac notation is used to describe the electronic states, |n,L>. The roots of the Bessel function are given below in terms of the n and L quantum numbers.
$\begin{array} & \text{L quantum number} \ Z = \begin{pmatrix} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & "n" \ 2.405 & 3.832 & 5.316 & 6.380 & 7.588 & 8.771 & 9.936 & 11.086 & 1 \ 8.654 & 10.173 & 11.620 & 13.015 & 14.373 & 15.700 & 17.004 & 18.288 & 3 \ 11.792 & 13.324 & 14.796 & 16.223 & 17.616 & 18.980 & 20.321 & 21.642 & 4 \ 14.931 & 16.471 & 17.960 & 19.409 & 20.827 & 22.218 & 23.586 & 24.935 & 5 \end{pmatrix} & \text{n quantum number} \end{array} \nonumber$
The manifold of allowed energy levels up to the LUMO is shown below and is populated with 6 π electrons. Note that the states with L > 0 are doubly degenerate.
The photon wavelength required for the first electronic transition involving the π electrons is now calculated. The ring circumference is approximated as six benzene carbon-carbon bond lengths.
$\begin{matrix} h = 6.6260755 (10^{-34}) \text{joule sec} & c = 2.99792458 (10^8) \frac{m}{sec} & m_e = 9.1093897 (10^{-31}) kg & pm = 10^{-12} m \end{matrix} \nonumber$
$\begin{matrix} C = 6(140) pm & \begin{array}{c|c} \frac{ \left( Z_{1,~1} \right)^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{ \left( Z_{1,~2} \right)^2 h^2}{2 m_e C^2} & _{ \text{solve, } \lambda}^{ \text{float, 3}} \rightarrow \frac{4.28e-8 kg~m^3}{joule~sec^2} = 42.8 nm \end{array} \end{matrix} \nonumber$
Benzene has a strong electronic transition at about 200 nm.
4.33: Modeling the Pi-electrons of Corannulene as Particles in a Ring
In this exercise the 20 π electrons of corannulene will be modeled as particles in a ring or circular corral. Corannulene is bowl-shaped not planar, so the model has some initial difficiencies.
Schrödinger's equation in plane polar coordinates and its energy eigenvalues are given below. R is the ring radius and C the ring circumference.
$\begin{matrix} \frac{-h^2}{8 \pi^2 m_e} \left( \frac{d^2}{dr^2} \Psi (r) + \frac{1}{r} \frac{d}{dr} \Psi (r) - \frac{L^2}{r^2} \Psi (r) \right) = E \Psi (r) & E_{n,~L} = \frac{ \left( Z_{n,~L} \right)^2 h^2}{8 \pi^2 m_e R^2} = \frac{ \left( Z_{n,~L} \right)^2 h^2}{2 m_e C^2} \end{matrix} \nonumber$
JL is the Lth order Bessel function, L is the angular momentum quantum number, n is the principle quantum number, Zn,L is the nth root of JL. Dirac notation is used to describe the electronic states, |n,L>. The roots of the Bessel function are given below in terms of the n and L quantum numbers.
$\begin{array} & \text{L quantum number} \ Z = \begin{pmatrix} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & "n" \ 2.405 & 3.832 & 5.316 & 6.380 & 7.588 & 8.771 & 9.936 & 11.086 & 1 \ 8.654 & 10.173 & 11.620 & 13.015 & 14.373 & 15.700 & 17.004 & 18.288 & 3 \ 11.792 & 13.324 & 14.796 & 16.223 & 17.616 & 18.980 & 20.321 & 21.642 & 4 \ 14.931 & 16.471 & 17.960 & 19.409 & 20.827 & 22.218 & 23.586 & 24.935 & 5 \end{pmatrix} & \text{n quantum number} \end{array} \nonumber$
The manifold of allowed energy levels up to the LUMO is shown below and is populated with 20 π electrons. Note that the states with L > 0 are doubly degenerate.
Corannulene has a strong electronic transition at 280 nm. This information will be used to calculate its circumference.
$\begin{matrix} h = 6.6260755 (10^{-34}) \text{joule sec} & c = 2.99792458 (10^8) \frac{m}{sec} & m_e = 9.1093897 (10^{-31}) kg & pm = 10^{-12} m \end{matrix} \nonumber$
$\begin{matrix} \lambda = 280 nm & \begin{array}{c|c} \frac{ \left( Z_{2,~1} \right)^2 h^2}{2 m_e C^2} + \frac{hc}{ \lambda} = \frac{ \left( Z_{1,~4} \right)^2 h^2}{2 m_e C^2} & _{ \text{solve, C}}^{ \text{float, 3}} \rightarrow \begin{pmatrix} - \frac{0.0000533 \sqrt{joule} \sqrt{nm} sec}{ \sqrt{kg} \sqrt{m}} \ \frac{0.0000533 \sqrt{joule} \sqrt{nm} sec}{ \sqrt{kg} \sqrt{m}} \end{pmatrix} = \begin{pmatrix} -1.685 \times 10^3 \ 1.685 \times 10^3 \end{pmatrix} pm \end{array} \end{matrix} \nonumber$
The result is reasonable given that there are 15 c-c bonds on the circumference:
$15 (140) pm = 2.1 \times 10^3 pm \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.31%3A_Calculating_the_Pi-electron_HOMO-LUMO_Electronic_Transition_for_Benzene.txt
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The first step in using Mathcad to do group theory is to enter the character table in matrix form for the symmetry under study. In this example we shall be looking at C60 and so the character table for the icosahedral point group has been entered in the matrix CIh below. The icosahedhral group has 120 symmetry operations which fall into ten classes. Ih is a vector containing the number of symmetry operations in each of the ten classes. Γuma (unmoved atoms) is a reducible representation showing the behavior of the atoms under the symmetry operations of the icosahedral group.2 In particular it records the number of atoms that are unmoved by the symmetry operations of the group. Because C60 is a cage molecule with no central atom, only the identity operation and the symmetry planes leave atoms unmoved. Obviously the identity operation leaves all sixty carbon atoms unmoved and inspection of the molecule shows that the symmetry planes each contain four atoms which are unmoved by reflection in the plane.
$\begin{matrix} \begin{array} E & E & C_5 & C_5^2 & & C_3& C_2& & i& & S_{10} & & S_{10}^3 & & S_6 & & \sigma \end{array} & ~ \ \text{CIh} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 \ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 \ 4 & -1 & -1 & 1 & 0 & 4 & -1 & -1 & 1 & 0 \ 5 & 0 & 0 & -1 & 1 & 5 & 0 & 0 & -1 & 1 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 & -3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & 1 \ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 & -3 & - \frac{1 - \sqrt{5}}{2} & - \frac{1 + \sqrt{5}}{2} & 0 & 1 \ 4 &-1 & -1 & 1 & - & -4 & 1 & 1 & -1 & 0 \ 5 & 0 & 0 & -1 & 1 & -5 & 0 & 0 & 1 & -1 \end{pmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 - x^2- y^2,~x^2 - y^2, xy, yz, xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} \end{matrix} \nonumber$
$\begin{matrix} Ih = \begin{pmatrix} 1 \ 12 \ 12 \ 20 \ 15 \ 1 \ 12 \ 12 \ 20 \ 15 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 60 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 4 \end{pmatrix} \end{matrix} \nonumber$
At this point it is necessary to identify the irreducible representations with a row in the matrix representing the character table for the icosahedral group. It should be noted that these irreducible representations could also have been entered individually as column or row vectors.
$\begin{matrix} Ag = \left( CIh^T \right)^{<1>} & T1g = \left( CIh^T \right)^{<2>} & T2g = \left( CIh^T \right)^{<3>} & Gg = \left( CIh^T \right)^{<4>} & Hg = \left( CIh^T \right)^{<5>} \ Ag = \left( CIh^T \right)^{<1>} \ Au = \left( CIh^T \right)^{<6>} & T1u = \left( CIh^T \right)^{<7>} & T2u = \left( CIh^T \right)^{<8>} & Gu = \left( CIh^T \right)^{<9>} & Hu = \left( CIh^T \right)^{<10>} \end{matrix} \nonumber$
The order of the group (the normalization constant) is the sum of the total number of symmetry operations.
$\begin{matrix} h = \sum Ih & h = 120 \end{matrix} \nonumber$
The irreducible representations are a set of orthonormal basis vectors that span the Ih space. This is demonstrated below for several cases.
$\begin{matrix} \frac{ \sum \overrightarrow{(Ih AgAg)}}{h} = 1 & \frac{ \sum \overrightarrow{(Ih T1g T1g)}}{h} = 1 & \frac{ \sum \overrightarrow{(Ih Ag T1u)}}{h} = 0 & \frac{ \sum \overrightarrow{(Ih Gu Hu)}}{h} = 0 \end{matrix} \nonumber$
These vector operations are the symmetry equivalent of the quantum mechanical overlap integrals.
Vibrational Spectroscopy
One of the major goals of a symmetry analysis is to determine how the individual degrees of freedom of the molecule behave under the symmetry operations of the group. This is particularly important in interpreting or predicting the infrared or Raman spectrum of a molecule. The easiest way to find the behavior (reducible representation) of C60's 180 degrees of freedom under the symmetry operations of the Ih group is to take the direct product of Γuma with the T1u irreducible representation for translation in the x, y, and z directions.2
$\begin{matrix} \Gamma_{tot} = \overrightarrow{ \left( \Gamma_{uma} T1u \right)} & \Gamma_{tot}^T = \begin{pmatrix} 180 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix} \end{matrix} \nonumber$
Now it is possible to find which irreducible representations contribute to Γtot by calculating the normalized vector sums shown below.3 The irreducible representations are the unit vectors of Ih space and any other vector in that space can be written as a linear combination of the unit vectors. Taking the product of each irreducible representation with the reducible representation, Γtot, yields the appropriate coefficient in the linear combination.
$\begin{matrix} \frac{ \sum \overrightarrow{(Ih Ag \Gamma_{tot})}}{h} - 2 & \frac{ \sum \overrightarrow{(Ih T1g \Gamma_{tot})}}{h} - 4 & \frac{ \sum \overrightarrow{(Ih Ag \Gamma_{tot})}}{h} - 4 & \frac{ \sum \overrightarrow{(Ih Gu \Gamma_{tot})}}{h} - 6 & \frac{ \sum \overrightarrow{(Ih Hg \Gamma_{tot})}}{h} - 8 \ \frac{ \sum \overrightarrow{(Ih Au \Gamma_{tot})}}{h} = 1 & \frac{ \sum \overrightarrow{(Ih T1u \Gamma_{tot})}}{h} = 5 & \frac{ \sum \overrightarrow{(Ih T2u \Gamma_{tot})}}{h} = 5 & \frac{ \sum \overrightarrow{(Ih Gu \Gamma_{tot})}}{h} = 6 & \frac{ \sum \overrightarrow{(Ih Hu \Gamma_{tot})}}{h} = 7 \end{matrix} \nonumber$
This result can also be computed more compactly as follows:
$\begin{matrix} i = 1 .. 10 & X_i = \frac{ \sum \left( \overrightarrow{ \left[ Ih \left( CIh^T \right)^{<i>} \Gamma_{tot} \right]} \right)}{h} & X^T = \begin{pmatrix} 2 & 4 & 4 & 6 & 8 & 1 & 5 & 5 & 6 & 7 \end{pmatrix} \end{matrix} \nonumber$
Of the 180 degrees of freedom that C60 has, three are due to translation of the center of mass in Cartesian space (T1u) and three are due to rotation about the Cartesian axes (T1g). This leaves 174 vibrational degrees of freedom. They have the following symmetry properties.
$\Gamma_{vib} = \Gamma_{tot} - T1u - T1g \nonumber$
$\begin{matrix} i = 1 .. 10 & X_i = \frac{ \sum \left( \overrightarrow{ \left[ Ih \left( CIh^T \right)^{<i>} \Gamma_{vib} \right]} \right)}{h} & X^T = \begin{pmatrix} 2 & 3 & 4 & 6 & 8 & 1 & 4 & 5 & 6 & 7 \end{pmatrix} \end{matrix} \nonumber$
One important consequence of the high symmetry of C60 is that many of the vibrational modes are degenerate. As the result above shows there are only 46 distinct vibrational modes. Another consequence of the symmetry of C60 is that most of these modes are not IR or Raman active.4 In fact only the four T1u vibrational modes are IR active and they appear at 528, 577, 1180, and 1430 cm‐1 with varying intensities. Sample calculations for the transition moment for IR transitions are shown below. Those with transition moments of zero are forbidden.
$\begin{matrix} \frac{ \sum \overrightarrow{(Ih T1g T1u Ag)}}{h} = 1 & \frac{ \sum \overrightarrow{(Ih T1u T1u Ag)}}{h} = 1 & \frac{ \sum \overrightarrow{(Ih Gg T1u Ag)}}{h} = 0 & \frac{ \sum \overrightarrow{(Ih Hu T1u Ag)}}{h} = 0 \end{matrix} \nonumber$
These vector multiplications are the symmetry representations of the quantum mechanical integrals for transition moments. For the transition moment to have a non‐zero value, the vector product must span the totally symmetric irreducible representation, Ag. In the examples shown above Ag is the symmetry of the ground vibrational state for all vibrational modes, T1u is the symmetry of the electric dipole operator, and T1g, T1u, Gg, and Hu are the irreducible representations for the vibrational states to be excited by the infrared radiation. As these calculations show, only the T1u vibrational modes have a non‐zero transition moment.
However, it is important to realize that the group theoretical analysis given above, and those that follow, can only decide which transitions are forbidden. They do not provide the transition probabilities of the transitions which are not forbidden. For example, the fact that the second vector sum above is 1 says nothing about the intensity of the actual absorption in the infrared, only that the vibrational modes with T1u symmetry are not forbidden.
There are ten Raman active modes, two modes with Ag symmetry and eight with Hg symmetry. They appear at the following frequencies: 273, 436, 496, 710, 773, 1100, 1250, 1435, 1470, and 1570 cm‐1. 4 The calculation is similar to that for IR activity except that the operators for the Raman interaction are the quadratic forms (the components of the polarizability tensor) which transform as Ag and Hg, as shown in the Ih character table.
$\begin{matrix} \frac{ \sum \overrightarrow{(Ih Ag Ag Ag)}}{h} = 1 & \frac{ \sum \overrightarrow{(Ih T1u Ag Ag)}}{h} = 0 & \frac{ \sum \overrightarrow{(Ih T1u Hg Ag)}}{h} = 0 & \frac{ \sum \overrightarrow{(Ih Hg Hg Ag)}}{h} = 1 \end{matrix} \nonumber$
In these sample calculations we see the basis for the widely used generalization that for a vibrational mode to be IR active it must have the same symmetry as one of the Cartesian coordinates and to be Raman active it must have the same symmetry as one of the components of the polarizability tensor. We also see quite clearly that for a molecule with a center of inversion there is no overlap between the IR and Raman active vibrational modes. In the appendix it is shown that three of the IR active modes and seven of the Raman active modes are stretching vibrations.
Electronic States of the π Electrons
To a first approximation the bonding in C60 might be described as follows: each carbon forms σ bonds to its three neighboring carbon atoms using sp2 hybridized orbitals with the remaining p orbital on each carbon available for π bonding. It is customary in approximate models to treat the π electrons independently of the σ electrons, just as chemists generally assume that they can treat valence electrons independently of the core electrons. The behavior of the π electrons under the symmetry operations of the icosahedral point group is easy to determine. Because the π orbitals are centered on the carbon atoms and are perpendicular to the sp2 "plane", they have the same symmetry properties as the carbon atoms.
$\Gamma \pi = \Gamma_{uma} \nonumber$
We now decompose Γπ into its irreducible representations using the method outlined above.
$\begin{matrix} \frac{ \sum \overrightarrow{Ih Ag \Gamma_{ \pi}}}{h} = 1 & \frac{ \sum \overrightarrow{Ih T1g \Gamma_{ \pi}}}{h} = 1 & \frac{ \sum \overrightarrow{Ih T2g \Gamma_{ \pi}}}{h} = 1 & \frac{ \sum \overrightarrow{Ih Ag \Gamma_{ \pi}}}{h} = 1 & \frac{ \sum \overrightarrow{Ih Gg \Gamma_{ \pi}}}{h} = 2 & \frac{ \sum \overrightarrow{Ih Hg \Gamma_{ \pi}}}{h} = 3 \ \frac{ \sum \overrightarrow{Ih Au \Gamma_{ \pi}}}{h} = 0 & \frac{ \sum \overrightarrow{Ih T1u \Gamma_{ \pi}}}{h} = 2 & \frac{ \sum \overrightarrow{Ih T2u \Gamma_{ \pi}}}{h} = 2 & \frac{ \sum \overrightarrow{Ih Gu \Gamma_{ \pi}}}{h} = 2 & \frac{ \sum \overrightarrow{Ih Hu \Gamma_{ \pi}}}{h} = 2 & \end{matrix} \nonumber$
Or calculated more compactly:
$\begin{matrix} i = 1 .. 10 & X_i = \frac{ \sum \left( \overrightarrow{ \left[ Ih \left( CIh^T \right)^{<i>} \Gamma_{uma} \right]} \right)}{} & X^T \end{matrix} \nonumber$
$Gamma_{uma} = Ag + T1g + T2g + 2Gg + 3Hg + 2T1u + 2T2u + 2Gu + 2Hu \nonumber$
Just as symmetry arguments cannot predict the actual intensities of allowed infrared or Raman transitions, they also can't predict the energy order of these π electron energy levels. However, group theory combined with a Huckel calculation yields the following order.5
Ag < T1u < Hg < T2u < Gu < (Gg + Hg) < Hu < T1u < T1g < Hg < T2u < Hu < Gg < Gu < T2g
With 60 π electrons to distribute in accordance with the Aufbau and Exclusion Principles, the Hu level is the HOMO, the T1u level is the LUMO, and the T1g level is the LUMO +1. This ordering of the levels is consistent with the basic magnetic and electronic properties of C60. For example, it is diagmagnetic with a completely filled HOMO level and it has a reasonably high electron affinity because of the low lying LUMO. The three‐fold orbital degeneracy of the LUMO is consistent with the fact that K3C60 and K6C60 are both known, and K3C60 is paramagnetic while K6C60 is diamagnetic. K4C60 is also known and is diamagnetic due to a Jahn‐Teller distortion that splits the T1u level into a lower two‐fold degenerate level and a higher singlet level.
Calculation of the transition moments for electronic transitions is similar to that for IR and Raman interactions with electromagnetic radiation. The HOMO ‐‐‐> LUMO electronic transition in C60 involves the following change in electronic structure: Hu10 ‐‐‐> Hu9 T1u1. The ground state, Hu10, has Ag symmetry as a fully filled level. The transition moment between this state and Hu9 T1u1 configuration is formulated as shown below because Hu9 has the same symmetry as Hu1. Starting at the right Ag is the symmetry of the ground electronic state, T1u is the symmetry of the electric dipole operator, (HuT1u) represents the symmetry of the manifold of excited electronic states associated with the Hu9T1u1 electronic configuration (see appendix), and Ih is, as mentioned earlier, the vector containing the number of symmetry operations in each symmetry class.
$\frac{ \sum \overrightarrow{ \left[ Ih (Hu T1u)T1u Ag \right]}}{h} = 0 \nonumber$
Thus the HOMO ‐‐‐> LUMO electronic transition is formally forbidden. However, the HOMO ‐‐‐‐> LUMO+1 is allowed as is shown below. In other words the transition moment for this electronic transition spans the Ag irreducible representation
$\frac{ \sum \overrightarrow{ \left[ Ih (Hu T1g)T1u Ag\right]}}{h} = 1 \nonumber$
The (Hu9T1g1) excited state consists of 15 microstates with symmetry T1u, T2u, Gu, and Hu (see appendix). Only the transition to the T1u state is allowed and it occurs at 408 nm.6 In calculating the transition moment for HOMO ‐‐> LUMO electronic transition it has been assumed that there was no change in the vibrational state of the molecule. However, it is possible for forbidden electronic transitions to become weakly allowed through coupling to changes in vibrational state. These are called vibronic transitions and they are allowed if the integral shown below is nonzero.
$\int \int \Psi ex \Psi vx \mu e \Psi eg \Psi vg ~ d \tau e ~ d \tau v \nonumber$
For example, if the formally forbidden HOMO ‐‐‐> LUMO electronic transition is accompanied by a vibrational transition of Au, T1u, T2u, Gu, or Hu symmetry the transition may be allowed (see appendix). The vibronic transition moment would be calculated as follows.
$\frac{ \sum \overrightarrow{ \left[ Ih (Hu T1u)Au T1u Ag Ag \right]}}{h} = 1 \nonumber$
Here (HuT1u) is the manifold of excited electronic states, Au is the excited vibrational state, T1u is the electric dipole operator, Ag is the ground electronic state, and Ag is the ground vibrational state.
Thus while formally forbidden, the HOMO ‐‐‐> LUMO transition is weakly allowed through vibronic coupling. This type of mechanism has been used to interpret the visible spectrum of C60 and recent analyses have put the lowest spin‐allowed, vibronically assisted HOMO ‐‐‐‐> LUMO transition at 620 nm.6
As can be seen, Mathcad provides an efficient programming environment for performing symmetry analyses with any of the finite point groups. Once a Mathcad worksheet is prepared for one molecule it serves as a template for all other molecules with that symmetry. The only changes that need to be made are to the vector Γuma which records how many atoms are unmoved by the symmetry operations of the group. Recently, Greathouse has demonstrated how molecules that belong to the infinite point groups can be analyzed numerically.7
Literature Cited:
1. Mathcad is a product of MathSoft Inc., 101 Main Street, Cambridge, MA 02142.
2. Harris, D. C.; Bertolucci, M. D. Symmetry and Spectroscopy: An Introduction to Vibrational and Electronic Spectroscopy, Dover Publications, Inc.: New York, 1989; Chater 3, pp. 140‐142.
3. Ibid, Chapter 1, page 56.
4. Bethume, D. S.; et. al. Chem. Phys. Lett. 1991, 179, 181‐186.
5. Hebard, A. F. Physics Today 1992, 45(11), 29.
6. Leach, S.; et, al. Chem. Phys. 1992, 160, 451‐466.
7. Greathouse, J. A. The Chemical Educator 1996, Vol. 1, No. 1.
Appendix
This is a demonstration that the Hu9 T1u1 configuration yields the manifold of states, T1g, T2g, Gg, and Hg.
$\begin{matrix} i = 1 .. 10 & X_i = \frac{ \sum \overrightarrow{ \left[ Ih \left(CIh^T \right)^{<i>} (Hu T1u) \right]}}{h} & X^T = \begin{pmatrix} 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{matrix} \nonumber$
This is a demonstration that the Hu9T1g1 configuration yields the manifold of states, T1u, T2u, Gu, and Hu.
$\begin{matrix} X_i = \frac{ \sum \overrightarrow{ \left[ Ih \left(CIh^T \right)^{<i>} (Hu T1g) \right]}}{h} & X^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \end{pmatrix} \end{matrix} \nonumber$
As shown earlier the vibrational modes span all of the irreducible representations. This calculation shows which of those modes provide vibrational coupling for the formally forbidden HOMO ‐‐‐> LUMO electronic transition.
$\begin{matrix} X_i = \frac{ \sum \overrightarrow{ \left[ Ih (Hu T1u) \left(CIh^T \right)^{<i>} T1u Ag Ag \right]}}{h} & X^T = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 3 & 4 \end{pmatrix} \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/04%3A_Spectroscopy/4.34%3A_The_Vibrational_and_Electronic_States_of_C60.txt
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Thumbnail: Diffraction pattern of red laser beam made on a plate after passing through a small circular aperture in another plate. Physical optics is used to explain effects such as diffraction. (CC BY-SA 3.0; Wisky).
05: Diffraction Phenomena
Abstract: Wave-particle duality, the superposition principle, the uncertainty principle, and single-particle interference are the most fundamental quantum mechanical concepts. The purpose of this paper is to demonstrate that these quantum mechanical principles are illuminated by a study of diffraction patterns created with laser light and a variety of two-dimensional masks.
The principles of X-ray crystallography are generally taught using the Bragg model in which X-rays irradiating a crystal behave as classical waves reflected by planes of atoms producing a characteristic diffraction pattern determined by the Bragg interference condition. Optical transform kits developed by the Institute for Chemical Education [1-4] and similar resources [5] have been used to teach these principles through classroom demonstrations of the diffraction patterns generated by laser light and two-dimensional masks. The purpose of this paper is to show that these same teaching aids can be used to teach quantum mechanical principles to chemistry majors in upper-division courses. The quantum concepts illuminated in this approach are
• wave-particle duality,
• the superposition principle,
• the uncertainty principle,
• single-particle interference,
• the quantum mechanical view of experimentation.
Our approach to using optical transforms is an extension of the method Marcella used recently to analyze the famous double-slit experiment [6, 7]. In what follows, the relationship of optical transforms to the quantum mechanical principles listed above is presented as concisely as possible while supporting information is relegated to footnotes and appendices.
Marcella identified three steps in a quantum mechanical experiment: (1) state preparation, (2) measurement of an observable property, and (3) comparison of experimental results with theoretical calculations. We will introduce Marcella's method with a review of the double-slit experiment and then extend this approach to optical transforms of other two-dimensional masks.
According to Richard Feynman the double-slit experiment is the paradigm for all quantum mechanics because it is a simple manifestation of the superposition principle, which is quantum theory is "only mystery" [8]. Stressing the fundamental importance of the double-slit experiment he said that any question in quantum mechanics could always be answered with the response, "You remember the experiment with the two holes? It is the same thing" [9].
In the double-slit experiment the apparatus consists of a source of particles (photons, electrons, neutrons, He atoms, C60 molecules, etc.; see reference 14 and papers cited therein), a slit-screen, and a detection screen. Wave-particle duality is manifested in this experiment as follows. The source creates particles and the detector registers particles, but between source and detector the behavior is wave-like. This is the underlying structure for all quantum mechanical experiments.
State preparation occurs at the slit screen where the particle is position is localized at holes or slits. According to quantum mechanical principles the coordinate-space wave function at the slit screen is a linear superposition describing the particle as present at both slits simultaneously. To begin we will treat the slits as point holes in a screen at (x1, y1) and (x2, y2). This view yields the following coordinate-space wave function [10].
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ x_1,~ y_1 \rangle + |x_2,~y_2 \rangle \right] = \frac{1}{ \sqrt{2}} \sum_{i=1}^2 |x_i,~y_i \rangle \nonumber$
In Marcella's quantum mechanical analysis of the double slit experiment, what is subsequently measured at the detection screen is actually the particle's momentum. In other words, the well-known diffraction pattern created by the double-slit geometry is the particle's momentum distribution in the plane of the detection screen. Therefore, to calculate the diffraction pattern one needs a momentum wave function, and this is obtained by a Fourier transform of eq 1 into momentum space. The primary dynamical variables of position and momentum have a conjugate relationship expressed by the Fourier transform (see Appendix A) shown in eq 2.
$\langle p| \Psi \rangle = \frac{1}{ \sqrt{2}} \sum_{i=1}^{2} \langle p_x | x_i \rangle \langle p_y | y_i \rangle = \frac{1}{ 2 \pi h \sqrt{2}} \sum_{i=1}^{2} exp \left( - \frac{ip_xx_i}{h} \right) exp \left( - \frac{ip_yy_i}{h} \right) \nonumber$
where $\langle p_q | q_i \rangle = \frac{1}{ \sqrt{2 \pi h}} exp \left( - \frac{ip_q q_i}{h} \right)$ are the position eigenfunctions in the momentum representation.
In summary, during propagation to the detection screen the particle is in a superposition of momentum states determined by the hole configuration of the slit screen. The square of the absolute magnitude of eq 2, the momentum probability distribution in the xy plane, is, therefore, the diffraction pattern. If this were calculated using eq 2, it would not resemble the actual diffraction because it has been assumed, initially, that the slits are infinitesimally small holes; however, if the holes are given a width d in both directions, the Fourier transform is given by eq 3.
$\langle p | \Psi \rangle = \frac{1}{2 \pi hd \sqrt{N}} \sum_{i=1}^{N} \int_{x_i - \frac{d}{2}} ^{x_i + \frac{d}{2}} exp \left[ - \frac{ip_xx_i}{ \hbar} \right] dx \int_{y_i - \frac{d}{2}}^{ y_i + \frac{d}{2}} exp \left[ - \frac{ip_y y_i}{ \hbar}\right] dy \nonumber$
In the present case, N = 2, and for two "holes" with dimension 0.3 a0 [11], separated by 1.0 a0 along the x axis, the diffraction pattern is shown in Figure 1. This pattern is in agreement with the diffraction from a two-hole mask as shown on plate 2, panel 3 of reference 5. It is interpreted here using both the uncertainty and superposition principles. The particle is localized in space by the holes in the slit screen, and this leads, according to the uncertainty principle, to a delocalization of the x and y components of the particle's momentum. The figure clearly shows this delocalization in the momentum components. In addition, because the particle is localized at two positions in the x direction, the superposition principle demands constructive and destructive interference in the x component of the momentum. This effect is also clearly shown in Figure 1.
The complementary relationship between position and momentum required by the uncertainty principle is illustrated further in Figure 2, where the position and momentum distributions are shown for holes of dimension 0.5 a0. Clearly, the decreased localization in coordinate space has led to an increased localization in momentum space. Similar quantitative arguments relating the uncertainty principle to slit width and momentum distribution have recently been made in both the pedagogical [12] and research [13] literature for single-slit diffraction phenomena.
We can change the two holes to two slits by increasing the y dimension of the holes. For example, holding the x dimension constant at 0.3 a0 while increasing the y dimension to 1.5 a0 yields the diffraction pattern shown in Figure 3. This result is in agreement with the diffraction pattern obtained using the multislit mask on panel b of the Discovery Slide of reference 2. Comparing Figures 1 and 3, nothing has changed in the x direction, but the particle is partially delocalized in the y direction at the slit screen in Figure 3. According to the uncertainty principle, its momentum in the y direction will become more localized the spread in observed momentum values will decrease. To summarize, spatial localization leads to a delocalized momentum distribution that is projected onto the detection screen and is calculated as $| \langle p | \Psi \rangle |^2$ using eq 3. Furthermore, spatial localization at more than one position in a particular direction leads to interference effects in the momentum distribution in that direction.
It should also be pointed out that the double-slit experiment has been carried out at low source intensity such that there is only one particle in the apparatus at a time [14]. The resulting diffraction pattern is the same as that observed with intense sources; it just takes longer to record to the same number of observations. This confirms the quantum mechanical view that wave functions 1 and 2 represent individual particles rather than an ensemble of particles. According to quantum mechanical principles the particles can be placed in a well defined state by the slit screen, but the outcome of a subsequent measurement, momentum in this case, is not uniquely determined. All we can do is calculate the probability that a particular momentum state will be observed. The mechanism, therefore, is single-particle, or self-interference; two particles do not interfere with each other [14, 15].
With this introduction it is easy to move to a general treatment of diffraction by two-dimensional masks. Following eq 1, for a mask consisting of N scatterers the coordinate wave function is given by eq 4 below, and the momentum wave function is given by eq 3.
$| \Psi \rangle = \frac{1}{ \sqrt{N}} \sum_{i=1}^N | x_i,~y_i \rangle \nonumber$
Figure 4 shows the diffraction pattern for a face-centered squaric arrangement of 25 holes (the two-dimensional analog of face-centered cubic), while Figure 5 was created with a simple squaric arrangement of 25 holes (the two-dimensional analog of simple cubic). Both calculated diffraction patterns are in excellent agreement with the experimental results shown in Figure 3a and b of reference 1. These examples show that the addition of a scatterer to the unit cell increases destructive interference and fewer momentum states are observed at the Figure 5. Hole dimensions are 0.5 a0 with horizontal and vertical detector.
Figures 6 and 7 show the calculated diffraction patterns for two concentric hexagons and two concentric pentadecagons, respectively. These calculated diffraction patterns are in excellent agreement with the experimental patterns found in Plates 10 and 12 in reference 5. Given the recently celebrated 50th anniversary of the publication of the DNA double helix structure [16] it is appropriate to draw special attention to references 3 and 4, which provide optical transforms that simulate DNA's experimental X-ray diffraction pattern.
Using the theoretical methods outlined above, it is possible to calculate in a rudimentary way several of the characteristic features of the DNA diffraction pattern. For example, representing DNA solely by a planar double strand of phosphate backbone groups as shown on the left side of Figure 8 generates the diffraction pattern shown on the right side. The essential features captured by this naive model are the characteristic X-shaped cross of the diffraction pattern and the missing fourth horizontal layer. Additional discussion on this rudimentary model is available in references 3 and 4.
The diffraction patterns presented in this paper have been calculated using Mathcad Plus 5.0. Naturally, once a calculation for one mask geometry has been programmed, it can serve as a template for other calculations with minor editing. Such a Mathcad file for calculating diffraction patterns for two-dimensional masks is available for download [17].
The conceptual and computational simplicity of the approach presented here makes it possible to bring experiment and theory together in the classroom in order to gain insight into the physical nature of diffraction phenomena from the quantum mechanical perspective.
Appendix A. The Fourier Transform
Wave-particle duality, as expressed by de Broglie's wave equation, is at the heart of quantum mechanics.
$\lambda = \frac{h}{p} \nonumber$
On the left side we have the wave property, wavelength, and on the right momentum, the particle property; thus, (A.1) shows wave and particle properties in one equation united by the ubiquitous Planck's constant.
The most general coordinate-space wave function for a free particle with wavelength λ is the complex Euler function shown below in Dirac notation.
$\langle x | \lambda \rangle = exp \left( i \frac{2 pi x}{ \lambda} \right) \nonumber$
Substitution of eq A.1 into eq A.2 yields the momentum wave function in the coordinate representation.
$\langle x | p \rangle = exp \left( \frac{ipx}{ \hbar} \right) \nonumber$
The complex conjugate of eq A.3, $\langle x | p \rangle = \langle p | x \rangle$, is the coordinate wave function in the momentum representation.
$\langle p|x \rangle = exp \left( - \frac{ipx}{ \hbar} \right) \nonumber$
These equations are extremely useful in quantum mechanics in translating coordinate-space information into the momentum representation, and vice versa.
For example, if a system is in a well-defined state, $| \Psi \rangle$, it might be expressed in coordinate language as $\langle x | \Psi \rangle$ or momentum language as $\langle p | \Psi \rangle$. If one is known, the other can be found by a Fourier transform because both representations contain the same state information. As shown below, the overlap integral between $\langle x | \Psi \rangle$ and $\langle p | x \rangle$ in coordinate space is the momentum wavefunction, and the overlap between $p | \Psi \rangle$ and $\langle x | p \rangle$ in momentum space is the position wavefunction.
$\langle p | \Psi \rangle = \int \langle p | x \rangle \langle x | \Psi \rangle dx = \frac{1}{ \sqrt{2 pi \hbar}} \int exp \left( - \frac{ipx}{ \hbar} \right) \Psi (x) dx \nonumber$
$\langle x | \Psi \rangle = \int \langle x | p \rangle \langle p | \Psi \rangle dx = \frac{1}{ \sqrt{2 pi \hbar}} \int exp \left( - \frac{ipx}{ \hbar} \right) \Psi (p) dp \nonumber$
In these equations $(2 \pi \hbar )^{ - \frac{1}{2}}$ is the normalization constant for $\langle x | p \rangle$ and $\langle p | x \rangle$.
In our analysis of diffraction phenomena, eq A.5 is used because state preparation occurs in coordinate space at the slit screen yielding $\langle x | \Psi \rangle$, which is then projected into momentum space, $\langle p | \Psi \rangle$, to calculate the diffraction pattern, $| \langle p | \Psi \rangle |^2$, at the detection screen.
Appendix B. Bragg's Law
The reader will have noticed that we only made brief mention of Bragg's Law in our presentation. This is because we prefer, for pedagogical reasons, Marcella's quantum mechanical approach with its roots in von Laue's early classical analysis of X-ray diffraction as a scattering phenomenon. The following quotation from French and Taylor's excellent text reveals the connection between von Laue and Bragg in the pre-quantum mechanical era [18].
The original analysis by von Laue treated the problem (X-ray diffraction), as basically one must, in terms of the scattering of an incident X-ray beam by all the individual atoms in the three-dimensional lattice making up the crystal. This is somewhat formidable, but very soon afterwards (in 1913), W. L. Bragg showed that the results of the analysis are the same as if one regarded the x-rays as being reflected at sets of parallel planes that include many atoms.
Reflection, therefore, is a conceptual device that simplifies analysis; diffraction is really a scattering phenomenon that behaves as if it were reflection for simple systems. While Bragg's approach simplifies analysis in simple cases, it does so at the expense of providing an incorrect physical picture of what is actually occurring. It has also led to abstract concepts such as the reciprocal lattice and Miller indices which students and teachers alike find difficult to understand. In our teaching we prefer the more direct appeal to the physical concepts of position and momentum, united as they always are by the uncertainty principle.
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Muiño has used single-slit diffraction to provide an introduction to the uncertainty principle suitable for an undergraduate physical chemistry course (1). His article provided both a theoretical analysis of single-slit diffraction and a lecture demonstration of the phenomenon using readily-available equipment. In the current research literature Nairz and colleagues have confirmed the uncertainty principle in a single-slit diffraction experiment with a beam of C70 molecules (2). And, quite recently, a research team led by Marcus Arndt and Anton Zeilinger performed multi-slit experiments demonstrating wave-particle behavior for tetraphenylporphyrin and a fluorinated fullerene, C60F48 (3). The significance of these results is that tetraphenylporphyrin is an important biomolecule and C60F48 is the most massive particle to demonstrate wavelike properties to date.
The purpose of this short article is to provide an alternative theoretical analysis of single-slit diffraction based on the Fourier transform between coordinate and momentum space. This approach was recently used by Marcella (4) to analyze single- and double-slit diffraction and has been extended by the author to more complicated slit geometries (5). Quantum experiments have three parts: (i) state preparation; (ii) subsequent measurement of an observable; and (iii) theoretical interpretation of experimental results. In diffraction experiments, passage through the slit screen represents a position measurement that establishes the state of the system in coordinate space. The coordinate-space wave function for a photon or massive particle that has passed a screen with a slit of width w centered at x = 0 is
$\Psi (x, w) = \begin{Bmatrix} 0 \text{ for } x < - \frac{w}{2} \ \frac{1}{ \sqrt{w}} \text{ for } - \frac{w}{2} \leq x \leq \frac{w}{2} \ 0 \text{ for } x > \frac{w}{2} \end{Bmatrix} \nonumber$
The quantum mechanical interpretation of diffraction is that the physical property recorded at the detection screen is the momentum distribution of the diffracted particle.
A Fourier transform of Ψ(x, w) into the momentum representation yields the momentum-space wave function
$\Phi (p_x,~w) = \int_{ - \frac{w}{2}}^{ \frac{w}{2}} \frac{1}{ \sqrt{2 pi \hbar}} exp \left( - \frac{i p_x x}{ \hbar} \right) \frac{1}{ \sqrt{w}} dx = \sqrt{ \frac{2 \hbar}{ \pi w}} \frac{ \sin \left( \frac{p_x w}{2 \hbar} \right)}{p_x} \nonumber$
where px is the x-direction momentum. Thus, according to quantum mechanics, the diffraction pattern observed is the square of the absolute magnitude of the momentum wave function, |Φ(px, w)|2 . This is shown in Figure 1 for two slit widths.
The uncertainty principle is clearly revealed—the narrow slit produces a broader momentum distribution. In other words, localization in coordinate space leads to delocalization in momentum space. However, we can also treat this effect in a more quantitative manner.
Following Muiño we will assume that the uncertainty in position is the slit width, w. The uncertainty in momentum is defined as half the width of the momentum distribution of the central diffraction band (2). As the momentum distribution is zero for
$\frac{ p_x w}{2 \hbar} = \pm n \pi \nonumber$
it is easy to show that using this criterion the uncertainty in momentum is $\frac{2 \pi \hbar}{w}$. Therefore, the product of the uncertainty in position and momentum is greater than \( \frac{ \hbar}{2}} as required by the uncertainty principle.
$\Delta x \Delta p_x = w \frac{2 pi \hbar}{w} = 2 \pi \hbar \nonumber$
The Fourier transform connecting complementary observables is ubiquitous in quantum theory (and, of course, in our laboratory instruments). Here the transform between position and momentum has been used to illuminate the intimate relationship between single-slit diffraction and the uncertainty principle.
Literature Cited
1. Muiño, P. L. J. Chem. Educ. 2000, 77, 1025–1027.
2. Nairz, O.; Arndt, M.; Zeilinger, A. Phys. Rev. A, 2002, 65, 032109.
3. Hackermüller, L.; Uttenhaler, S.; Hornberger, K.; Reiger, E.; Brezger, B.; Zeilinger, A.; Arndt, M. Phys. Rev. L. 2003, 91, 090408.
4. Marcella, T. V. Eur. J. Phys. 2002, 23, 615–621. Also see: French, A. P.; Taylor, E. F. An Introduction to Quantum Physics; W. W. Norton and Co. Inc.: New York, 1978; pp 331–336.
5. Rioux, F. Eur. J. Phys. 2003, 24, N1–N3
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Diffraction has a simple quantum mechanical interpretation based on the uncertianty principle. Or we could say diffraction is an excellent way to illustrate the uncertainty principle.
A screen with a single slit of width, w, is illuminated with a coherent photon or particle beam. The normalized coordinate‐space wave function at the slit screen is,
$\begin{matrix} w = 1 & \Psi (x, w) = \text{if} \left[ \left( x \geq - \frac{w}{2} \right) \left( x \leq \frac{w}{2} \right),~ \frac{1}{ \sqrt{w}},~0 \right] & x = \frac{-w}{2},~\frac{+w}{2} + 0.005 .. \frac{w}{2} \end{matrix} \nonumber$
The coordinate‐space probability density, $| \Psi (x,~w)|^2$, is displayed for a slit of unit width below
The slit‐screen measures position, it localizes the incident beam in the x‐direction. According to the uncertainty principle, because position and momentum are complementary, or conjugate, observables, this measurement must be accompanied by a delocalization of the x‐component of the momentum. This can be seen by a Fourier transform of Ψ(x,w) into momentum space to obtain the momentum wave function, Φ(px,w).
$\Psi (p_x,~w) = \frac{1}{ \sqrt{2 \pi}} \int_{ - \frac{w}{2}}^{ \frac{w}{2}} exp (-i~ p_x~ x) \frac{1}{ \sqrt{w}} dx ~ \text{simplify } \rightarrow \frac{ \sqrt{2} \sin \left( \frac{p_x w}{2} \right)}{ \sqrt{ \pi} p_x \sqrt{w}} \nonumber$
It is the momentum distribution, $| \Phi (p_x,~w)|^2$, shown histographically below that is projected onto the detection screen. Thus, a position measurement at the detection screen is also effectively a measure of the x‐component of the particle momentum.
In this figure we see the spread in momentum required by the uncertainty principle, plus interference fringes due to the fact that the incident beam can imerge from any where within the slit, allowing for constructive and destructive interference at the detection screen. If the slit width is decreased the position is more precisely known and the uncertainty principle demands a broadening in the momentum distribution as shown below.
Equating uncertainty in position with slit width and uncertainty in momentum with the width of the intense center of the diffraction pattern, we have in atomic units: ΔxΔpx = 12.6. If the slit width is decreased the position is more precisely known and the uncertainty principle demands a broadening in the momentum distribution as shown below. For slit width 0.5 we again find the product of the uncertainties is 12.6.
Naturally if the slit width is increased to 2.0 the position uncertainty increases and the uncertainty in momentum decreases yielding again ΔxΔpx = 12.6.
The x‐direction momentum can be expressed in terms of the wavelength of the illuminating beam and the diffraction angle using the following sequence of equations of which the second is the de Broglie relation in atomic units (h = 2π).
$\begin{matrix} p_x = p \sin( \Theta) & p = \frac{2 \pi}{ \lambda} & p_x = \frac{2 \pi}{ \lambda} \sin ( \Theta) \end{matrix} \nonumber$
$\Phi ( \Theta_x,~w,~\lambda ) = \sqrt{ \frac{2}{ \pi w}} \frac{ \sin \left( \frac{ \pi w}{ \lambda} \sin ( \Theta_x) \right)}{ \frac{2 \pi}{ \lambda} \sin ( \Theta_x)} \nonumber$
This allows one to explore the effect of the wavelength of the illuminating beam on the diffraction pattern. The figure below shows that a short wavelength (high momentum) illuminating beam gives rise to a narrower diffraction pattern.
The method used here to calculate single‐slit diffraction patterns (momentum‐space distribution functions) is easily extended to multiple slits, and also to diffraction at two‐dimensional masks with a variety of hole geometries.
Relevant literature:
Primary source: ʺQuantum interference with slits,ʺ Thomas Marcella which appeared in European Journal of Physics 23, 615‐621 (2002).
See also: ʺCalculating diffraction patterns,ʺ F. Rioux in European Journal of Physics, 24, N1‐N3 (2003). “Using Optical Transforms to Teach Quantum Mechanics,” F. Rioux; B. J. Johnson, The Chemical Educator, 9, 12‐16 (2004). ʺSingle‐slit Diffraction and the Uncertainty Prinicple,ʺ F. Rioux in Journal of Chemical Education, 82, 1210 (2005).
ʺExperimental verification of the Heisenberg uncertainty principle for hot fullerene moleculesʺ, O. Nairz, M. Arndt, and A. Zeilinger, Phys. Rev. A, 65, 032109 (2002).
ʺIntroducing the Uncertainty Principle Using Diffraction of Light Waves,ʺ Pedro L. Muino, Journal of Chemical Education, 77, 1025‐1027 (2000).
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.03%3A_Single-slit_Diffraction_and_the_Uncertainty_Principle_%28Mathcad_Version%29.txt
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The publication of the DNA double-helix structure by x-ray diffraction in 1953 is one the most significant scientific events of the 20th century (1). Therefore, it is important that science students and their teachers have some understanding of how this great achievement was accomplished. X-ray diffraction is conceptually simple: a source of X-rays illuminates a sample which scatters the x-rays, and a detector records the arrival of the scattered x-rays (diffraction pattern). However, the mathematical analysis required to extract from diffraction pattern the molecular geometry of the sample that caused the diffraction pattern is quite formidable. Therefore, the purpose of this tutorial is to illustrate some of the elements of the mathematical analysis required to solve a structure.
The famous X-ray diffraction pattern obtained by Rosalind Franklin is shown below (2).
This X-ray picture stimulated Watson and Crick to propose the now famous double-helix sturcture for DNA. It was surely fortuitous that Crick had recently completed an unrelated study of the diffraction patterns of helical molecules (3).
To gain some understanding of how the experimental pattern led to the hypothesis of a double-helical structure we will work in reverse. We will assume the double-helix structure, calculate the diffraction pattern, and compare it with the experimental result. This, therefore, is a deductive exercise as opposed to the brilliant inductive accomplishment of Watson and Crick in determining the DNA structure from Franklin's experimental X-ray pattern.
The experimental pattern will be simulated by modeling DNA solely as a planar double strand of sugar-phosphate backbone groups shown below. Reference 4 provides the justification and the limitations in using two-dimensional models for three-dimensional structures when simulating X-ray diffraction experiments.
The double-strand geometry shown below was created using the following mathematics. Calculations are carried out in atomic units.
$\begin{matrix} \text{Sugar-phosphate groups per strand:} & A = 20 & \text{Strand radius:} & R = 1 & \text{Phase difference between strands:} & 0.8 \pi \end{matrix} \nonumber$
$\begin{matrix} \text{First strand:} & m = 1 .. A & \Theta_m = \frac{4 \pi m}{A} & y_m = m & x_m = R \cos \left( \Theta_m \right) \ \text{Second strand:} & m = 21 .. 40 & \Theta_m = \frac{4 \pi (m - A)}{A} & y_m = (m-A) & x_m = R \cos \left( \Theta_m + 0.8 \pi \right) \ m = 1 .. 20 & n = 21 .. 40 \end{matrix} \nonumber$
According to quantum mechanical principles, the photons illuminating this geometrical arrangement interact with all its members simultaneously thus being cast into the spatial superposition, Ψ, given below.
$| \Psi \rangle = \frac{1}{ \sqrt{N}} \sum_{i = 1}^N |x_i,~y_i \rangle \nonumber$
This spatial wave function is then projected into momentum space by a Fourier transform to yield the theoretical diffraction pattern. What is measured at the detector according to quantum mechanics is the two-dimensional momentum distribution created by the spatial localization that occurs during illumination of the structure. If the sugar-phosphate groups are treated as point scatterers the momentum wave function is given by the following Fourier transform.
$\Theta (p_x,~p_y) = \frac{1}{2 \pi} \sum_{m = 1}^{40} exp (-i~p_x~x_m ) exp( -i~p_y~y_m) \nonumber$
The theoretical diffraction pattern can now be displayed as the absolute magnitude squared of the momentum wave function.
$\begin{matrix} \Delta = 8 & N = 200 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\text{Diffraction pattern}_{j,~k} = \left( \left| \Phi \left( px_j,~y_k \right) \right| \right)^2 \nonumber$
Clearly the naive model diffraction pattern presented here captures several important features of the experimental diffraction pattern. Among those are the characteristic X-shaped cross of the diffraction pattern and the missing fourth horizontal layer (indicated by arrows).
Lucas, Lisensky, and co-workers (4, 5) have simulated the DNA diffraction pattern using the optical transform method. This tutorial might therefore be considered to be a theoretical companion to their more empirical approach to the subject.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.04%3A_Simulating_DNA%27s_Diffraction_Pattern.txt
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The publication of the DNA double‐helix structure by x‐ray diffraction in 1953 is one the most significant scientific events of the 20th century (1). Therefore, it is important that science students and their teachers have some understanding of how this great achievement was accomplished. X‐ray diffraction is conceptually simple: a source of X‐rays illuminates a sample which scatters the x‐rays, and a detector records the arrival of the scattered x‐rays (diffraction pattern). However, the mathematical analysis required to extract from diffraction pattern the molecular geometry of the sample that caused the diffraction pattern is quite formidable. Therefore, the purpose of this tutorial is to illustrate some of the elements of the mathematical analysis required to solve a structure.
The famous X‐ray diffraction pattern obtained by Rosalind Franklin is shown below (2).
This X‐ray picture stimulated Watson and Crick to propose the now famous double‐helix sturcture for DNA. It was surely fortuitous that Crick had recently completed an unrelated study of the diffraction patterns of helical molecules (3).
To gain some understanding of how the experimental pattern led to the hypothesis of a double‐helical structure we will work in reverse. We will assume the double‐helix structure, calculate the diffraction pattern, and compare it with the experimental result. This, therefore, is a deductive exercise as opposed to the brilliant inductive accomplishment of Watson and Crick in determining the DNA structure from Franklinʹs experimental X‐ray pattern.
The experimental pattern will be simulated by modeling DNA solely as a planar double strand of sugar‐phosphate backbone groups shown below. Reference 4 provides the justification and the limitations in using two‐dimensional models for three‐dimensional structures when simulating X‐ray diffraction experiments.
The double‐strand geometry shown below was created using the following mathematics. Calculations are carried out in atomic units.
$\begin{matrix} \text{Sugar-phosphate groups per strand:} & A = 20 & \text{Strand radius:} & R = 1 & \text{Phase difference between strands:} & 0.8 \pi \end{matrix} \nonumber$
$\begin{matrix} \text{First strand:} & m = 1 .. A & \Theta_m = \frac{4 \pi m}{A} & y_m = m & x_m = R \cos \left( \Theta_m \right) \ \text{Second strand:} & m = 21 .. 40 & \Theta_m = \frac{4 \pi (m - A)}{A} & y_m = (m-A) & x_m = R \cos \left( \Theta_m + 0.8 \pi \right) \ m = 1 .. 20 & n = 21 .. 40 \end{matrix} \nonumber$
According to quantum mechanical principles, the photons illuminating this geometrical arrangement interact with all its members simultaneously thus being cast into the spatial superposition, Ψ, given below.
$| \Psi \rangle = \frac{1}{ \sqrt{N}} \sum_{i = 1}^{N} | x_i,~y_i \rangle \nonumber$
This spatial wave function is then projected into momentum space by a Fourier transform to yield the theoretical diffraction pattern. What is measured at the detector according to quantum mechanics is the two‐dimensional momentum distribution created by the spatial localization that occurs during illumination of the structure. If the sugar‐phosphate groups are treated as scatterers of dimension d the momentum wave function is given by the following Fourier transform.
$\begin{matrix} d = 0.4 & \Phi (p_x,~p_y) = \sum_{m = 1}^{40} \left( \int_{y_m - \frac{d}{2}}^{y_m + \frac{d}{2}} \exp (-i~p_y~y) dy \right) \end{matrix} \nonumber$
The theoretical diffraction pattern can now be displayed as the absolute magnitude squared of the momentum wave function.
$\begin{matrix} \Delta = 8 & N = 200 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\text{Diffraction Pattern}_{j,~k} = \left( \left| \Phi ( px_j,~py_k ) \right| \right)^2 \nonumber$
Clearly the naive model diffraction pattern presented here captures several important features of the experimental diffraction pattern. Among those are the characteristic X‐shaped cross of the diffraction pattern and the missing fourth horizontal layer (indicated by arrows). Lucas, Lisensky, and co‐workers (4, 5) have simulated the DNA diffraction pattern using the optical transform method. This tutorial might therefore be considered to be a theoretical companion to their more empirical approach to the subject.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.05%3A_Simulating_DNA%27s_Diffraction_Pattern_with_a_More_Realistic_Model.txt
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$\begin{matrix} \text{Sugar-phosphate groups per strand:} & A = 20 & \text{Strand radius:} & R = 1 & \text{Phase difference between strands:} & 0.8 \pi \end{matrix} \nonumber$
$\begin{matrix} \text{First strand:} & m = 1 .. A & \Theta_m = \frac{4 \pi m}{A} & y_m = m & x_m = R \cos \left( \Theta_m \right) \ \text{Second strand:} & m = 21 .. 40 & \Theta_m = \frac{4 \pi (m - A)}{A} & y_m = (m-A) & x_m = R \cos \left( \Theta_m + 0.8 \pi \right) \ m = 1 .. 20 & n = 21 .. 40 \end{matrix} \nonumber$
Coordinate space wave function:
$| \Psi \rangle = \frac{1}{ \sqrt{N}} \sum_{i = 1}^{N} | x_i,~y_1 \rangle \nonumber$
Momentum space wave function:
$\Phi (p_x,~p_y) = \frac{1}{ 2 \pi} \sum_{m=1}^{40} exp (-i p_x x_m) exp( -i p_y y_m) \nonumber$
$\begin{matrix} \Delta = 8 & N = 200 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\text{Diffraction Pattern}_{j,~k} = \left( \left| \Phi (px_j,~py_k ) \right| \right)^2 \nonumber$
5.07: A Model Graphene Diffraction Pattern
The purpose of this tutorial is to model graphene as seven fused benzene rings (see below) and use a Fourier transform of the atomic positions to calculate its diffraction pattern.
$\begin{matrix} \text{Number of atoms:} & A = 24 & \text{Atomic dimension:} & d = .25 & \text{Atomic positions:} \ x_1 = 0 & y_1 = 1.386 & x_2 = 0 & y_2 = -1.386 & x_{15} = 0 & y_{15} = 2.772 \ x_3 = -1.2 & y_3 = 0.693 & x_4 = 1.2 & y_4 = 0.693 & x_{16} = 1.2 & y_{16} = 3.465 \ x_5 = 1.2 & y_5 = -0.693 & x_6 = -1.2 & y_6 = -0.693 & x_{17} = 2.4 & y_{17} 2.772 \ x_7 = 2.4 & y_7 = 1.386 & x_8 = 3.6 & y_8 = 0.693 & x_{18} = 0 & y_{18} = -2.772 \ x_9 = 3.6 & y_9 = -0.693 & x_{10} = 2.4 & y_{10} = -1.386 & x_{19} = 1.2 & y_{19} = -3.465 \ x_{11} = -2.4 & y_{11} = 1.368 & x_{12} = -3.6 & y_{12} = 0.693 & x_{20} = 2.4 & y_{20} = -2.772 \ x_{13} = -3.6 & y_{13} = -0.693 & x_{14} = -2.4 & y_{14} = -1.386 & x_{21} = -2.4 & y_{21} = 2.772 \ x_{22} = -1.2 & y_{22} = 3.465 & x_{23} = -2.4 & y_{23} = -2.772 & x_{24} = -1.2 & y_{24} = -3.465 \end{matrix} \nonumber$
The diffraction pattern is the Fourier transform of the atomic positions into momentum space.
$\begin{matrix} \Delta = 20 & N = 200 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\begin{matrix} \Psi (p_x,~p_y) = \sum_{m=1}^{A} \left( \int_{x_m - \frac{d}{2}} ^{x_m + \frac{d}{2}} exp (-i p_x x)dx \int_{y_m - \frac{d}{2}} ^{y_m + \frac{d}{2}} exp(-i p_y y) dx \right) & p_{j,~k} = \left( \left| \Psi ( px_j,~py_j) \right| \right)^2 \end{matrix} \nonumber$
$i = 1 .. A$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.06%3A_Simulating_DNA%27s_Diffraction_Pattern_-_Short_Version.txt
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A two-dimensional Fibonacci lattice lacks translational periodicity but has a discrete diffraction pattern, just like a quasicrystal. However, it does not fit the definition of a quasilattice because it does not posses one of the 'forbidden' n-fold rotational symmetries (n = 5 or greater than 6) that are characteristic of quasicrystals and incompatible with translational periodicity. R. Lifshitz1, therefore, recommends that the symmetry requirement be relaxed so that two- and three-dimensional Fibonacci lattices can have quasilattice stature.
A one-dimensional Fibonacci grid consists of a sequence of long (L) and short (S) segments such as LSLLSLSLLS.... with L/S = 1.618, the golden ratio. A two-dimensional array is created by superimposing two such grids at a 90o angle and placing atomic scatterers at the vertices.
$\begin{matrix} \text{Dimension of grid:} & A = 10 & m = 1 .. A & n = 1 .. A & \tau = \frac{1 + \sqrt{5}}{2} \end{matrix} \nonumber$
Calculate the coordinates of the Fibonacci vertices in a two-dimensional lattice (see Lifshitz).
$\begin{matrix} x_m = \text{floor} \left( \frac{m}{ \tau} \right) \tau + \left( m - \text{floor} \left( \frac{m}{ \tau} \right) \right) -1 & y_n = \text{floor} \left( \frac{n}{ \tau} \right) \tau + \left( n - \text{floor} \left( \frac{n}{ \tau} \right) \right) - 1 \end{matrix} \nonumber$
$\begin{matrix} x^T = \begin{pmatrix} 0 & 1.618 & 2.618 & 4.236 & 5.854 & 6.854 & 8.472 & 9.472 & 11.09 & 12.708 \end{pmatrix} \ y^T = \begin{pmatrix} 0 & 1.618 & 2.618 & 4.236 & 5.854 & 6.854 & 8.472 & 9.472 & 11.09 & 12.708 \end{pmatrix} \end{matrix} \nonumber$
Display the two-dimensional Fibonacci array:
2D Fibonacci Lattice
$\begin{matrix} \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \end{matrix} \nonumber$
The diffraction pattern is the Fourier transform of the spatial Fibonacci array into the momentum representation.
Calculate momentum-space wave function:
$\Phi (p_x,~p_y) = \frac{1}{2 \pi} \sum_{m = 1}^A exp (-i p_x x_m) \sum_{n = 1}^A exp ( -i p_y y_n) \nonumber$
Display momentum-space distribution function (diffraction pattern).
$\begin{matrix} \Delta = 10 & N = 200 & j = 1 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 1 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\text{Diffraction Pattern}_{j,~k} = \left( \left| \Phi (px_j,~py_k ) \right| \right)^2 \nonumber$
1. R. Lifshitz, "The square Fibonacci tiling," Journal of Alloys and Compounds 342, 186-190 (2002).
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.08%3A_Is_a_Two-dimensional_Fibonacci_Array_a_Quasilattice.txt
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I wish to describe a simple extension of Marcella’s [1] recent analysis of the double-slit experiment to two dimensions.
The essential point Marcella makes in his unique treatment of this well-known experiment is that the diffraction pattern at the detection screen is actually a measurement of the momentum distribution of the diffracted particles. Therefore the calculated diffraction pattern is simply obtained from the Fourier transform of the coordinate space wave function (the double-slit geometry) into momentum space. Marcella considered two spatial models: (1) infinitesimally thin slits represented by Dirac delta functions, and (2) slits of finite width.
About sixty years ago Sir Lawerence Bragg [2] proposed the optical transform as an aid in the interpretation of the x-ray diffraction patterns of crystals. This required the fabrication of two-dimensional masks of various crystal or molecular geometries and the generation of the diffraction pattern using visible electromagnetic radiation. Present day laser technology has made the generation of such diffraction patterns routine, even in the classroom.
In addition, Marcella’s computational approach makes calculating the diffraction patterns conceptually and mathematically straightforward. If one considers the mask as consisting of point scatterers (model 1), the coordinate space wave function is a linear superposition of the scattering positions,
$| \Psi \rangle = \frac{1}{ \sqrt{N}} \sum_{i = 1}^N |x_i,~y_1 \rangle \nonumber$
where N is the number of point scatterers.
The Fourier transform into the momentum representation yields,
$\langle p | \Psi \rangle = \frac{1}{ \sqrt{N}}_{i=1}^N \langle p_x | x_i \rangle \langle p_y | y_i \rangle = \frac{1}{2 \pi h \sqrt{N}} \sum_{i = 1}^N exp \left[ - \frac{i}{h} (p_xx_i + p_yy_i) \right] \nonumber$
For model 2, which assumes finite-sized scatterers, equation (2) becomes,
$\langle p | \Psi \rangle = \frac{1}{2 \pi h r \sqrt{N}} \sum_{i = 1}^N \int_{x_i - \frac{r}{2}}^{x_i + \frac{r}{2}} exp \left[ - \frac{ip_xx}{h} \right] dx \int_{y_i - \frac{r}{2}}^{y_i + \frac{r}{2}} exp \left[ - \frac{ip_y y}{h} \right] dy \nonumber$
where r is the spatial dimension of the scatterers. In the interest of mathematical simplicity, the scatterers are assumed to be small squares rather than circles.
Figure 1 shows the diffraction pattern, $| \langle p | \Psi \rangle |^2$, for a hexagonal arrangement of six point scatterers calculated using equation (2), while figure 2 shows the pattern obtained with equation (3) for six finite hexagonal scatterers. The calculated diffraction pattern shown in figure 2 is in excellent agreement with the experimental diffraction pattern available in the literature [3]. The calculations [4] of the diffraction patterns were carried out in atomic units (h = 2π) so that positions are given in a0 and momenta in a0 -1. The distance between adjacent scatters is 1.4 a0 and their spatial dimension is 0.3 a0 on a side.
In addition to showing the interference effects accompanying scattering from multiple positions, the figures also illustrate the uncertainty principle. Figure 1 shows no attenuation of the diffraction pattern at extreme values of momentum because the point scatterers sharply localize the particle being scattered in coordinate space, leading to a delocalized momentum distribution as required by the position-momentum uncertainty relation. By comparison the finite scatterers of figure (2) lead to some uncertainty in position and, therefore, less uncertainty in momentum. Therefore the diffraction pattern is attenuated at large values for both the x- and y-momentum components.
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In this tutorial a model diffraction pattern for C60 is calculated and compared with the experimental diffraction pattern.
Solid C60 has the cubic close-packed crystal structure. For the sake of computational convenience this structure will be modeled as a planar close-packed array consisting of seven C60 molecules represented by squares (see the mask geometry shown below). According to quantum mechanical principles, radiat illuminating this geometrical arrangement interacts with all its members simultaneously thus being cas the spatial superposition, Ψ, given below. This spatial wave function is then projected into momentum space by a Fourier transform. Therefore what is measured at the detector can be interpreted as the two-dimensional momentum distribution created by the spatial localization that occurs at the mask up illumination.
$\begin{matrix} \text{Create mask geometry:} & A = 6 & R = 1.4 & m = 1 .. A & \Theta_m = \frac{2 \p m}{A} \ x_m = R \sin ( \Theta_m) & y_m = R \cos ( \Theta_m) & x_7 = 0 & y_7 = 0 & \text{Molecule size: } d = 3 \end{matrix} \nonumber$
$\begin{matrix} \text{Display coordinate wave function (mask geometry):} & m = 1 .. 7 \end{matrix} \nonumber$
Fourier transform the coordinate wave function into the momentum representation:
$\Phi (p_x,~p_y) = \sum_{m = 1}^{A + 1} \int_{x_m - \frac{d}{2}}^{x_m + \frac{d}{2}} exp ( -p~p_x~x) dx \int_{y_m - \frac{d}{2}}^{y_m + \frac{d}{2}} exp (-i~p_y~y) dy \nonumber$
Display diffraction pattern:
$\begin{matrix} \Delta = 15 & N = 100 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\text{Diffraction Pattern}_{j,~k} = \left( \left| \Phi (px_j,~py_k ) \right| \right)^2 \nonumber$
The calculated diffraction pattern above compares favorably with the experimental diffraction pattern shown below, indicating that the simple planar model proposed for C60 captures the essential element the actual solid structure. See page 154 of Perfect Symmetry by Jim Baggot for further information.
5.11: Diffraction Pattern for Pentagonal Point Scatterers
Establish mask geometry:
$\begin{matrix} R = 2 & m = 1 .. A & \Theta_m = \frac{2 \pi m}{A} & x_m = R \sin ( \Theta_m) & y_m = R \cos ( \Theta_m) \end{matrix} \nonumber$
Fourier transform of position wave function (mask geometry) into the momentum representation:
$\Phi (p_x,~p_y) = \frac{1}{ 2 \pi \sqrt{A}} \sum_{m = 1}^{A} ( exp(-ip_xx_m) exp (-i p_y y_m)) \nonumber$
Display mask geometry and diffraction pattern: $A \equiv 5$
$\begin{matrix} N = 100 & \Delta p = 12 & j = 0 .. N & k = 0 .. N & px_j = - \Delta p + \frac{2 \Delta p j}{N} & py_k = - \Delta p + \frac{2 \Delta p~ k}{N} \end{matrix} \nonumber$
$\text{Diffraction Pattern}_{j,~k} = \left( \left| \Phi (px_j,~py_k ) \right| \right)^2 \nonumber$
5.12: Diffraction Pattern for Pentagonal Finite Point Scatterers
Establish mask geometry:
$\begin{matrix} R = 2 & m = 1 .. A & \Theta_m = \frac{2 \pi m}{A} & x_m = R \sin ( \Theta_m) & y_m = R \cos ( \Theta_m) \end{matrix} \nonumber$
Fourier transform of position wave function (mask geometry) into the momentum representation:
$\Phi (p_x,~p_y) = \frac{1}{2 \pi d \sqrt{A}} \left[ \sum_{m = 1}^{A} \left( \int_{x_m - \frac{d}{2}}^{ x_m + \frac{d}{2}} exp(-i p_x x) dx \int_{y_m - \frac{d}{2}}^{y_m + \frac{d}{2}} exp (-i p_y y) dy \right) \right] \nonumber$
Display mask geometry and diffraction pattern: $\begin{matrix} A = 5 & d = .3 \end{matrix}$
$\begin{matrix} N = 100 & \Delta p = 10 & j = 0 .. N & k = 0 .. N & px_j = - \Delta + \frac{2 \Delta p j}{N} & py_k = - \Delta p + \frac{2 \Delta p k}{N} \end{matrix} \nonumber$
$\text{Diffraction pattern}_{j,~k} = \left( \left| \Phi ( px_j,~py_k ) \right| \right)^2 \nonumber$
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Establish mask geometry:
$\begin{matrix} R= 2 & m = 1 .. 5 & \Theta_m = \frac{2 \pi m}{5} & x_m = R \sin ( \Theta_m ) & y_m = R \cos ( \Theta_m) \ R = 0.75 & m = 6 .. 10 & \Theta_m = \frac{2 \pi (m - 0.5)}{5} & x_m = R \sin ( \Theta_m) & y_m = R \cos ( \Theta_m) \end{matrix} \nonumber$
Fourier transform of position wave function (mask geometry) into the momentum representation: $m = 1 .. 10$
$\begin{matrix} d = .15 & \Phi (p_x,~p_y) = \frac{1}{2 \pi d \sqrt{10}} \left[ \sum_{m = 1} \left( \int_{x_m - \frac{d}{2}}^{ x_m + \frac{d}{2}} exp(-i p_x x) dx \int_{y_m - \frac{d}{2}}^{y_m + \frac{d}{2}} exp (-i p_y y) dy \right) \right] \end{matrix} \nonumber$
Display mask geometry and diffraction pattern:
$\begin{matrix} N = 100 & \Delta p = 20 & j = 0 .. N & k = 0 .. N & px_j = - \Delta + \frac{2 \Delta p j}{N} & py_k = - \Delta p + \frac{2 \Delta p k}{N} \end{matrix} \nonumber$
$\text{Diffraction pattern}_{j,~k} = \left( \left| \Phi ( px_j,~py_k ) \right| \right)^2 \nonumber$
5.14: Model Diffraction Pattern for Napthalene
Ten holes in a two-dimensional mask are used to model the diffraction pattern for napthalene. It is assumed that only the carbon atoms scatter radiation.
Establish mask geometry:
$\begin{matrix} R= 2 & m = 1 .. 6 & \Theta_m = \frac{2 \pi m}{6} & x_m = R \sin ( \Theta_m ) - \sqrt{3} & y_m = R \cos ( \Theta_m) \ ~ & m = 7 .. 10 & \Theta_m = \frac{2 \pi (m - 0.5)}{6} & x_m = R \sin ( \Theta_m) + \sqrt{3} & y_m = R \cos ( \Theta_m - \frac{ \pi}{ \sqrt{3}} ) \end{matrix} \nonumber$
Fourier transform of position wave function (mask geometry) into the momentum representation. Initially the carbon atoms are considered to be point scatterers.
$\begin{matrix} m = 1 .. 10 & \Phi (p_x,~p_y) = \frac{1}{2 \pi d \sqrt{10}} \left[ \sum_{m = 1} \left( \int_{x_m - \frac{d}{2}}^{ x_m + \frac{d}{2}} exp(-i p_x x) dx \int_{y_m - \frac{d}{2}}^{y_m + \frac{d}{2}} exp (-i p_y y) dy \right) \right] \end{matrix} \nonumber$
Display mask geometry and diffraction pattern:
$\begin{matrix} N = 100 & \Delta p = 15 & j = 0 .. N & k = 0 .. N & px_j = - \Delta + \frac{2 \Delta p j}{N} & py_k = - \Delta p + \frac{2 \Delta p k}{N} \end{matrix} \nonumber$
$\text{Diffraction pattern}_{j,~k} = \left( \left| \Phi ( px_j,~py_k ) \right| \right)^2 \nonumber$
This calculation illustrates the uncertainty principle. There is no attenuation in the diffraction pattern (momentum distribution) because there is no uncertainty in the carbon atom positions.
Now the diffraction pattern is recalculated assuming an arbitrary finite dimension for the carbon atoms.
$\begin{matrix} r = .15 & \Phi (p_x,~p_y) = \frac{1}{2 \pi d \sqrt{10}} \left[ \sum_{m = 1} \left( \int_{x_m - r}^{ x_m + r} exp(-i p_x x) dx \int_{y_m - r}^{y_m + r} exp (-i p_y y) dy \right) \right] \end{matrix} \nonumber$
$\text{Diffraction pattern}_{j,~k} = \left( \left| \Phi ( px_j,~py_k ) \right| \right)^2 \nonumber$
There is attenuation in the momentum distribution (less uncertainty) because there is more uncertainty in the carbon atom positions.
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The Airy diffraction pattern is created by illuminating a screen containing a circular hole with photons. The experiment can be performed with weak sources such that there is only one photon interacting with the screen at a time. This photon-screen interaction constitutes a position measurement.
The position wave function has a constant amplitude within the area of the hole and is shown to be normalized.
$\begin{matrix} \Psi (x,~y) = \frac{1}{ \sqrt{ \pi R^2}} & R^2 = x^2 + y^2 & \int_{-R}^{R} \int_{- \sqrt{R^2 - x^2}}^{ \sqrt{R^2 - x^2}} \Psi (x,~y)^2 dy dx = 1 \end{matrix} \nonumber$
The Airy diffraction pattern is the Fourier transform of the position wave function into the momentum representation. In other words, the interference pattern at the detection screen actually represents a momentum measurement. The following calculations are carried out in atomic units using a hole radius of 0.2.
Hole radius: R = .2
Calculate the Airy diffraction pattern:
$\begin{matrix} \Delta = 100 & N = 80 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
$\begin{matrix} \Phi (p_x,~p_y) = \frac{1}{ \pi} \int_{R}^R \int_{- \sqrt{R^2 - x^2}}^{ \sqrt{R^2 -x^2}} \frac{1}{ \sqrt{ \pi R^2}} exp(-i p_x x) exp(-i p_y y) dy dx & P_{j,~k} = \left( \left| \Phi (px_j,~py_k ) \right| \right)^2 \end{matrix} \nonumber$
Display the Airy diffraction pattern.
Truncating the high intensity central disk provides a better picture of the outer maxima and minima.
Examining a radial slice of the Airy diffraction pattern provides a simple illustration of the uncertainty principle. Assume that the position uncertainty is given by the diameter of the hole and that the momentum uncertainty is given by the momentum range of the central disk.
$\begin{matrix} py = 0 & px = -100,~-99, .. 100 & \Phi (px,~py~R) = \frac{1}{ \pi} \int_{-R}^R \int_{- \sqrt{R^2 - x^2}}^{ \sqrt{R^2 - x^2}} \frac{1}{ \sqrt{ \pi R^2}} exp (-i px x)exp(-i py y) dy dx \end{matrix} \nonumber$
For a diameter of 0.4the position-momentum uncertainty product is: $0.4 ~ 38.5 = 15.4$
For a diameter of 0.2 the position-momentum uncertainty product is: $.2 ~77.0 = 15.4$
The reciprocal relationship between the uncertainty in position and momentum is clearly revealed in this example.
5.16: Diffraction Pattern for Two Concentric Rings
Create hole positions:
$\begin{matrix} A = 32 & R = 1.2 & m = 1 .. 16 & \Theta_m = \frac{2 \pi m}{16} & x_m = R \sin ( \Theta_m) & y_m = R \cos ( \Theta_m) \ ~ & R = .9 & m = 17 .. A & \Theta_m = \frac{2 \pi m}{16} & x_m = R \sin ( \Theta_m) & y_m = R \cos ( \Theta_m ) \end{matrix} \nonumber$
Display coordinate-space wave function (mask geometry): $m = 1 .. A$
Fourier transform position wave function into the momentum representation:
$\begin{matrix} \Delta = 30 & N = 200 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
Hole dimension:
$\begin{matrix} d = .1 & \Psi (p_x,~p_y) = \frac{1}{2 \pi d \sqrt{A}} \sum_{m = 1}^A \left( \int_{x_m - \frac{d}{2}}^{x_m + \frac{d}{2}} exp(-i p_x x)dx \int_{y_m - \frac{d}{2}}^{y_m + \frac{d}{2}} exp(-i p_y y) dy \right) \end{matrix} \nonumber$
Display diffraction pattern:
$P_{j,~k} = \left( \left| \Psi (x_j,~Py_k ) \right| \right)^2 \nonumber$
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A sharply focused particle beam (photons, electrons, molecules, etc.) is incident on a screen with two slits. According to quantum mechanics the individual particles are represented by a coherent superposition of being simultaneously at both slits.
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |1 \rangle + |2 \rangle \right] \nonumber$
In the interest of mathematical simplicity, 1 and 2 label slits that are infinitesimally narrow in the x-direction and infinitely long in the y-direction. The density operator for this state is
$\hat{ \rho} = | \Psi \rangle \langle \Psi | = \frac{1}{2} \left[ |1 \rangle + |2 \rangle \right] \left[ \langle 1 | + \langle 2 | \right] = \frac{1}{2} \left[ |1 \rangle \langle 1 | + |1 \rangle \langle 2 | + |2 \rangle \langle 1 | + |2 \rangle \langle 2 | \right] \nonumber$
The expectation value for the arrival of a particle at position x on the detection screen is
$\langle x | \hat{ \rho} | x \rangle = \frac{1}{2} \left[ \langle x |1 \rangle \langle 1 | x \rangle + \langle x | 1 \rangle \langle 2 | x \rangle + \langle x | 2 \rangle \langle 1 | x \rangle + \langle x | 2 \rangle \langle 2 | x \rangle \right] \nonumber$
Rearrangement yields,
$\langle x | \hat{ \rho} | x \rangle = \frac{1}{2} \left[ | \langle x | 1 \rangle |^2 + | \rangle x |2 \rangle |^2 + \langle x | 1 \rangle \langle x | 2 \rangle* + \langle x | 2 \rangle \langle x | 1 \rangle* \right] \nonumber$
The probability amplitudes in this equation represent the phase of a particle on arrival at position x from slits 1 and 2. For example, using Euler’s equation we calculate the phase of a particle arriving at x from slit 1 as follows,
$\langle x | 1 \rangle = \frac{1}{ \sqrt{2 \pi r}} exp \left( i 2 \pi \frac{ \delta x_1}{ \lambda} \right) \nonumber$
where δx1 is the distance from slit 1 to position x on the detection screen and λ is the de Broglie wavelength of the particle.
Using this form for the probability amplitudes we can write the expectation value in terms of the distances to x from slits 1 and 2.
$\langle x | \hat{ \rho} | x \rangle = \frac{1}{4 \pi} \left[ 2 + exp \left( i 2 \pi \frac{( \delta x_1 - \delta x_2)}{ \lambda} \right) + exp \left( -i2 \pi \frac{( \delta x_1 - \delta x_2)}{ \lambda} \right) \right] = \frac{1}{2 \pi} \left[ 1 + \cos \left( \frac{( \delta x_1 - \delta x_2)}{ \lambda} \right) \right] \nonumber$
Clearly $\frac{( \delta x_1 - \delta x_2)}{ \lambda}$ will vary continuously along the x-axis of the detector from large negative values at one end to large positive values at the other end leading to minima and maxima in the cosine term and therefore $\langle x | \hat{ \rho} | x \rangle$, thereby yielding the well-known interference fringes associated with the double-slit experiment. Naturally a more realistic slit geometry will lead to a mathematically more complicated expression for the expectation value.
If one takes a classical view of the double-slit experiment that assumes the particle goes through one slit or the other, and has a 50% chance of going through either slit, the coherent superposition,
$\hat{ \rho} = \frac{1}{2} |1 \rangle \langle 1 | + \frac{1}{2} |2 \rangle \langle 2 | \nonumber$
The expectation value for the arrival of the particle at x on the detection screen is now,
$\langle x | \hat{ \rho}_{cl} | x \rangle = \frac{1}{2} \langle x | 1 \rangle \langle 1 | x \rangle + \frac{1}{2} \langle x | 2 \rangle \langle 2 | x \rangle = \frac{1}{2} | \langle x | 1 \rangle |^2 + \frac{1}{2} | \langle x | 2 \rangle |^2 = \text{ constant} \nonumber$
which has a constant value with no oscillations in arrival probability as a function of x. In other words, no interference fringes.
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Illumination of a double-slit screen with a coherent particle beam leads to a Schrödinger "cat state" that can be represented by a linear superposition (unnormalized) of two Gaussian wavepackets. The probability distribution function in coordinate space, |Y(x)|2, at the slit-screen for this "cat state" is shown below.
$\begin{matrix} x = -10, ~ -9.99 .. 10 & \Psi (x) = exp \left[ -(x-5)^2 \right] + exp \left[ - (x + 5)^2 \right] & P(x) = \left( \left| \Psi (x) \right| \right)^2 \end{matrix} \nonumber$
The slits localize the particle in the x-direction which leads to a spread in the x-component of the momentum required by the uncertainty principle, DxDpx > h/4p. The momentum wave function is obtained by a Fourier transform of the coordinate-space wave function.
$\Phi (p_x) = \frac{1}{ \sqrt{2 \pi}} \left[ \int_{- \infty}^{ \infty} exp (-i p_x x) \left[ exp \left[ - (x-5)^2 \right] + exp \left[ - (x+5)^2 \right] \right] dx \right] \nonumber$
Evaluation of the integral yields,
$\Phi (p_x ) = \frac{1}{ \sqrt{2}} \left[ exp \left[ \frac{-1}{4} p_x (p_x + 20i) \right] + exp \left[ \frac{-1}{4} p_x (p_x - 20i \right] \right] \nonumber$
The momentum probability function in the x-direction is |F(px)|2 and simplifies to the expression given below when evaluated.
$P(p_x) = 2 exp \left( \frac{-1}{2} p_x^2 \right) \cos (5 p_x )^2 \nonumber$
This momentum probability function is displayed below.
$p_x = -4, -3.99 .. 4 \nonumber$
Because the arrival at position x on the detection screen is proportional to px it is also proportional to |F(px)|2. In other words, the particle distribution at the detection screen is determined by the momentum distribution at the slit screen. This means the position measurement at the detection screen is effectively a measurement of the px. Therefore, the particle distribution at the detector screen will have the same shape as shown in the figure above.
In summary, the double-slit experiment clearly reveals the three essential steps in a quantum mechanical experiment:
1. State preparation (interaction of the incident beam with the slit-screen)
2. Measurement of an observable (arrival of scattered beam at the detection screen)
3. Calculation of expected results of the measurement step
*The preparation of this tutorial was stimulated by reading "Quantum interference with slits" by Thomas Marcella which appeared in European Journal of Physics 23, 615-621 (2002). This paper offers a lucid and novel quantum mechanical analysis of a very important experiment.
Additional references:
R. P. Feynman, R. B. Leighton, and M.Sands, The Feynman Lectures on Physics, Volume 3; Addison-Wesley; Reading, 1965, Chapters 1 and 3.
R. P. Feynman, The Character of Physical Law; MIT Press: Cambridge, 1967, Chapter 6.
A. Tonomura, J. Endo, T. Matsuda, T. Kawasaki, and H. Exawa, "Demonstration of single-electron buildup of an interference pattern" Am. J. Phys. 57, 117-120 (1989).
D. Leibfried, T. Pfau, and C. Monroe, "Shadows and Mirrors: Reconstructing Quantum States of Atom Motion" Phys. Today 51(4), 22-28 (1998).
The double-slit experiment with single electrons was recently selected (informally) as physics most beautiful experiment. The following web reference traces the history of double-slit interference experiments from the time of Thomas Young to the present, presenting numerous literature references in the process: http://physicsweb.org/article/world/15/9/1.
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Diffraction has a simple quantum mechanical interpretation based on the uncertianty principle. Or we could say diffraction is an excellent way to illustrate the uncertainty principle.
A screen with a single slit of width, w, is illuminated with a coherent photon or particle beam. The normalized coordinate‐space wave function at the slit screen is,
$\Psi (x,~w) = \left| \begin{matrix} \frac{1}{ \sqrt{w}} \text{if} \left( x \geq - \frac{w}{2} \right) \left( x \leq \frac{w}{2} \right) \ 0 \text{ otherwise} \end{matrix} \right. \nonumber$
The coordinate‐space probability density, |Ψ(x,w)|2, is displayed for a slit of unit width below
The slit‐screen measures position, it localizes the incident beam in the x‐direction. According to the uncertainty principle, because position and momentum are complementary, or conjugate, observables, this measurement must be accompanied by a delocalization of the x‐component of the momentum. This can be seen by a Fourier transform of Ψ(x,w) into momentum space to obtain the momentum wave function, Φ(px,w).
$\Phi (p_x,~w) = \frac{1}{ \sqrt{2 \pi}} \int_{ - \frac{w}{2}}^{ \frac{w}{2}} exp ( -i p_x x) \frac{1}{w} dx \text{ simplify} \rightarrow \frac{ \sqrt{2} \sin \left( \frac{p_x w}{2}}{ \sqrt{ \pi} p_x \sqrt{w}} \nonumber$
It is the momentum distribution, |Ψ(px ,w)|2, shown below that is projected onto the detection screen. Thus, a position measurement at the detection screen is also effectively a measure of the x‐component of the particle momentum.
In this figure we see the spread in momentum required by the uncertainty principle, plus interference fringes due to the fact that the incident beam can imerge from any where within the slit, allowing for constructive and destructive interference at the detection screen. If the slit width is decreased the position is more precisely known and the uncertainty principle demands a broadening in the momentum distribution as shown below.
Equating uncertainty in position with slit width and uncertainty in momentum with the width of the intense center of the diffraction pattern, we have in atomic units: ΔxΔpx = 12.6. If the slit width is decreased the position is more precisely known and the uncertainty principle demands a broadening in the momentum distribution as shown below. For slit width 0.5 we again find the product of the uncertainties is 12.6.
Naturally if the slit width is increased to 2.0 the position uncertainty increases and the uncertainty in momentum decreases yielding again ΔxΔpx = 12.6.
The x‐direction momentum can be expressed in terms of the wavelength of the illuminating beam and the diffraction angle using the following sequence of equations of which the second is the de Broglie relation in atomic units (h = 2π).
$\begin{matrix} p_x = p \sin ( \Theta) & p = \frac{2 \pi}{ \lambda} & p_x = \frac{2 \pi}{ \lambda} \sin ( \Theta) \end{matrix} \nonumber$
$\Phi ( \Theta_x,~w,~ \lambda) = \sqrt{ \frac{2}{ \pi w}} \frac{ \sin \left( \frac{ \pi w}{ \lambda} \sin ( \Theta_w ) \right)}{ \frac{2 \pi}{ \lambda} \sin ( \Theta_w )} \nonumber$
This allows one to explore the effect of the wavelength of the illuminating beam on the diffraction pattern. The figure below shows that a short wavelength (high momentum) illuminating beam gives rise to a narrower diffraction pattern.
The method used here to calculate single‐slit diffraction patterns (momentum‐space distribution functions) is easily extended to multiple slits, and also to diffraction at two‐dimensional masks with a variety of hole geometries.
Primary source: ʺQuantum interference with slits,ʺ Thomas Marcella which appeared in European Journal of Physics 23, 615‐621 (2002).
See also: "Calculating diffraction patterns,ʺ F. Rioux in European Journal of Physics, 24, N1‐N3 (2003). ʺSingle‐slit Diffraction and the Uncertainty Principle,ʺ F. Rioux in Journal of Chemical Education, 82, 1210 (2005).
ʺExperimental verification of the Heisenberg uncertainty principle for hot fullerene moleculesʺ, O. Nairz, M. Arndt, and A. Zeilinger, Phys. Rev. A, 65, 032109 (2002).
ʺIntroducing the Uncertainty Principle Using Diffraction of Light Waves,ʺ Pedro L. Muino, Journal of Chemical Education, 77, 1025‐1027 (2000).
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The American Journal of Physics published a translation of Claus Jonsson's paper "Electron Diffraction at Multiple Slits" in American Journal of Physics 42, 4-11 (1974). The following calculations are in agreement with the diffraction patterns reported by Jonsson.
$\begin{matrix} \text{Number of slits:} & n = 6 & \text{Slit width:} & w = .5 \ \text{Slit locations:} & s_1 = 0 & s_2 = 2 & s_3 = 4 & s_4 = 6 & s_5 = 8 & s_6 = 10 \end{matrix} \nonumber$
Normalized coordinate-space wave function at the slit screen:
$\Psi (x) = \frac{1}{ \sqrt{N}} \left| \begin{matrix} \frac{1}{ \sqrt{w}} \text{ if } \sum_{j = 1}^{n} \left[ ( x \geq - s_j ) ( x \leq s_j + w ) \right] \ 0 \text{otherwise} \end{matrix} \right. \nonumber$
Fourier transform the position wave function into the momentum representation:
$\Phi (p_x) = \frac{1}{ \sqrt{2 \pi}} \int_0 ^{s_n + w} exp(-i p_x x) \Psi (x) dx \nonumber$
5.21: Multiple Slit Diffraction and the Fourier Transform
The American Journal of Physics published a translation of Claus Jonssonʹs paper ʺElectron Diffraction at Multiple Slitsʺ in American Journal of Physics 42, 4‐11 (1974). The following calculation is in agreement with the diffraction pattern reported by Jonsson.
A four slit geometry is created. This represents the coordinate space wave function. It is Fourier transformed into the momentum representation to generate its diffraction pattern.
$\begin{matrix} \text{Number of slits:} & n = 4 & \text{Slit positions:} & j = 1 .. n & x_j = j & \text{Slit width:} & \delta = .2 \end{matrix} \nonumber$
$\Phi (p) = \frac{ \sum_{j=1}^n \int_{x_j - \frac{ \delta}{2}}^{ x_j + \frac{ \delta}{2}} \frac{1}{ \sqrt{2 \pi}} exp (-i px) \frac{1}{ \sqrt{ \delta}} dx}{ \sqrt{n}} \nonumber$
The momentum wave function is Fourier transformed back to coordinate space to generate the spatial wave function or slit geometry.
$\begin{matrix} x = 0, .01 .. 5 & \Psi (x) = \frac{1}{ \sqrt{2 \pi}} \int_{-30}^{30} exp (ipx) \Phi (p) dp \end{matrix} \nonumber$
5.22: The Double-Slit Experiment with C60 Molecules
This is an attempt to capture with a simple model the behavior of C60 molecules in a double-slit experiment as reported in Nature Magazine (14 OCT 1999, pp 680-682). See Eur. J. Phys. 23, 615-621 (2002) for the origin of the model.
The C60 wave function at the slit screen is shown below. This coordinate-space wave function is Fourier transformed into momentum space to obtain the momentum distribution at the slit screen. This momentum distribution is ultimately projected onto the detection screen.
In the work reported in Nature the slit width and the distance between slits was equal (50 nm). Arbitrary distance units are used here to illustrate the effect. The slit screen is actually a diffraction grating, but the researchers assumed the interference effect was caused by adjacent slits.
$\begin{matrix} x = -20,-19.99, .. 20 & \Psi (x) = \text{if} \left[ ( x \geq -7.5) (x \leq -2.5) + (x \geq 2.5) (x \leq 7.5), ~ \frac{1}{ \sqrt{10}},~0 \right] \end{matrix} \nonumber$
$\begin{matrix} P(p_x) = \left( \left| \frac{1}{ \sqrt{2 \pi}} \int_{-10}^{10} exp (-i p_x x) \Psi (x) dx \right| \right)^2 & p_x = -3,-2.99, .. 3 \end{matrix} \nonumber$
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Prior to 1991 crystals were defined to be solids having only 2-, 3-, 4- and 6-fold rotational symmetry because only these rotational symmetries have the required translational periodicity to build the long-range order of a crystalline solid. Long-range order is synonymous with periodicity, requiring some unit structure which repeats itself by translation in all directions infinitely. It is easy to demonstrate that a pentagon, with 5-fold rotational symmetry cannot be used as a unit cell to create long-range order in a plane or in three-dimensions.
The justification for this definition was that solid structures with 2-, 3-, 4- and 6-fold rotational symmetry yield discrete diffraction patterns that also have translational periodicity. Another way to put this is to say that solid structures with 2-, 3-, 4- and 6-fold rotational symmetry have reciprocal lattices that also have translational periodicity. Yet, another way to put this, of course, is that the Fourier transforms of geometries with 2-, 3-, 4- and 6-fold rotational symmetry yield lattice-like momentum distributions with translational periodicity. This latter statement is preferred by the author because it emphasizes that diffraction patterns are actually the momentum distributions of the diffracted particles.
The key in this latter interpretation is that diffraction experiments involve an initial spatial localization of the radiation through interaction with the crystal lattice, followed as required by the uncertainty principle, a delocalization of the momentum distribution in the detection plane.
Let’s look at some examples. First we examine the Fourier transforms of two mini-lattices with two- and three-fold rotational symmetry.
Clearly both diffraction patterns exhibit translational periodicity, their repeating units being a 90 degree rotation of the spatial structure. Next we look at four-fold rotational symmetry and see that the unit cell is obvious.
Six-fold rotational symmetry is more interesting than the previous three examples, but again the unit cell is easy to find.
Now look at what happens when we consider 5-fold symmetry – the diffraction pattern generated by a pentagon.
The unit cell, the universal repeating unit, is gone. The diffraction pattern is well-defined, it has rotational symmetry and it is appealing, but it does not satisfy the criterion for translational periodicity. That’s why 5-fold rotational symmetry is excluded from the list of symmetries that can generate diffraction patterns that have translational periodicity, and why by definition crystalline solids are not supposed to have 5-fold axes, or rotational axes greater than order six.
However, in 1984 an international research team consisting of D. Shechtman, I. Blech, D. Gratias and J. W. Cahn, published “Metallic phase with long-range orientational order and no translational symmetry” in Physical Review Letters 53, 1951-1953 (1984). The crystalline metallic phases they studied produced discrete diffraction patterns that were characteristic of the 5- and 10-fold rotational symmetry axes that were prohibited by the accepted definition of a crystalline solid.
In the face of this contradictory evidence, 5-fold rotational symmetry and a well-defined diffraction pattern, the International Union of Crystallography in 1991 redefined crystal to mean any solid having a discrete diffraction pattern. However, the solid phases discovered by Shechtman and his co-workers go by the name quasicrystals, indicating that they don’t quite have the same stature as those that don’t violate the rotational symmetry rule.
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According to Richard Feynman1 the double-slit experiment is the paradigm for all of quantum mechanics because it is a simple manifestation of the superposition principle, which is quantum theory’s “only mystery.” In an interesting quantum mechanical analysis of Young’s double-slit experiment, Marcella2 writes the coordinate wave function of the particle interacting with the slit-screen as a linear superposition of being at both slits,
$| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ |x_1 \rangle + | x_2 \rangle \right] \nonumber$
where x1 and x2 are the positions of two infinitesimally thin slits. The diffraction pattern at the detection screen is determined by the momentum distribution at the slit screen. The momentum wave function is obtained by projecting equation (1) into momentum space,
$\langle p_x | \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \langle p_x | x_1 \rangle + \langle p_x | x_2 \rangle \right] = \frac{1}{2 \sqrt{ \pi h}} \left[ exp \left( - \frac{ip_x x_1}{h} \right) + exp \left( - \frac{ip_x x_2}{h} \right) \right] \nonumber$
where the $\langle p_x | x_1 \rangle$ are the position eigenfunctions in the momentum representation. The square of the absolute magnitude of this function is shown in Figure 1.
Like the double-slit phenomenon, x-ray diffraction by crystals is also based on the linear superposition principle. For the sake of simplicity attention will be restricted to two-dimensional crystals. Also, following Marcella’s first model, the atoms or ions occupying the lattice sites are considered to be point scatterers. Under these assumptions, the coordinate-space wave function of a photon illuminating a crystal is a superposition of the atomic scattering positions.
$| \Psi = \frac{1}{ \sqrt{N}} \sum_{i=1}^N | x_i,~y_i \rangle \nonumber$
As we wish to know the diffraction pattern that will be found at the detector, we express this superposition in the momentum representation as follows.
$\langle p | \Psi \rangle = \frac{1}{ \sqrt{N}} \sum_{i=1}^N \langle p_x | x_i \rangle \langle p_y | y_i \rangle \nonumber$
Employing the expression for the position wave function in the momentum representation we can convert equation (4) to
$\langle p | \Psi \rangle = \frac{1}{ \sqrt{N}(2 \pi h)} \sum_{i=1}^N f_i \text{ exp} \left[ - \frac{i}{h} (p_xx_i + p_yy_i ) \right] \nonumber$
where the fi are the atomic scattering factors. A face-centered “squaric” arrangement of five atomic point scatterers yields the following diffraction pattern, 2 p Ψ .
The Mathcad document used to generate this diffraction pattern is shown in the Appendix.
Literature cited:
1. R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Vol. 3; Addison-Wesley: Reading, 1965.
2. T. V. Marcella, “Quantum interference with slits,” European Journal of Physics 23, 615-621 (2002).
Appendix
$\begin{matrix} \text{Number of atoms: } A = 5 & \text{Atomic positions and relative scattering factors:} \end{matrix} \nonumber$
$\begin{matrix} x_1 = 0 & y_1 = 0 & f_1 = \frac{1}{4} & ~ & x_2 = 0 & y_2 = 1 & f_2 = \frac{1}{4} \ x_3 = 1 & y_3 = 0 & f_3 = \frac{1}{4} & ~ & x_4 = 1 & y_4 = 1 & f_4 = \frac{1}{4} \ x_5 = \frac{1}{2} & y_5 = \frac{1}{2} & f_5 = \frac{1}{2} \end{matrix} \nonumber$
$\begin{matrix} \Delta = 10 & N = 200 & j = 0 .. N & px_j = - \Delta + \frac{2 \Delta j}{N} & k = 0 .. N & py_k = - \Delta + \frac{2 \Delta k}{N} \end{matrix} \nonumber$
Fourier transform of position wave function into the momentum representation:
$\begin{matrix} \Psi (p_x,~p_y) = \sum_{m=1}^A f_m exp(-i p_x x_m -ip_yy_m ) & P_{j,~k} = \left( \left| \Psi \left( px_j,~py_k \right) \right| \right)^2 \end{matrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.24%3A_X-ray_Crystallography_from_a_Quantum_Mechanical_Perspective.txt
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Recently I was reading an encyclopedia article [McMillan Encyclopedia of Physics, John Rigden, Editor, page 716] about holography...
The word "holography" is derived from Greek roots that literally mean "entire picture." The act of doing holograhpy is to make holograms. There are many different types of holograms, but they all share one common distinguishing feature, they recreate truly three-dimensional imgages of original objects.
The field of holography was originally discovered by Dennis Gabor in 1947. He was awarded the Nobel Prize for physics in 1971. In the early 1960s, Emmett N. Leith and Juris Upatinieks of the United States and Yu. Denisyuk of Russia independently discovered additional methods in using laser light to make holograms.
What is a hologram? A hologram is a recording on a lightsensitive medium (e.g. photographic emulsion) of interference patterns formed between two or more beams of light derived from the same laser.
In holography, as the figure below shows, a laser beam is split by a beam splitter with one arm going to the object and the other going to a mirror. The beam arms recombine at the detector, in this case a photographic plate, and constructive or destructive interference occurs.
Although it isn't stated explicitly above, I assume that holography is an example of single-photon interference, and that the same can be said about any diffraction phenomenon - X-ray, electron, neutron, etc. To support this assumption I quote the celebrated remark by Dirac, "Each photon then interferes only with itself. Interference between two different photons never occurs." [I am aware of Roy Glauber's cautionary remarks about the second sentence - American Journal of Physics 63, 12 (1995). For example, two-photon interference can be observed if the photons are created in an entangled state.]
According to this single-particle mechanism, after the beam splitter each photon is considered to be in a linear superposition of being transmitted (toward object) and reflected (toward mirror). At the detector the probability amplitudes for these paths interfere (constructively or destructively) giving an interference pattern characteristic of the object. In Dirac notation, the probability of detection at some point x on the photographic film is,
$P(x) = | \langle \text{Detector(x) | Mirror} \rangle \langle \text{Mirror|Source} \rangle + \langle \text{Detector(x)|Object} \rangle \langle \text{Object|Source} \rangle |^2 \nonumber$
This is a clear example of quantum mechanical wave-particle duality; the photon is created by the source as a particle, behaves like a wave, and is detected at the photographic plate as a particle.
I would like to add further support the underlying assumption of this tutorial that holography is an example of single-particle interference with the following excerpt from Volume 23 of the Encyclopedia Britannica, 15th Edition, pages 20- 21.
It was at one time suggested that interference and diffraction phenomena are caused by collisions between photons considered as small particles, or at least to some kind of complex interaction between photons. A simple experiment performed by a British physicist, Geoffrey Ingram Taylor, in 1908 excludes this possibility. He photographed the diffraction pattern of a needle and reduced the illumination until long exposures were needed to obtain an image. When the chance of two or more energy quanta passing through the apparatus simultaneously was made extremely small, the diffraction pattern was exactly the same as that obtained with a strong source of light.
This experiment is supported by many later experiments, showing that interference and diffraction are to be associated with single photons. The discussion in the section on coherence above implies that when ordinary sources of light are used each photon can interfere only with itself and not with any other photon. An interference pattern can, of course, be photographed only by recording the effects of many photons because one photon can activate only one grain on the photographic plate, but it does not matter whether the photons all arrive over a time span of a microsecond or of several weeks.
The photograph of an interference pattern is both a wave phenomenon because it shows the characteristic spatial periodicity and a quantum phenomenon because the whole energy of a photon can be used to activate a single grain. It is not possible to trace the path of one photon (regarded as a particle) through the apparatus and, at the same time, obtain the interference pattern. In Young's experiment there is no way of finding out through which slit a given photon passes--except by covering one slit and thus losing the interference fringes.
For a discussion of the related double-slit experiment see the related tutorial: Calculating the Double-Slit Interference Pattern. To learn about contemporary applications of holographic techniques at the atomic level see "Optics: Holograms of Atoms," Nature 410, 1037-1040 (2001) and references cited therein.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.25%3A_Holography_Involves_Single_Photon_Interference.txt
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This tutorial on X‐ray diffraction was stimulated by an appendix in the first edition of Peter Atkinsʹ celebrated physical chemistry text (pp. 734‐739) titled ʺDiffraction as a particle property.ʺ In exploring this approach we will work in one dimension in the interests of mathematical and conceptual simplicity.
Modeling the periodicity of the electron density in a one‐dimensional atomic lattic requires the superposition of many electron wavelengths and, therefore, by the de Broglie relation (λ = h/p) many electron momentum states. For example, a superposition of eight cosine functions yields the sharply defined electron density distribution with lattice spacing d shown below.
$\begin{matrix} d=1 & \Psi (x) = \sum_{n=1}^8 \cos \left( n 2 \pi \frac{x}{d} \right) \end{matrix} \nonumber$
where d is the lattice spacing and λ = d/n with n = 1, 2, 3...
According to the quantum mechanical interpretation of diffraction, when this one‐dimensional electron density distribution interacts with an X‐ray source the individual photons are temporarily localized simultaneously at all lattice sites. The uncertainty principle requires that spatial localization leads to a delocalization (scattering) of the momentum distribution and the appearance of interference fringes because the spatial localization occurs at multiple sites. The resulting momentum distribution is the diffraction pattern and is calculated by a Fourier transform of the electron density distribution, Ψ(x)2, into momentum space as is shown below (in atomic units: h = 2π).
$\Phi (p) = \frac{1}{ \sqrt{2 \pi}} \int_{-.5}^{4.5} \Psi (x)^2 exp(ipx)dx \nonumber$
The approach adopted by Atkins is to model diffraction as an elastic particle‐like collision between the X‐ray photon and the electron distribution of the lattice. In this view one asks what momenta the lattice can transfer to the incident particle (photon). The quantum mechanical answer to this question is provided by the scattering amplitude or structure factor shown in the following equation.
$F(p') = \int \langle p-p' | \Psi \rangle \langle \Psi | p \rangle dp \nonumber$
Reading the Dirac brackets above from right to left, this expression is the scattering amplitude or the ʺpossibilityʺ that an electronic state Ψ with momentum p can transfer momentum pʹ to the scattered particle. Integrating the brackets over p yields all the ways that the electron distribution represented by Ψ can transfer momentum pʹ to the incident photon.
The structure factor can be cast into the coordinate representation by applying the completeness relation
$\int | x \rangle \langle x | dx = 1 \nonumber$
to the left bracket in the integral followed by rearrangement.
$F(p') = \int \int \langle p-p' |x \rangle \langle x | \Psi \rangle \langle \Psi | p \rangle dxdp \nonumber$
Using the momentum eigenfunction in the coordinate representation and the position eigenfunction in the momentum representation (normalization constants are ignored throughout this treatment),
$\langle x|p \rangle = \langle p|x \rangle* = \text{exp} \left( \frac{ipx}{ \hbar} \right) \nonumber$
we see that the middle bracket on the right can be transformed as follows,
$\langle p - p" | x \rangle = \text{exp} \left( \frac{-i (p-p')x}{ \hbar} \right) = \text{exp} \left( - \frac{ipx}{ \hbar} \right) \text{exp} \left( \frac{ip'x}{ \hbar} \right) = \langle p | x \rangle \langle x | p' \rangle \nonumber$
F(p') can now be written as
$F(p') = \int \int \langle \Psi | p \rangle \langle p|x \rangle \langle x | p' \rangle \langle x| \Psi \rangle dxdp \nonumber$
Using the momentum-space completeness relation,
$\int |p \rangle \langle p | dp = 1 \nonumber$
we find,
$F(p') = \int \langle \Psi | x \rangle \langle x | p' \rangle \langle x | \Psi \rangle dx \nonumber$
Expressing F(p') in traditional notation yields,
$F(p') = \int \Psi* (x) \text{exp} \left( \frac{ip'x}{ \hbar} \right) \Psi (x) dx = \int \rho (x) \text{exp} \left( \frac{ip'x}{ \hbar} \right) dx \nonumber$
Note that we have reached the same point as in our initial approach: Fourier transform the electron density into the momentum representation to calculate the diffraction pattern.
We conclude by recovering the traditional form of the structure factor found in all physical chemistry texts and books dealing with X‐ray crystallography. The momenta, pʹ, that the electron distribution can transfer to the incident photon are the momenta that it has.
Earlier it was noted that to build a one‐dimensional electron density distribution with lattice spacing d requires the superposition of wavelengths restricted by the relation λ = d/n, where n = 1,2,3, ... This requires, by de Broglieʹs wave equation, a superposition of quantized momentum states given by pʹ = nh/d. Substitution of this equation into F(pʹ) yields
$F(n) = \int \rho \text{exp} \left( 2 \pi i \frac{nx}{d} \right) dx \nonumber$
where n is the Miller index that normally goes by the designation h. The extension to three dimensions using the traditional (h, k, l) designation for the Miller indices, and a, b and c for the lattice constants is straightforward.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/05%3A_Diffraction_Phenomena/5.26%3A_X-ray_Diffraction.txt
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Symmetry Analysis for H2O
Water belongs to the C2v symmetry group and has the following symmetry elements: E, C2z, σxz, and σyz. Its character table is shown below.
$\begin{array}{|c|c|c|c|c|c|} \hline \text{C}_{2v} & \text{E} & \text{C}_2^z & \sigma_{xz} & \sigma_{yz} & \text{h = 4} \ \hline \text{A}_1 & 1 & 1 & 1 & 1 & z,~ x^2,~y^2,~z^2 \ \hline \text{A}_2 & 1 & 1 & -1 & -1 & \text{xy, R}_z \ \hline \text{B}_1 & 1 & -1 & 1 & -1 & \text{x, xz, R}_y \ \hline \text{B}_2 & 1 & -1 & -1 & 1 & \text{y, yz, R}_x \ \hline \Gamma_{tot} & -9 & -1 & 1 & 3 \ \hline \end{array} \nonumber$
The C2v symmetry group has four symmetry elements and four associated symmetry operations. It can be thought of as a four dimensional space with the A1, A2, B1, and B2 irreducible representations playing the role of unite vectors of vector algebra. The irreducible representations span the space and any other vector or representation in that space can be written as a linear combination of them. For simple groups like this character table can be generated by examining how translations along the x-, y-, and z-axes and rotations about these axes transform under the symmetry operations of the group. Using the figure show below you should be able to confirm the designations shown in the right-hand column of the character table. That is the translation in the z-direction transforms like A1, rotation about the z-axis transforms like A2, translation in the x-direction and rotation about the y-axis have symmetry properties represented by B1, and translation in the y-direction and rotation about the x-axis have B2 symmetry. In doing this note (see Atkins) that you record +1 if something is transformed into itself, -1 if it is transformed into minus itself, and 0 if it is transformed into something else under the symmetry operations of the group.
Note that just like the unit vectors in the Cartesian coordinate system, these irreducible representations are orthogonal. That is they have zero overlap.
$\begin{matrix} \text{A}_1 \text{B}_2 = \dfrac{[(-1)(1) + (1)(-1) + (1)(-1) + (1)(1)]}{4} = 0 \ \text{A}_2 \text{B}_1 = \dfrac{[(1)(1) + (1)(-1) + (-1)(1) + (-1)(-1)]}{4} = 0 \end{matrix} \nonumber$
Dividing by 4, the order of the group, normalizes the calculation. We can also show that all of the irreducible representation are normalized or have an overlap of 1.
$\text{A}_2 \text{A}_2 = \dfrac{[(1)(1) + (1)(1) + (-1)(-1) + (-1)(-1)]}{4} = 1 \nonumber$
Note that these vector operations are exactly equivalent to evaluating quantum mechanical overlap integrals $\int \Psi_1 \Psi_2 d \tau$.
Since H2O is a triatomic molecule it has a total of 9 degrees of freedom - three for each atom in the molecule. These are the x-, y-, and z-coordinates of the individual atoms and are shown as small vectors located on each of the atoms in the figure below.
A symmetry analysis for water begins by determining how these 9 coordinates behave under the symmetry operations of the C2v group. You should be able to show that this generates the reducible representation Γtot, which is given in the last row of the character table shown above. Γtot can also be calculated as Γumax + Γy + Γz), where Γuma is the behavior of the atoms under the symmetry operations of the group. Of the 9 degrees of freedom possessed by the water molecule, three are for translation of the center of mass in the x-, y- and z-directions, and three are related to rotation about the x-, y-, and z-axes. This leaves three vibrational degrees of freedom. To determine the symmetry of the vibrational modes we decompose Γtot into the unit vectors or irreducible representations of the C2v character table. This involves taking the dot product of Γtot with each of the irreducible representations of the C2v symmetry.
$\begin{matrix} \Gamma_{tot} \text{A}_1 = \dfrac{[(9)(1) + (-1)(1) + (1)(1) + (3)(1)]}{4} = 3 \ \Gamma_{tot} \text{A}_2 = \dfrac{[(9)(1) + (-1)(1) + (1)(-1) + (3)(-1)]}{4} = 1 \ \Gamma_{tot} \text{B}_1 = \dfrac{[(9)(1) + (-1)(-1) + (1)(1) + (3)(-1)]}{4} = 2 \ \Gamma_{tot} \text{B}_2 = \dfrac{[(9)(1) + (-1)(-1) + (1)(-1) + (3)(1)]}{4} = 3 \end{matrix} \nonumber$
This procedure has revealed the that the reducible representation, Γtot, is composed of the following irreducible representations:
$\Gamma_{tot} = 3 \text{A}_1 + \text{A}_2 + 2 \text{B}_1 + 3 \text{B}_2 \nonumber$
From the right side of the character table we see that translation and rotation have the following symmetry properties:
$\begin{matrix} \Gamma_{trans} = \text{A}_1 + \text{B}_1 + \text{B}_2 \ \Gamma_{rot} = \text{A}_2 + \text{B}_1 + \text{B}_2 \end{matrix} \nonumber$
From this information we can determine the symmetry properties of the vibrational modes of the water molecule.
$\Gamma_{vib} = \Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot} = 2 \text{A}_1 + \text{B}_2 \nonumber$
There are three vibrational fundamentals. And, since there are two bonds there will be two stretching vibrations and one bending vibration. To determine which symmetry classification the bend belongs to, examine how the H-O-H bond angle transforms under the symmetry operations of the C2v group.
The vibrational modes are shown above. Convince yourself that their symmetry properties are captured in the table below.
$\begin{array}{|c|c|c|c|c|c|} \hline \text{C}_{2v} & \text{E} & \text{C}_2^z & \sigma_{xz} & \sigma_{yz} \ \hline \Gamma_{bend} & 1 & 1 & 1 & 1 & \text{A}_1 \ \hline \Gamma_{stretch} & 1 & 1 & 1 & 1 & \text{A}_1 \ \hline \Gamma_{stretch} & 1 & -1 & -1 & 1 & \text{B}_2 \ \hline \end{array} \nonumber$
Whether any of these modes is active in the infrared region of the spectrum is another question. This is determined by evaluating the quantum mechanical transition probability integral,
$P = e \int \Psi^f (vib) \hat{q} \Psi^i (vib) d \tau \nonumber$
where q is the spatial coordinate and is either x, y, or z. If this integral is non-zero the transition is allowed. Group theory enables us to determine if the integral is non-zero as follows. We evaluate the direct product (note the similarity to the quantum mechanical transition probability integral) ΓvibfΓqΓvibi and if it turns out to be equal to, or contain, the irreducible representation A1 then the transition is allowed. If the direct product isn't equal to or contain A1 then the transition is forbidden.
Γvibi, the representation for the ground state is always equal to A1 and Γvibf is always equal to A1 and Γvibf has the symmetry of the vibrational mode being excited, which in our case is either A1 or B2. Thus we can see that a vibrational mode will be infrared active if it belongs to the same symmetry species as one of the Cartesian coordinates. You should verify that this is correct and also be able to show that all three vibrational modes of the water molecule are infrared active.
Group theory is also very helpful in constructing molecular orbitals from linear combinations of atomic orbitals. For the water molecule the atomic orbitals are the hydrogen 1s orbitals and the 2s and three 2p orbitals on the oxygen atom. We proceed by examining how the orbitals transform under the symmetry operations of the group. This is shown in the table below.
$\begin{array}{|c|c|c|c|c|c|} \hline \text{C}_{2v} & \text{E} & \text{C}_2^z & \sigma_{xz} & \sigma_{yz} \ \hline \text{H}_1 & \text{H}_1 & \text{H}_2 & \text{H}_2 & \text{H}_1 \ \hline \text{O}_{2s} & \text{O}_{2s} & \text{O}_{2s} & \text{O}_{2s} & \text{O}_{2s} \ \hline \text{O}_{2px} & \text{O}_{2px} & - \text{O}_{2px} & \text{O}_{2px} & - \text{O}_{2px} \ \hline \text{O}_{2py} & \text{O}_{2py} & - \text{O}_{2py} & - \text{O}_{2py} & \text{O}_{2py} \ \hline \text{O}_{2pz} & \text{O}_{2pz} & \text{O}_{2pz} & \text{O}_{2pz} & \text{O}_{2pz} \ \hline \text{A}_1 & 1 & 1 & 1 & 1 \ \hline \text{A}_2 & 1 & 1 & -1 & -1 \ \hline \text{B}_1 & 1 & -1 & 1 & -1 \ \hline \text{B}_2 & 1 & -1 & -1 & 1 \ \hline \end{array} \nonumber$
To find the molecular orbital with A1 symmetry take the sum of the dot product of the A1 irreducible representation with each of the representations in the top half of the table. This is 2(H1 + H2) + 4O2s + 4O2pz. This indicates that the molecular orbital will consist of a linear combination of four of the six orbitals shown in the table above.
$\Psi (A_1) = c_h \left[ \psi (H_1) + \psi (H_2) \right] + c_o \psi (O_{2s}) + c_o \psi (O_{2pz}) \nonumber$
When this is done for A2 symmetry it is found that the sum is zero, indicating that there is no molecular orbital with A2 symmetry. Repeating this process for B1 symmetry yields 4O2px. This means Ψ(B2) = Ψ(O2px) and we say that the oxygen 2px orbital is non-bonding because there are hydrogen orbitals with the same symmetry. This is easily seen by inspection of the water molecule when placed in a Cartesian coordinate system as shown in the figure above.
For B2 symmetry the result is 2(H1 - H2) + 4O2py. This is clearly another bonding molecular orbital.
$\Psi (B_2) = c_h \left[ \psi (H_1) - \psi (H_2) \right] + c_o \psi (O_{2py}) \nonumber$
The next step is to use the molecular orbitals from this symmetry analysis as the basis for a self-consistent field calculation on the water molecule. This will yield a set of coefficients for the molecular orbitals and a molecular orbital energy level diagram. This results are given below. You should compare these results with the traditional valence bond (Lewis structure) formulation of the bonding in H2O.
$\begin{matrix} \Psi (A_1) = 0.15 \left[ \psi (H_1) + \psi (H_2) \right] + 0.82 \psi (O_{2s}) + 0.13 \psi (O_{2pz}) \ \Psi (B_2) = 0.42 \left[ \psi (H_1) - \psi (H_2) \right] + 0.62 \psi (O_{2py}) \ \Psi (A_1) = 0.26 \left[ \psi (H_1) + \psi (H_2) \right] - 0.50 \psi (O_{2s}) + 0.79 \psi (O_{2pz}) \ \Psi (B_1) = \Psi (O_{2px}) \ \Psi (A_1) = 0.75 \left[ \psi (H_1) + \psi (H_2) \right] - 0.84 \psi (O_{2s}) - 0.70 \psi (O_{2pz}) \ \Psi (B_2) = 0.89 \left[ \psi (H_1) - \psi (H_2) \right] - 0.99 \psi (O_{2py}) \end{matrix} \nonumber$
The 1a1 MO is, as would be expected, the oxygen non-valence 1s orbital. The 2a1 MO is mainly a linear combination of the hydrogen 1s and the 2s on oxygen. The 1b2 MO is entirely the oxygen 2py and the hydrogen 1s atomic orbitals. The 3a1 MO and the 1b1 MO correspond to the two long pairs. The 3a1 MO and the 1b1 MO correspond to the two long pairs. The 3a1 MO consists mainly of the oxygen 2s and 2pz and the hydrogen 1s atomic orbitals and has its maximum value in the +z direction. The 1b1 is entirely oxygen 2px and thus is directed at right angles to the plane of the molecule; the lone pairs thus have their maxima in the xz plane.
The first two electronic transitions in water occur at 170 and 130 nm. The molecular orbital diagram obtained previously and the principles of group theory can be used to intepret the electronic spectrum as is shown in the figure below.
Molecular orbital scheme and state diagram for H2O.
By analogy with IR spectroscopy those electronic transitions are allowed for which the transition moment integral is non-zero.
$M = e \int \Psi^f (elec) \hat{q} \Psi^i (elec) d \tau \nonumber$
As in the case of IR spectroscopy, this integral can be non-zero if and only if ΓvibfΓqΓvibi is equal to or contains the totally symmetric irreducible representation, A1. The figure shows that the electronic ground state has A1 symmetry while the electric dipole operator (μ=eq) has the symmetry of the Cartesian coordinates (A1, B1, B2). Thus ΓvibfΓqΓvibi will equal A1 if the excited state has A1, B1, or B2 symmetry. If this is the case then the transition is said to be orbitally allowed. Note that several orbitally allowed transitions are spin-forbidden (involve a change in spin multiplicity) and therefore do not occur.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.01%3A_Group_Theory_Principles_Applied_to_H2O.txt
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When Paquette's group synthesized dodecahedrane, C20H20, they measured its infrared and Raman spectra (JACS 1983, 105, 5446-5450).
Dodecahedrane: (Public Domain; Yikrazuul ).
They found three IR active bands at 2945, 1298, and 728 cm-1 and eight Raman frequencies at 2924, 2938, 1324, 1164, 1092, 840, 676, and 480 cm-1. Use group theory to show that these data are consistent with the fact that dodecahedrane has icosahedral symmetry.
$\begin{matrix} \begin{array} E & E & C_5 & C_5^2 & & C_3& C_2& & i& & S_{10} & & S_{10}^3 & & S_6 & & \sigma \end{array} & ~ \ \text{CIh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 \ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 \ 4 & -1 & -1 & 1 & 0 & 4 & -1 & -1 & 1 & 0 \ 5 & 0 & 0 & -1 & 1 & 5 & 0 & 0 & -1 & 1 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 & -3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & 1 \ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 & -3 & - \frac{1 - \sqrt{5}}{2} & - \frac{1 + \sqrt{5}}{2} & 0 & 1 \ 4 &-1 & -1 & 1 & - & -4 & 1 & 1 & -1 & 0 \ 5 & 0 & 0 & -1 & 1 & -5 & 0 & 0 & 1 & -1 \end{bmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 - x^2- y^2,~x^2 - y^2, xy, yz, xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} \end{matrix} \nonumber$
$\begin{matrix} \text{Ih} = \begin{pmatrix} 1 & 12 & 12 & 20 & 15 & 1 & 12 & 12 & 20 & 15 \end{pmatrix} & \text{Ih = Ih}^T & \Gamma \text{uma} = \begin{pmatrix} 40 & 0 & 0 & 4 & 0 & 0 & 0 & 0 & 0 & 8 \end{pmatrix} & \Gamma \text{uma} = \Gamma \text{uma}^T \end{matrix} \nonumber$
$\begin{matrix} \text{Ag} = ( \text{CIh}^T )^{<1>} & \text{T1g} = ( \text{CIh}^T )^{<2>} & \text{T2g} = ( \text{CIh}^T )^{<3>} & \text{Gg} = ( \text{CIh}^T )^{<4>} & \text{Hg} = ( \text{CIh}^T )^{<5>} \ \text{Au} = ( \text{CIh}^T )^{<6>} & \text{T1u} = ( \text{CIh}^T )^{<7>} & \text{T2u} = ( \text{CIh}^T )^{<8>} & \text{Gu} = ( \text{CIh}^T )^{<9>} & \text{Hu} = ( \text{CIh}^T )^{<10>} \end{matrix} \nonumber$
$\begin{matrix} h = \sum Ih & h = 120 & \Gamma_{tot} = \overrightarrow{( \Gamma T1u)} & \Gamma_{vib} = \Gamma_{tot} - T1g - T1u & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & i = 1 .. 10 \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{[ \overrightarrow{ \text{Ih} ( \text{CIh}^T)^{ <i>} \Gamma \text{vib}}]}{h} & \text{Stretch}_i = \frac{[ \overrightarrow{ \text{Ih} ( \text{CIh}^T)^{ <i>} \Gamma \text{stretch}}]}{h} & \text{Bend}_i = \frac{[ \overrightarrow{ \text{Ih} ( \text{CIh}^T)^{ <i>} \Gamma \text{bend}}]}{h} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib} = \begin{bmatrix} 2 \ 1 \ 2 \ 4 \ 6 \ 0 \ 3 \ 4 \ 4 \ 4 \end{bmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 - x^2- y^2,~x^2 - y^2, xy, yz, xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} & \text{Stretch} = \begin{bmatrix} 2 \ 0 \ 0 \ 2 \ 3 \ 0 \ 2 \ 2 \ 2 \ 1 \end{bmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 - x^2- y^2,~x^2 - y^2, xy, yz, xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} \end{matrix} \nonumber$
$\begin{matrix} \text{Bend} = \begin{bmatrix} 0 \ 1 \ 2 \ 2 \ 3 \ 0 \ 1 \ 2 \ 2 \ 3 \end{bmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 - x^2- y^2,~x^2 - y^2, xy, yz, xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} \end{matrix} \nonumber$
According to the usual selection rules only the three T1u vibrations are IR active. The Ag and Hg vibrations are Raman active giving a total of eigth frequencies in the Raman spectra. This analysis is in agreement with the experimental spectroscopic results quoted by Paquette. Note also that, as is usual for molecules with a center of inversion, there are no coincidences between the IR and Raman active modes.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.02%3A_Dodecahedrane.txt
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Xenon tetrafluoride is a chemical compound with chemical formula $\ce{XeF4}$. It was the first discovered binary compound of a noble gas. It is produced by the chemical reaction of xenon with fluorine, $\ce{F2}$, according to the chemical equation:
$\ce{Xe + 2 F2 → XeF4} \nonumber$
The infrared spectrum of $\ce{XeF4}$ has absorptions at 161, 291, and 586 cm-1 (two bends, one stretch), while the Raman spectrum has peaks at 218, 524, and 554 cm-1 (one bend, two stretches). Is its molecular structure tetrahedral or square planar?
Tetrahedral Symmetry - Td
$\begin{matrix} \begin{array} E & E & C_3 & C_2 & & S_4 & \sigma \end{array} & ~ \ \text{CIh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 \ 2 & -1 & 2 & 0 & 0 \ 3 & 0 & -1 & 1 & -1 \ 3 & 0 & -1 & -1 & 1 \end{bmatrix} & \begin{array} \text{A}_1: ~ x^2 + y^2 + z^2 \ \text{A}_2 \ \text{E}:~2z^2-x^2y^2, x^2-y^2 \ \text{T}_1:~(R_x,~R_y,~R_z) \ \text{T}_2: \text{ (x, y, z)(xy, xz, yz)} \end{array} & \text{Td} = \begin{bmatrix} 1 \ 8 \ 3 \ 6 \ 6 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 5 \ 2 \ 1 \ 1 \ 3 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{Td}^T)^{<1>} & A_2 = ( C_{Td}^T )^{<2>} & E = ( C_{Td}^T )^{<3>} & T_1 = (C_{Td}^T )^{<4>} \ T_2 = (C_{Td}^T )^{<5.} & \Gamma_{tot} = \overrightarrow{ \Gamma_{uma} T_2} & h = \sum Td & \Gamma_{tot}^T = \begin{pmatrix} 15 & 0 & -1 & -1 & 3 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{vib} = \Gamma_{tot} - T_1 - T_2 & \text{Vib}_i = \frac{ \sum \overrightarrow{ [Td (C Td^T)^{<i>} \Gamma_{vib}]}}{h} & \text{Vib} = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \ 2 \end{bmatrix} \begin{array} \text{A}_1:~ x^2 + y^2 +z^2 \ \text{A}_2 \ \text{E: } 2z^2 -x^2 - y^2, x^2 - y^2 \ \text{T}_1:~ (R_x,~R_y,~R_z) \ \text{T}_2:~ (x,~y,~z)(xy,~xz,~yz) \end{array} \end{matrix} \nonumber$
This analysis predicts two IR active modes (2T2) and four Raman active modes (A1, E, 2T2). It also suggests that the IR and Raman should have the two T2 modes in common. Thus, tetrahedral geometry is not consistent with the spectroscopic data. Further detail can be obtained by noting that the stretching vibrations have the same symmetry properties as the chemical bonds. This allows the vibrational modes to be decomposed further into the symmetry of the stretches and bends.
Square Planar Geometry - D4H
$\begin{matrix} \begin{array} E & & E & C_4 & C_2 & C_2' & C_2"& i& S_4& & \sigma_h & \sigma_v & \sigma_v & \end{array} & ~ \ \text{CIh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 \ 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 \ 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 \ 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1\ 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 \ 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \ 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 \end{bmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g: Rz} \ \text{B1g: } x^2-y^2\ \text{B2g: xy} \ \text{Eg: (Rx, Ry), xz, yz)} \ \text{A1u} \ \text{A2u: z} \ \text{B1u} \ \text{B2u} \ \text{Eu: (x, y)} \end{array} & \text{D4h} = \begin{bmatrix} 1 \ 2 \ 1 \ 2 \ 2 \ 1 \ 2 \ 1 \ 2 \ 2 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 5 \ 1 \ 1 \ 3 \ 1 \ 1 \ 1 \ 5 \ 3 \ 1 \end{bmatrix} & \Gamma_{bonds} = \begin{bmatrix} 4 \ 0 \ 0 \ 2 \ 0 \ 0 \ 0 \ 4 \ 2 \ 0 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{A}_{1g} = ( \text{C}_{D4h}^T )^{<1>} & \text{A}_{2g} = ( \text{C}_{D4h}^T )^{<2>} & \text{B}_{1g} = ( \text{C}_{D4h}^T )^{<3>} & \text{B}_{2g} = ( \text{C}_{D4h}^T )^{<4>} & \text{E}_g = ( \text{C}_{D4h}^T )^{<5>} \ \text{A}_{1u} = ( \text{C}_{D4h}^T )^{<6>} & \text{A}_{2u} = ( \text{C}_{D4h}^T )^{<7>} & \text{B}_{1u} = ( \text{C}_{D4h}^T )^{<8>} & \text{B}_{2u} = ( \text{C}_{D4h}^T )^{<9>} & \text{E}_u = ( \text{C}_{D4h}^T )^{<10>} \end{matrix} \nonumber$
$\begin{matrix} h = \sum D4h & \Gamma_{trans} = A_{2u} + E_u & \Gamma_{rot} = A_{2g} + E_g & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} \Gamma_{trans})} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{vib} = \Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot} & \Gamma_{stretch} = \Gamma_{bonds} & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & i = 1 .. 10 \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{[ D4h ( C_{D4h}^T )^{<i>} \Gamma_{vib}]}}{h} & \text{Stretch}_i = \frac{ \sum \overrightarrow{[ D4h ( C_{D4h}^T )^{<i>} \Gamma_{stretch}]}}{h} & \text{Bend}_i = \frac{ \sum \overrightarrow{[ D4h ( C_{D4h}^T )^{<i>} \Gamma_{bend}]}}{h} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib} = \begin{bmatrix} 1 \ 0 \ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 2 \end{bmatrix} \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g: Rz} \ \text{B1g: } x^2-y^2\ \text{B2g: xy} \ \text{Eg: (Rx, Ry), xz, yz)} \ \text{A1u} \ \text{A2u: z} \ \text{B1u} \ \text{B2u} \ \text{Eu: (x, y)} \end{array} & \Gamma_{uma} = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \end{bmatrix} \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g: Rz} \ \text{B1g: } x^2-y^2\ \text{B2g: xy} \ \text{Eg: (Rx, Ry), xz, yz)} \ \text{A1u} \ \text{A2u: z} \ \text{B1u} \ \text{B2u} \ \text{Eu: (x, y)} \end{array} & \Gamma_{bonds} = \begin{bmatrix} 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \end{bmatrix} \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g: Rz} \ \text{B1g: } x^2-y^2\ \text{B2g: xy} \ \text{Eg: (Rx, Ry), xz, yz)} \ \text{A1u} \ \text{A2u: z} \ \text{B1u} \ \text{B2u} \ \text{Eu: (x, y)} \end{array} \end{matrix} \nonumber$
This analysis predicts three IR active modes (A2u, 2Eu) and three Raman active modes (A1g, B1g, B2g). It also indicates that there are no coincidences between the IR and Raman spectra. Thus, square planar geometry is consistent with the spectroscopic data. Examining the symmetry of the stretches and bends adds further support to this conclusion.
This analysis also predicts two Raman (A1g, B1g) and one IR (Eu) active stretches, and two IR (A2u, Eu) and one Raman (B2g) active bends. This is also consistent with the experiemental data.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.03%3A_Xenon_Tetrafluoride.txt
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Diborane - D2h Symmetry
Diborane has 18 vibrational degrees of freedom. Nine modes are Raman active and eight are IR active. The experimental results are provided in the table below. Do a symmetry analysis to confirm the assignments given below, and identify stretches and bends.
$\begin{pmatrix} D_{2h} & A_g & A_g & A_g & A_g & B_{1g} & B_{1g} & B_{2g} & B_{2g} & B_{3g} \ \frac{ \text{Raman}}{ \text{cm}} & 2524 & 2104 & 1180 & 794 & 1768 & 1035 & 2591 & 920 & 1012 \ D_{2h} & A_u & B_{1u} & B_{1u} & B_{1u} & B_{2u} & B_{2u} & B_{3u} & B_{3u} & B_{3u} \ \frac{ \text{IR}}{ \text{cm}} & 0 & 2612 & 950 & 368 & 1915 & 973 & 2525 & 1606 & 1177 \end{pmatrix} \nonumber$
$\begin{matrix} \begin{array} E & & & E & C_2^z & C_2^y & C_2^x & i& \sigma_{xy} & \sigma_{xz} & \sigma_{yz} & \end{array} & ~ \ \text{C}_{D2h} = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \ 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \ 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \ 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 \ 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 \ 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 \ 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 \end{pmatrix} & \begin{array} \text{A}_g:~ x^2,~ y^2,~ z^2 \ \text{B}_{1g}:~ R_x,~xy \ \text{B}_{2g}:~ R_y,~xz \ \text{B}_{3g}:~ R_x,yx \ \text{A}_u \ \text{B}_{1u}:~z \ \text{B}_{2u}:~y \ \text{B}_{3u}:~x \end{array} & \text{D2h} = \begin{pmatrix} 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 8 \ 0 \ 2 \ 2 \ 0 \ 4 \ 6 \ 2 \end{pmatrix} & \Gamma_{bonds} = \begin{pmatrix} 8 \ 0 \ 0 \ 0 \ 0 \ 4 \ 4 \ 0 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{A}_{g} = ( \text{C}_{D4h}^T )^{<1>} & \text{B}_{2g} = ( \text{C}_{D4h}^T )^{<2>} & \text{B}_{2g} = ( \text{C}_{D4h}^T )^{<3>} & \text{B}_{3g} = ( \text{C}_{D4h}^T )^{<4>} & ~ \ \text{A}_{u} = ( \text{C}_{D4h}^T )^{<5>} & \text{B}_{1u} = ( \text{C}_{D4h}^T )^{<6>} & \text{B}_{2u} = ( \text{C}_{D4h}^T )^{<7>} & \text{B}_{3u} = ( \text{C}_{D4h}^T )^{<8>} & h = \sum \text{D2h} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{trans} = B_{1u} + B_{2u} + B_{3u} & \Gamma_{tot} = B_{1g} + B_{2g} + B_{3g} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} \Gamma_{trans})} \ \Gamma_{vib} = \Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot} & \Gamma_{vib}^T = \begin{pmatrix} 18 & 2 & 0 & 0 & 0 & 4 & 6 & 2 \end{pmatrix} & i = 1 .. 8 \ \Gamma_{stretch} = \Gamma_{bonds} & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{[ D2h ( C_{D2h}^T )^{<i>} \Gamma_{vib}]}}{h} & \text{Stretch}_i = \frac{ \sum \overrightarrow{[ D2h ( C_{D2h}^T )^{<i>} \Gamma_{stretch}]}}{h} & \text{Bend}_i = \frac{ \sum \overrightarrow{[ D2h ( C_{D2h}^T )^{<i>} \Gamma_{bend}]}}{h} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib} = \begin{pmatrix} 4 \ 2 \ 2 \ 1 \ 1 \ 3 \ 2 \ 3 \end{pmatrix} \begin{array} \text{A}_g:~ x^2,~ y^2,~ z^2 \ \text{B}_{1g}:~ R_x,~xy \ \text{B}_{2g}:~ R_y,~xz \ \text{B}_{3g}:~ R_x,yx \ \text{A}_u \ \text{B}_{1u}:~z \ \text{B}_{2u}:~y \ \text{B}_{3u}:~x \end{array} & \text{Stretch} = \begin{pmatrix} 2 \ 1 \ 1 \ 0 \ 0 \ 1 \ 1 \ 2 \end{pmatrix} \begin{array} \text{A}_g:~ x^2,~ y^2,~ z^2 \ \text{B}_{1g}:~ R_x,~xy \ \text{B}_{2g}:~ R_y,~xz \ \text{B}_{3g}:~ R_x,yx \ \text{A}_u \ \text{B}_{1u}:~z \ \text{B}_{2u}:~y \ \text{B}_{3u}:~x \end{array} & \text{Bend} = \begin{pmatrix} 2 \ 1 \ 1 \ 1 \ 1 \ 2 \ 1 \ 1 \end{pmatrix} \begin{array} \text{A}_g:~ x^2,~ y^2,~ z^2 \ \text{B}_{1g}:~ R_x,~xy \ \text{B}_{2g}:~ R_y,~xz \ \text{B}_{3g}:~ R_x,yx \ \text{A}_u \ \text{B}_{1u}:~z \ \text{B}_{2u}:~y \ \text{B}_{3u}:~x \end{array}\end{matrix} \nonumber$
This analysis is in agreement with the experimental data. There are 9 Raman active modes and 8 IR active modes. Furthermore there are 4 Raman stretches at 2524 (Ag), 2104 (Ag), 1768 (B1g), and 2591 (B2g). The five Raman bends occur at 1180 (Ag), 794 (Ag), 1035 (B1g), 920 (B2g), and 1012 (B3g).
The 4 IR stretches occur at 2612 (B1u), 1915 (B2u), 2525 (B3u), and 1606 (B3u). The bends appear at 950 (B1u), 368 (B1u), 973 (B2u), 1177 (B3u).
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.04%3A_Diborane.txt
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Cubane, C8H8, has 42 vibrational degrees of freedom, but only three IR active modes. Cubane belongs to the octahedral point group. Show that group theory predicts three IR active modes. Determine how many vibrational modes will be Raman active. Will there be any coincidences between the IR and Raman active modes? The synthesis and characterization of cubane was reported in 1964 by Philip Eaton and Thomas Cole in JACS 1964, 86, 3157-3158. They reported three IR bands at 3000, 1231, and 851 cm-1.
$\begin{matrix} \begin{array} E & & E & C_3 & C_2 & C_4 & C_2" & i & S_{4} & S_{6} & \sigma_h & \sigma_d \end{array} & ~ \ \text{C}_{Oh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & 1 & 1 \ 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 \ 3 & 0 & -1 & 1 & -1 & 3 & 1 & 0 & -1 & 0 \ 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 \ 1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 \ 2 & -1 & 0 & 0 & 2 & -2 & 0 & 1 & -2 & 0 \ 3 & 0 & -1 & 1 & -1 & -3 & -1 & 0 & 1 & 1 \ 3 & 0 & 1 & -1 & -1 & -3 & 1 & 0 & 1 & -1 \end{bmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2,~x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g: }xy,~xz,~yz \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} \end{matrix} \nonumber$
$\begin{matrix} \text{Ag} = ( \text{C}_{Oh}^T )^{<1>} & \text{A}_{2g} = ( \text{C}_{Oh}^T )^{<2>} & \text{E}_{g} = ( \text{C}_{Oh}^T )^{<3>} & \text{T}_{1g} = ( \text{C}_{Oh}^T )^{<4>} & \text{T}_{2g} = (C_{Oh}^T)^{<5>} \ \text{A}_{1u} = ( \text{C}_{Oh}^T )^{<6>} & \text{A}_{2u} = ( \text{C}_{Oh}^T )^{<7>} & \text{E}_{u} = ( \text{C}_{Oh}^T )^{<8>} & \text{T}_{1u} = ( \text{C}_{Oh}^T )^{<9>} & \text{T}_{2u} = (C_{Oh}^T)^{<10>} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{[ Oh (C_{Oh}^T)^{<i>} \Gamma_{vib} ]}}{h} & \text{Stretch}_i = \frac{ \sum \overrightarrow{[ Oh (C_{Oh}^T)^{<i>} \Gamma_{stretch} ]}}{h} & \text{Bend}_i = \frac{ \sum \overrightarrow{[ Oh (C_{Oh}^T)^{<i>} \Gamma_{bend} ]}}{h} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib} = \begin{bmatrix} 2 \ 0 \ 2 \ 1 \ 4 \ 0 \ 2 \ 2 \ 3 \ 2 \end{bmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2, x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} & \text{Stretch} = \begin{bmatrix} 2 \ 0 \ 1 \ 0 \ 2 \ 0 \ 1 \ 0 \ 2 \ 1 \end{bmatrix} \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2, x^2 - y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} & \text{Bend} = \begin{bmatrix} 0 \ 0 \ 1 \ 1 \ 2 \ 0 \ 1 \ 2 \ 1 \ 1 \end{bmatrix} \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2, x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} \end{matrix} \nonumber$
Only the three T1u vibrational modes are IR active, which is consistent with the spectroscopic data. The character table indicates that the A1g, Eg, and T2g modes are Raman active. Thus there should be 8 Raman active modes. These have been observed at 2996, 2978, 1185, 1081, 1002, 912, 826, and 665 cm-1 (J. Phys. Chem. 1981, 85, 2186). There should be no coincidences between the IR and Raman modes because cubane has a center of inversion.
The vibrational modes can be sorted into stretches and bends by determining how the chemical bonds transform under the symmetry operations of the octahedral group. The symmetry of the stretching modes is the same as the symmetry of the bonds.
This analysis tells us that there are two IR active stretches (2T1u) and five Raman active stretches (2A1g, Eg, and 2T2g). This is consistent with the experimental spectra in that stretches generally occur at a higher frequency than bends.
6.05: Cubane
C60 Has Icosahedral Symmetry
Buckminsterfulerene has four IR active vibrational modes (528, 577, 1180, 1430 cm-1) and ten Raman active modes (273, 436, 496, 710, 773, 110, 1250, 1435, 1470, 1570 cm-1). Demonstrate that the assumption of icosahedral symmetry for C60 is consistent with this data.
$\begin{matrix} \begin{array} E & & E & C_5 & & C_5^2 & C_3 & & C_2 & i & & S_{10} & S_{10}^3 & & S_6 & \sigma \end{array} & ~ \ \text{CIh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1- \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1- \sqrt{5}}{2} & \frac{1+ \sqrt{5}}{2} & 0 & -1 \ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1+ \sqrt{5}}{2} & 0 & -1 & 3 & \frac{1+ \sqrt{5}}{2} & \frac{1- \sqrt{5}}{2} & 0 & -1 \ 4 & -1 & -1 & 1 & 0 & 4 & -1 & -1 & 1 & 0 \ 5 & 0 & 0 & -1 & 1 & 5 & 0 & 0 & -1 & 1 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 3 & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} & 0 & -1 & -3 & - \frac{1 - \sqrt{5}}{2} & - \frac{1 + \sqrt{5}}{2} & 0 & 1 \ 3 & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & 0 & -1 & -3 & - \frac{1 + \sqrt{5}}{2} & - \frac{1 - \sqrt{5}}{2} & 0 & 1 \ 4 & -1 & -1 & 1 & 0 & -4 & 1 & 1 & -1 & 0 \ 5 & 0 & 0 & -1 & 1 & -5 & 0 & 0 & 1 & -1 \end{bmatrix} & \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{Eg: } 2z^2-x^2-y^2,x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 -x^2-y^2,~x^2-y^2, ~xy,~yz,~xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} \end{matrix} \nonumber$
$\begin{matrix} \text{Ih:} \begin{pmatrix} 1 & 12 & 12 & 20 & 15 & 1 & 12 & 12 & 20 & 15 \end{pmatrix} & \text{Ih = Ih}^T & \Gamma_{uma} = \begin{pmatrix} 60 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 \end{pmatrix} & \Gamma_{uma} = \Gamma_{uma}^T \ \Gamma_{bonds} = \begin{pmatrix} 90 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 8 \end{pmatrix} & \Gamma_{bonds} = \Gamma_{bonds}^T & \Gamma_{stretch} = \Gamma_{bonds} \end{matrix} \nonumber$
$\begin{matrix} \text{Ag} = ( \text{CIh}^T )^{<1>} & \text{T}_{1g} = ( \text{CIh}^T )^{<2>} & \text{T}_{2g} = ( \text{CIh}^T )^{<3>} & \text{G}_{g} = ( \text{CIh}^T )^{<4>} & \text{H}_{g} = (\text{CIh}^T)^{<5>} \ \text{A}_{u} = ( \text{CIh}^T )^{<6>} & \text{T}_{1u} = ( \text{CIh}^T )^{<7>} & \text{A}_{u} = ( \text{CIh}^T )^{<8>} & \text{G}_{u} = ( \text{CIh}^T )^{<9>} & \text{H}_{u} = (\text{CIh}^T)^{<10>} \end{matrix} \nonumber$
$\begin{matrix} h = \sum \text{Ih} & h = 120 & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} T1u)} & \Gamma_{vib} = \Gamma_{tot} - T1g - T1u & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & i = 1 .. 10 \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{\sum \overrightarrow{[\text{Ih} ( \text{CIh}^T)^{<i>} \Gamma_{vib}]}}{h} & \text{Vib} = \begin{bmatrix} 2 \ 3 \ 4 \ 6 \ 8 \ 1 \ 4 \ 5 \ 6 \ 7 \end{bmatrix} \begin{array} \text{Ag: }x^2 + y^2 + z^2 \ \text{Eg: } 2z^2-x^2-y^2,x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g} \ \text{Gg} \ \text{Hg: }2z^2 -x^2-y^2,~x^2-y^2, ~xy,~yz,~xz \ \text{Au} \ \text{T1u: x, y, z} \ \text{T2u} \ \text{Gu} \ \text{Hu} \end{array} \end{matrix} \nonumber$
The 4 T1u modes are IR active and the 2 Ag and 8 Hg modes are Raman active. Also there are no coincidences. Thus the assumption of icosahedral symmetry is consistent with the spectroscopic data.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.05%3A_Cubane/6.5.01%3A_Buckminsterfulerene.txt
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C3v Symmetry - BCl3
The infrared spectrum of BCl3 shows vibrational bands at 995, 480, and 244 cm-1, while Raman bands appear at 995, 471, and 244 cm-1. Is the geometry of the molecule trigonal pyramid (C3v) or trigonal planar (D3h)? Is your answer to this question consistent with chemical bonding principles (VSEPR)? Assign symmetry labels to the vibrational bands and identify the stretches and bends.
$\begin{matrix} ~ & \begin{array} E & C_3 & \sigma_v \end{array} \ C_{C3v} = & \begin{pmatrix} 1 & 1 & 1 \ 1 & 1 & -1 \ 2 & -1 & 0 \end{pmatrix} & \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \ A_2:~R_z \ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} & C3v = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 4 \ 1 \ 2 \end{pmatrix} & \Gamma_{bonds} = \begin{pmatrix} 3 \ 0 \ 1 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{C3v}^T)^{<1>} & A_2 = (C_{C3v}^T)^{<2>} & E = (C_{C3v}^T)^{<3>} & h = \sum C3v \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{tot} = \overrightarrow{[\Gamma_{uma} (A_1 + E)]} & \Gamma_{tot}^T = \begin{pmatrix} 12 & 0 & 2 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - A_1 - A_2 - 2E \end{matrix} \nonumber$
$\begin{matrix} i = 1 .. 3 & \text{Vib}_i = \frac{ \sum \overrightarrow{ [C3v (C_{C3v}^T)^{<i>} \Gamma_{vib} ]}}{h} & \text{Vib} = \begin{pmatrix} 2 \ 0 \ 2 \end{pmatrix} \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \ A_2:~R_z \ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} \ \Gamma_{stretch} = \Gamma_{bonds} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ [C3v (C_{C3v}^T)^{<i>} \Gamma_{stretch} ]}}{h} & \text{Stretch} = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \ A_2:~R_z \ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} \ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{ \sum \overrightarrow{ [C3v (C_{C3v}^T)^{<i>} \Gamma_{bend} ]}}{h} & \text{Bend} = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \begin{array} A_1:~ z,~x^2 + y^2,~z^2 \ A_2:~R_z \ E:~(x,~y)(R_x,~R_y),~(x^2 + y^2,~xy)(xz,~yz) \end{array} \end{matrix} \nonumber$
This analysis predicts that there should be 4 IR and 4 Raman active vibrations. It also predicts that the IR and Raman vibrations should be coincident. This is not in agreement with the experimental data, so BCl3 does not have trigonal pyramidal geometry.
This analysis also predicts that the stretches should have A1 and E symmetry, but since it has already been concluded that the molecule does not have C3v symmetry this result will not be discussed further.
D3h Symmetry - BCl3
$\begin{matrix} ~ & \begin{array} E & C_3 & C_2 & \sigma_h & S_3 & \sigma_v \end{array} \ C_{C3v} = & \begin{pmatrix} 1 & 1 & 1 1 & 1 & 1 \ 1 & 1 & -1 & 1 & 1 & -1 \ 2 & -1 & 0 & 2 & -1 & 0 \ 1 & 1 & 1 & -1 & -1 & -1 \ 1 & 1 & -1 & -1 & 1 \ 2 & -1 & 0 & -2 & 1 & 0 \end{pmatrix} & \begin{array} A_1':~ x^2 + y^2,~z^2 \ A_2':~Rz \ E':~(x,~y), ~(x^2 + y^2,~xy) \ A1": \ A2":~z \ \text{E": (Rx, Ry), (xz, yz)} \end{array} & D3h = \begin{bmatrix} 1 \ 2 \ 3 \ 1 \ 2 \ 3 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 4 \ 1 \ 2 \ 4 \ 1 \ 2 \end{bmatrix} & \Gamma_{bonds} = \begin{bmatrix} 3 \ 0 \ 1 \ 3 \ 0 \ 1 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{D3h}^T)^{<1>} & A_2 = (C_{D3h}^T)^{<2>} & E = (C_{D3h}^T)^{<3>} & A_{11} = (C_{D3h}^T)^{<4>} \ A_{21} = (C_{D3h}^T )^{<5>} & E_1 = (C_{D3h}^T )^{<6>} & h = \sum D3h & \Gamma_{tot} = \overrightarrow{[ \Gamma_{uma} (A_{21} + E)]} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{tot}^T = \begin{pmatrix} 12 & 0 & -2 & 4 & -2 & 2 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - A_2 - E - A_{21} - E_1 & i = 1 .. 6 \ \Gamma_{stretch} = \Gamma_{bonds} & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{ [D3h (C_{D3h}^T)^{<i>} \Gamma_{vib} ]}}{h} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ [D3h (C_{D3h}^T)^{<i>} \Gamma_{stretch} ]}}{h} & \text{Bend}_i = \frac{ \sum \overrightarrow{ [D3h (C_{D3h}^T)^{<i>} \Gamma_{bend} ]}}{h} \ \text{Vib} = \begin{bmatrix} 1 \ 0 \ 2 \ 0 \ 1 \ 0 \end{bmatrix} \begin{array} A_1':~ x^2 + y^2,~z^2 \ A_2':~Rz \ E':~(x,~y), ~(x^2 + y^2,~xy) \ A1": \ A2":~z \ \text{E": (Rx, Ry), (xz, yz)} \end{array} & \text{Stretch} = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \ 0 \ 0 \end{bmatrix} \begin{array} A_1':~ x^2 + y^2,~z^2 \ A_2':~Rz \ E':~(x,~y), ~(x^2 + y^2,~xy) \ A1": \ A2":~z \ \text{E": (Rx, Ry), (xz, yz)} \end{array} \ \text{Bend} = \begin{bmatrix} 0 \ 0 \ 1 \ 0 \ 1 \ 0 \end{bmatrix} \begin{array} A_1':~ x^2 + y^2,~z^2 \ A_2':~Rz \ E':~(x,~y), ~(x^2 + y^2,~xy) \ A1": \ A2":~z \ \text{E": (Rx, Ry), (xz, yz)} \end{array} \end{matrix} \nonumber$
This analysis shows that there are 3 IR active modes (2E', A2") and 3 Raman active modes (A1', E'). It also show that there are two coincidences (2E'). Thus, D3h symmetry for BCl3 is consistent with the spectroscopic data.
Thus the specific assignments are: A1' = 471; A2" = 480; E' = 995; E' = 244. From the analysis above it can be seen that the stretches occur at 995 and 471, and the bends are at 480 and 244 cm-1.
VSEPR theory would predict a trigonal planar arrangement of chlorine atoms around the electron deficient boron, so the symmetry analysis and the bonding theory are in agreement with each other and the spectroscopic measurements.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.06%3A_BCl3.txt
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This exercise has to do with the interpretation of the visible spectrum Ti(H2O)63+ which is shown below.
The analysis will begin by assuming that Ti(H2O)63+ has octahedral symmetry. This assumption accounts for the gross features of the spectrum, but does not explain the shoulder that appears on the main absorption peak. The hexaaquatitanium(III) complex is orbitally degenerate and is therefore subject to a Jahn-Teller distortion which reduces the symmetry to square planar. It will be shown that D4h symmetry is fully consistent with the experimental spectrum. The d-orbital energy level diagrams for Oh and D4h symmetry are shown below and will be referred to later in the analysis.
On the basis of these energy level diagrams we would make the following predictions. For octahedral symmetry there is one electronic transition in the visible region. For square planar symmetry there are two electronic transitions in the visible region and one in the infrared. In the analysis that follows it will be shown that none of the electronic transitions is orbitally allowed, but they are allowed through vibronic coupling. Further more the square planar geometry is in better agreement with the experimental spectrum than octahedral symmetry.
Ti(H2O)63+ - Octahedral Symmetry
$\begin{matrix} \begin{array} E & & E & C_3 & C_2 & C_4 & C_2" & i & S_{4} & S_{6} & \sigma_h & \sigma_d \end{array} & ~ \ \text{C}_{Oh} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & 1 & 1 \ 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 \ 3 & 0 & -1 & 1 & -1 & 3 & 1 & 0 & -1 & 0 \ 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 \ 1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 \ 2 & -1 & 0 & 0 & 2 & -2 & 0 & 1 & -2 & 0 \ 3 & 0 & -1 & 1 & -1 & -3 & -1 & 0 & 1 & 1 \ 3 & 0 & 1 & -1 & -1 & -3 & 1 & 0 & 1 & -1 \end{bmatrix} & \begin{array} \text{A1g: }x^2 + y^2 + z^2 \ \text{A2g} \ \text{Eg: } 2z^2-x^2-y^2,~x^2-y^2 \ \text{T1g: Rx, Ry, Rz} \ \text{T2g: }xy,~xz,~yz \ \text{A1u:} \ \text{A2u} \ \text{Eu} \ \text{T1u: x, y, z} \ \text{T2u} \end{array} & \text{Oh} = \begin{bmatrix} 1 \ 8 \ 6 \ 6 \ 3 \ 1 \ 6 \ 8 \ 3 \ 6 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 7 \ 1 \ 1 \ 3 \ 3 \ 1 \ 1 \ 1 \ 5 \ 3 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{Ag} = ( \text{C}_{Oh}^T )^{<1>} & \text{A}_{2g} = ( \text{C}_{Oh}^T )^{<2>} & \text{E}_{g} = ( \text{C}_{Oh}^T )^{<3>} & \text{T}_{1g} = ( \text{C}_{Oh}^T )^{<4>} \ \text{T}_{2g} = (C_{Oh}^T)^{<5>} & \text{A}_{1u} = ( \text{C}_{Oh}^T )^{<6>} & \text{A}_{2u} = ( \text{C}_{Oh}^T )^{<7>} & \text{E}_{u} = ( \text{C}_{Oh}^T )^{<8>} \ \text{T}_{1u} = ( \text{C}_{Oh}^T )^{<9>} & \text{T}_{2u} = (C_{Oh}^T)^{<10>} & h = \sum \text{Oh} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} T1u)}\end{matrix} \nonumber$
$\begin{matrix} \Gamma_{tot}^T = \begin{pmatrix} 21 & 0 & -1 & 3 & -3 & -3 & -1 & 0 & 5 & 3 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - T1u - T1g \end{matrix} \nonumber$
Determine which irreducible representations contribute to Γvib:
$\begin{matrix} i = 1 .. 10 & X_i = \frac{ \sum \overrightarrow{[ Oh (COh^T)^{<i>} \Gamma_{vib}]}}{h} & X^T = \begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 2 & 1 \end{pmatrix} \end{matrix} \nonumber$
Thus we see that: $\Gamma_{vib} = A1g + Eg + T2g + 2T1u + 2T2u$
Inspection of the character table shows that in octahedral symmetry the d-orbitals are split into a lower T2g (dxy, dxz, dyz) level and an upper Eg (dz2, dx2-y2) level. Ti3+ has one d-electron in the T2g level. As the spectrum below shows, the complex absorbs in the visible region at 20,000 cm-1 (500 nm). However the T2g ---> Eg transition is orbitally forbidden as is shown below.
$\begin{matrix} \int \Psi_{ex} \mu_{e} \Psi_{eg} d \tau_e = 0 & \frac{ \sum \overrightarrow{(Oh~ Eg~ T1u ~T2g)}}{h} = 0 \end{matrix} \nonumber$
In calculating the transition moment for T2g ---> Eg electronic transition it has been assumed that there was no change in the vibrational state of the molecule. However, it is possible for formally forbidden electronic transitions to become allowed through coupling to changes in vibrational state. In other words pure electronic transitions do not actually occur, because the vibrational (and rotational) states of the molecule change at the same time. These are called vibronic transitions and they are allowed if the integral shown below is nonzero.
$\int \int ( \Psi_{ex} \Psi_{vx} \mu_e \Psi_{eg} \Psi_{vg} ) d \tau_e d \tau v \nonumber$
The calculations below show that vibronic calculations involving the T1u and T2u vibrational modes are allowed because the transition moment is not zero.
$\begin{matrix} \frac{ \sum \overrightarrow{(Oh~T1u~Eg~T1u~T2g~A1g)}}{h} = 2 & \frac{ \sum \overrightarrow{(Oh~T1u~Eg~T1u~T2g~A1g)}}{h} = 2\end{matrix} \nonumber$
At this point we have shown that the vibrationally assisted T2g ---> Eg electronic transition is allowed. However, the shoulder on the experimental spectrum suggest that more than one electronic transition is occuring. In the next section we will see that a reduction to square planar symmetry under the Jahn-Teller effect leads to an d-orbital energy level diagram that is consistent with the experimental spectrum.
Ti(H2O)63+ - Square Planar Symmetry
The Jahn-Teller effect predicts a tetragonal distortion of the octahedral complex to the lower D4h square planar symmetry. The energy level diagram is shown above - essentially the ligands on the z-axis move in toward the titanium ion.
$\begin{matrix} ~ & \begin{array} E & C_4 & C_2 & C_2' & C_2" & i & S_4 & \sigma_h & \sigma_v & \sigma_d \end{array} \ CD4h = & \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 \ 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 \ 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 \ 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 \ 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \ 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 \ 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 \ 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \ 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 \end{bmatrix} & \begin{array} A_{1g}:~x^2 + y^2,~z^2 \ A_{2g}:~Rz \ B_{1g}:~ x^2-y^2 \ B_{2g}:~ xy \ E_{g}:~ \text{(Rx, Ry), (xz, yz)} \ A_{1u}: \ A_{2u}:~z \ B_{1u} \ B_{2u} \ E_{u}:~ \text{x, y)} \end{array} & \text{D4h} = \begin{bmatrix} 1 \ 2 \ 1 \ 2 \ 2 \ 1 \ 2 \ 1 \ 2 \ 2 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} \ 1 \ 1 \ 3 \ 1 \ 1 \ 1 \ 5 \ 3 \ 1 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} \text{A1g} = ( \text{CD4h}^T )^{<1>} & \text{A2g} = ( \text{CD4h}^T )^{<2>} & \text{B1g} = ( \text{CD4h}^T )^{<3>} & \text{B2g} = ( \text{CD4h}^T )^{<4>} & \text{Eg} = (\text{CD4h}^T)^{<5>} \ \text{A1u} = ( \text{CD4h}^T )^{<6>} & \text{A2u} = ( \text{CD4h}^T )^{<7>} & \text{B1u} = ( \text{CD4h}^T )^{<8>} \ \text{B2u} = ( \text{CD4h}^T )^{<9>} & \text{Eu} = ( \text{CD4h}^T)^{<10>} \end{matrix} \nonumber$
$\begin{matrix} h = \sum \text{D4h} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} (A2u + Eu))} & \Gamma_{tot}^T = \begin{pmatrix} 15 & 1 & -1 & -3 & -1 & -3 & -1 & 5 & 3 & 1 \end{pmatrix} \end{matrix} \nonumber$
Symmetry of the vibrational modes.
$\begin{matrix} \Gamma_{vib} = \Gamma_{tot} - A2g - Eg - A2u - Eu & i = 1 .. 10 & Y_i = \frac{ \sum \overrightarrow{[D4h (CD4h^T )^{<i>} \Gamma_{vib}]}}{h} \ Y^T = \begin{pmatrix} 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 2 \end{pmatrix} & ~ & \Gamma_{vib} = A1g + B1g + B2g + A2u + B2u + 2Eu \end{matrix} \nonumber$
The energy level diagram above shows three possible electronic transitions, one IR transition and two transitions in visible region of the spectrum. The calculations below show that all three are formally forbidden. Vibronic coupling is invoked to explain the appearance of the two electronic transitions in the visible region.
$\begin{matrix} \frac{ \sum \overrightarrow{(D4h~Eg(A2u + Eu)B2g))}}{h} = 0 & \frac{ \sum \overrightarrow{(D4h~B1g(A2u + Eu)B2g))}}{h} = 0 & \frac{ \sum \overrightarrow{(D4h~A1g(A2u + Eu)B2g))}}{h} = 0 \end{matrix} \nonumber$
A A1u or Eu vibrational mode can provide vibronic coupling for the B2g --> B1g transition.
$\begin{matrix} \frac{ \sum \overrightarrow{(D4h~A1u~B1g(A2u + Eu)B2g~A1g))}}{h} = 1 & \frac{ \sum \overrightarrow{(D4h~Eu~B1g(A2u + Eu)B2g~A1g))}}{h} = 1 \end{matrix} \nonumber$
A B1u or Eu vibrational mode can provide vibronic coupling for the B2g --> A1g transition.
$\begin{matrix} \frac{ \sum \overrightarrow{(D4h~B1u~A1g(A2u + Eu)B2g~A1g))}}{h} = 1 & \frac{ \sum \overrightarrow{(D4h~Eu~A1g(A2u + Eu)B2g~A1g))}}{h} = 1 \end{matrix} \nonumber$
A close examination of the experimental spectrum indicates the presence of two electronic transitions of similar energy (shoulder). So the energy level diagram and the vibronic analysis are consistent with the actual spectrum.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.07%3A_Ti%28H2O%296_3.txt
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Tetrahedral Symmetry for Methane
The infrared spectrum of methane shows two absorptions: a bend at 1306 cm-1 and a stretch at 3019 cm-1. Demonstrate that a symmetry analysis assuming tetrahedral symmetry for methane is consistent with this spectroscopic data. Also predict how many Raman active modes methane should have.
$\begin{matrix} ~ & \begin{array} E E & C_3 & C_2 & S_4 & \sigma \end{array} \ C_{Td} = & \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 \ 2 & -1 & 2 & 0 & 0 \ 3 & 0 & -1 & 1 & -1 \ 3 & 0 & -1 & -1 & 1 \end{bmatrix} & \begin{array} A A_1:~x^2 + y^2 + z^2 \ A_2 \ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} & Td = \begin{bmatrix} 1 \ 8 \ 3 \ 6 \ 6 \end{bmatrix} & \Gamma_{uma} = \begin{bmatrix} 5 \ 2 \ 1 \ 1 \ 3 \end{bmatrix} & \Gamma_{bonds} = \begin{bmatrix} 4 \ 1 \ 0 \ 0 \ 2 \end{bmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{Td}^T)^{<1>} & A_2 = (C_{Td}^T )^{<2>} & E = (C_{Td}^T )^{<3>} & T_1 = ( C_{Td}^T )^{<4>} \ T_2 = (C_{Td}^T )^{<5>} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} T_2)} & h = \sum Td & \Gamma_{tot}^T = \begin{pmatrix} 15 & 0 & -1 & -1 & 3 \end{pmatrix} & i = 1 .. 5 \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{vib} = \Gamma_{tot} - T_1 - T-2 & \text{Vib}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{vib} \right] }}{h} & \text{Vib} = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \ 2 \end{bmatrix} \begin{array} A A_1:~x^2 + y^2 + z^2 \ A_2 \ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \ \Gamma_{stretch} = \Gamma_{bonds} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{stretch} \right] }}{h} & \text{Stretch} = \begin{bmatrix} 1 \ 0 \ 0 \ 0 \ 1 \end{bmatrix} \begin{array} A A_1:~x^2 + y^2 + z^2 \ A_2 \ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{bend} \right] }}{h} & \text{Bend} = \begin{bmatrix} 0 \ 0 \ 1 \ 0 \ 1 \end{bmatrix} \begin{array} A A_1:~x^2 + y^2 + z^2 \ A_2 \ E:~ 2z^2 - x^2 - y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \end{matrix} \nonumber$
Thus the vibrational modes have A1, E, and T2 symmetry. Only the two T2 modes are infrared active which is consistent with the experimental data quoted above. One of the T2 modes is a stretch (3019 cm-1) and the other is a bend (1306 cm-1).
This symmetry analysis predicts that all of vibrational modes are Raman active - one singly degenerate mode (A1), one doubly degenerate mode (E), and two triply degenerate modes (T2). Indeed four Raman active modes are found at 3019, 2917, 1534, and 1306 cm-1. Note, as expected from the symmetry analysis, there are two coincidences between the IR and Raman spectra.
6.09: P
Td - Tetrahedral Symmetry for P4
The following Raman and IR frequencies have been observed for the tetrahedral P4 molecule. Is the assignment of tetrahedral geometry to this molecule in agreement with the spectroscopic data? Explain.
$\begin{pmatrix} R & R & R,IR \ \frac{614}{ \text{cm}} & \frac{372}{ \text{cm}} & \frac{466}{ \text{cm}} \end{pmatrix} \nonumber$
$\begin{matrix} ~ & \begin{array} E E & C_3 & C_2 & S_4 & \sigma \end{array} \ C_{Td} = & \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 \ 2 & -1 & 2 & 0 & 0 \ 3 & 0 & -1 & 1 & -1 \ 3 & 0 & -1 & -1 & 1 \end{pmatrix} & \begin{array} A A_1:~x^2 + y^2+z^2 \ A_2 \ E:~2z^2-x^2-y^2,~x^-y^2 \ T_1:~(R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} & Td = \begin{pmatrix} 1 \ 8 \ 3 \ 6 \ 6 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 4 \ 1 \ 0 \ 0 \ 2 \end{pmatrix} & \Gamma_{bonds} = \begin{pmatrix} 6 \ 0 \ 2 \ 0 \ 2 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{Td}^T)^{<1>} & A_2 = (C_{Td}^T)^{<2>} & E = (C_{Td}^T)^{<3>} & T_1 = (C_{Td}^T)^{<4>} \ T_2 = (C_{Td}^T)^{<5>} & \Gamma_{tot} = \overrightarrow{( \Gamma_{uma} T_2)} & h = \sum Td & \Gamma_{tot}^T = \begin{pmatrix} 12 & 0 & 0 & 0 & 2 \end{pmatrix} & i = 1 .. 5 \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{vib} = \Gamma_{tot} - T_1 - T_2 & \text{Vib}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{vib} \right]}}{h} & \text{Vib} = \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \end{pmatrix} \begin{array} A A_1:~x^2 + y^2+z^2 \ A_2 \ E:~2z^2-x^2-y^2,~x^-y^2 \ T_1:~(R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \ \Gamma_{stretch} = \Gamma_{bonds} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{stretch} \right]}}{h} & \text{Stretch} = \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 1 \end{pmatrix} \begin{array} A A_1:~x^2 + y^2+z^2 \ A_2 \ E:~2z^2-x^2-y^2,~x^-y^2 \ T_1:~(R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{ \sum \overrightarrow{ \left[ Td (C_{Td}^T )^{<i>} \Gamma_{bend} \right]}}{h} & \text{Bend} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \begin{array} A A_1:~x^2 + y^2+z^2 \ A_2 \ E:~2z^2-x^2-y^2,~x^-y^2 \ T_1:~(R_x,~R_y,~R_z) \ T_2:~ (x,~y,~z),~(xy,~xz,~yz) \end{array} \end{matrix} \nonumber$
The group theoretical analysis assuming tetrahedral geometry is in excellent agreement with the spectroscopic data. Group theory predicts one IR active mode, and that it is coincident with a Raman frequency. This is observed with the T2 vibration. In addition theory predicts that there are two additional Raman active modes A1 and E.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.08%3A_CH.txt
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Symmetry Analysis for Tetrahedrane
Tetrahedrane, C4H4, belongs to the Td point group. Use group theory to predict the number of IR and Raman acitive vibrational modes it has. To date tetrahedrane has not been synthesized.
$\begin{matrix} ~ & \begin{array} E E & C_3 & C_2 & S_4 & \sigma \end{array} \ C_{Td} = & \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 1 & 1 & -1 & -1 \ 2 & -1 & 2 & 0 & 0 \ 3 & 0 & -1 & 1 & -1 \ 3 & 0 & -1 & -1 & 1 \end{pmatrix} & \begin{array} A_1:~x^2+y^2+z^2 \ A_2 \ E:~ 2z^2-x^2-y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~(x,~y,~z),~(xy,~xz,~yz) \end{array} & Td = \begin{pmatrix} 1 \ 8 \ 3 \ 6 \ 6 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 8 \ 2 \ 0 \ 0 \ 4 \end{pmatrix} & \Gamma_{bonds} = \begin{pmatrix} 10 \ 1 \ 2 \ 0 \ 4 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{Td}^T )^{<1>} & A_2 = ( C_{Td}^T )^{<2>} & E = ( C_{Td}^T )^{<3>} & T_1 = ( C_{Td}^T )^{<4>} \ T_2 = ( C_{Td}^T )^{<5>} & \Gamma_{tot} = \overrightarrow{ ( \Gamma_{uma} T_2)} & h = \sum Td & \Gamma_{tot}^T = \begin{pmatrix} 20 & 0 & 0 & 0 & 4 \end{pmatrix} \ i = 1 .. 5 \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{vib} = \Gamma_{tot} - T_1 - T_2 & \text{Vib}_i = \frac{ \sum \overrightarrow{[Td \left( C_{Td}^T \right)^{<i>} \Gamma_{vib}]}}{h} & \text{Vib} = \begin{pmatrix} 2 \ 0 \ 2 \ 1 \ 3 \end{pmatrix} & \begin{array} A_1:~x^2+y^2+z^2 \ A_2 \ E:~ 2z^2-x^2-y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~(x,~y,~z),~(xy,~xz,~yz) \end{array} \ \Gamma_{stretch} = \Gamma_{bonds} & \text{Stretch}_i = \frac{ \sum \overrightarrow{[Td \left( C_{Td}^T \right)^{<i>} \Gamma_{stretch}]}}{h} & \text{Stretch} = \begin{pmatrix} 2 \ 0 \ 1 \ 0 \ 2 \end{pmatrix} & \begin{array} A_1:~x^2+y^2+z^2 \ A_2 \ E:~ 2z^2-x^2-y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~(x,~y,~z),~(xy,~xz,~yz) \end{array} \ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{ \sum \overrightarrow{[Td \left( C_{Td}^T \right)^{<i>} \Gamma_{bend}]}}{h} & \text{Bend} = \begin{pmatrix} 0 \ 0 \ 1 \ 1 \ 1 \end{pmatrix} & \begin{array} A_1:~x^2+y^2+z^2 \ A_2 \ E:~ 2z^2-x^2-y^2,~x^2-y^2 \ T_1:~ (R_x,~R_y,~R_z) \ T_2:~(x,~y,~z),~(xy,~xz,~yz) \end{array} \end{matrix} \nonumber$
According to the selection rules, tetrahedrane should have three IR active modes (3T2) and seven Raman active modes (2A1 + 2E + 3T2). Two of the IR modes are stretches, while five of the Raman modes are stretches.
6.11: PH3
C3v Symmetry - PH3
PH3 (pyramidal) has IR and Raman active vibrations at 2421, 2327, 1121, and 991 cm-1. Make assignments in terms of stretches and bends.
$\begin{array} ~ & \begin{array} E E & C_3 & \sigma_v \end{array} \ C_{3v} = & \begin{pmatrix} 1 & 1 & 1 \ 1 & 1 & -1 \ 2 & -1 & 0 \end{pmatrix} \begin{array} A_1: ~z,~x^2 + y^2,~z^2 \ A_2: ~R_z \ E:~(x,~y),~(R_x,~R_y),~(x^2+y^2,~xy)(xz,~yz) \end{array} & C_{3v} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 4 \ 1 \ 2 \end{pmatrix} & \Gamma_{bonds} = \begin{pmatrix} 3 \ 0 \ 1 \end{pmatrix} \end{array} \nonumber$
$\begin{matrix} A_1 = (C_{C3v}^T )^{<1>} & A_2 = (C_{C3v}^T )^{<2>} & E = (C_{C3v}^T )^{<3>} & h = \sum C3v \ \Gamma_{tot} = \overrightarrow{ \left[ \Gamma_{uma} (A_1 + E) \right] } & \Gamma_{tot}^T = \begin{pmatrix} 12 & 0 & 2 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - A_1 - A_2 - 2E \end{matrix} \nonumber$
$\begin{matrix} i = 1 .. 3 & \text{Vib}_i = \frac{\sum \overrightarrow{ \left[ C3v (C_{C3v}^T )^{<i>} \Gamma_{vib} \right]}}{h} & \text{Vib} = \begin{pmatrix} 2 \ 0 \ 2 \end{pmatrix} \begin{array} A_1: ~z,~x^2 + y^2,~z^2 \ A_2: ~R_z \ E:~(x,~y),~(R_x,~R_y),~(x^2+y^2,~xy)(xz,~yz) \end{array} \ \Gamma_{stretch} = \Gamma_{bends} & \text{Stretch}_i = \frac{\sum \overrightarrow{ \left[ C3v (C_{C3v}^T )^{<i>} \Gamma_{stretch} \right]}}{h} & \text{Stretch} = \begin{pmatrix} 2 \ 0 \ 2 \end{pmatrix} \begin{array} A_1: ~z,~x^2 + y^2,~z^2 \ A_2: ~R_z \ E:~(x,~y),~(R_x,~R_y),~(x^2+y^2,~xy)(xz,~yz) \end{array} \ \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} & \text{Bend}_i = \frac{\sum \overrightarrow{ \left[ C3v (C_{C3v}^T )^{<i>} \Gamma_{bend} \right]}}{h} & \text{Bend} = \begin{pmatrix} 1 \ 0 \ 1 \end{pmatrix} \begin{array} A_1: ~z,~x^2 + y^2,~z^2 \ A_2: ~R_z \ E:~(x,~y),~(R_x,~R_y),~(x^2+y^2,~xy)(xz,~yz) \end{array} \end{matrix} \nonumber$
Group theory predicts two singly degenerate (A1) vibrational modes and two doubly degenerate (E) vibrational modes. This is consistent with the appearance of four fundamentals in the experimental IR and Raman spectra. It further predicts that there are two stretches and two bends. This is also consistent with the experimental data.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.10%3A_Tetrahedrane.txt
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D3h Symmetry - C3H6
The following IR and Raman spectroscopic data is available for cyclopropane, C3H6. Demonstrate that this data is consistent with a D3h symmetry assignment for cyclopropane.
$\begin{pmatrix} \text{Frequency} & 3038 & 1479 & 1188 & 3025 & 1438 & 1029 & 866 & 3103 & 854 & 3082 & 1188 & 734 \ \text{Activity} & R & R & R & R, IR & R,IR & R, IR & R,IR & IR & IR & R & R & R \ \text{Type} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
$\begin{matrix} ~ & \begin{array} E & C_3 & C_2 & \sigma_h & S_3 & \sigma_v \end{array} \ E_{D3h} = & \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \ 1 & 1 & -1 & 1 & 1 & -1 \ 2 & -1 & - & 2 & -1 & 0 \ 1 & 1 & 1 & -1 & -1 & -1 \ 1 & 1 & -1 & -1 & -1 & 1 \ 2 & -1 & 0 & -2 & 1 & 0 \end{pmatrix} & \begin{array} A1':~x^2 + y^2,~z^2 \ A2':~Rz \ E':~(x,~y),~(x^2-y^2,~xy) \ A1":~ \ A2":~z \ E':~(Rx, Ry),~(xz,~yz) \end{array} & D3h = \begin{pmatrix} 1 \ 2 \ 3 \ 1 \ 2 \ 3 \end{pmatrix} & \Gamma_{uma} = \begin{pmatrix} 9 \ 0 \ 1 \ 3 \ 0 \ 3 \end{pmatrix} & \Gamma_{bounds} = \begin{pmatrix} 9 \ 0 \ 1 \ 3 \ 0 \ 3 \end{pmatrix} \end{matrix} \nonumber$
$\begin{matrix} A_1 = (C_{D3h}^T )^{<1>} & A_2 = (C_{D3h}^T )^{<2>} & E = (C_{D3h}^T )^{<3>} & A_{11} = (C_{D3h}^T )^{<4>} \ A_{21} = (C_{D3h}^T )^{<5>} & E_1 = (C_{D3h}^T )^{<6>} & h = \sum D3h & \Gamma_{tot} \overrightarrow{ \left[ \Gamma_{uma} (A_{21} + E) \right]} \end{matrix} \nonumber$
$\begin{matrix} \Gamma_{tot}^T = \begin{pmatrix} 27 & 0 & -1 & 3 & 0 & 3 \end{pmatrix} & \Gamma_{vib} = \Gamma_{tot} - A_2 - E - A_{21} - E_1 & i = 1 .. 6 \ \Gamma_{stretch} = \Gamma_{bonds} & \Gamma_{bend} = \Gamma_{vib} - \Gamma_{stretch} \end{matrix} \nonumber$
$\begin{matrix} \text{Vib}_i = \frac{ \sum \overrightarrow{ \left[ D3h (C_{D3h}^T )^{<i>} \Gamma_{vib} \right]}}{h} & \text{Stretch}_i = \frac{ \sum \overrightarrow{ \left[ D3h (C_{D3h}^T )^{<i>} \Gamma_{bend} \right]}}{h} & \text{Bend}_i = \frac{ \sum \overrightarrow{ \left[ D3h (C_{D3h}^T )^{<i>} \Gamma_{bend} \right]}}{h} \ \text{Vib} = \begin{pmatrix} 3 \ 1 \ 4 \ 1 \ 2 \ 3 \end{pmatrix} \begin{array} A1':~x^2 + y^2,~z^2 \ A2':~Rz \ E':~(x,~y),~(x^2-y^2,~xy) \ A1":~ \ A2":~z \ E':~(Rx, Ry),~(xz,~yz) \end{array} & \text{Stretch} = \begin{pmatrix} 2 \ 0 \ 2 \ 0 \ 1 \ 1 \end{pmatrix} \begin{array} A1':~x^2 + y^2,~z^2 \ A2':~Rz \ E':~(x,~y),~(x^2-y^2,~xy) \ A1":~ \ A2":~z \ E':~(Rx, Ry),~(xz,~yz) \end{array} \ \text{Bend} = \begin{pmatrix} 1 \ 1 \ 2 \ 1 \ 1 \ 2 \end{pmatrix} \begin{array} A1':~x^2 + y^2,~z^2 \ A2':~Rz \ E':~(x,~y),~(x^2-y^2,~xy) \ A1":~ \ A2":~z \ E':~(Rx, Ry),~(xz,~yz) \end{array} \end{matrix} \nonumber$
$\begin{pmatrix} \text{Frequency} & 3038 & 1479 & 1188 & 3025 & 1438 & 1029 & 866 & 3103 & 854 & 3082 & 1188 & 734 \ \text{Activity} & R & R & R & R, IR & R,IR & R, IR & R,IR & IR & IR & R & R & R \ \text{Type} & S & S & B & S & S & B & B & S & B & S & B & B \ \text{Symmetry} & A1' & A1' & A1' & E' & E' & E' & E' & A2" & A2" & E" & E" & E" \end{pmatrix} \nonumber$
There are 9 Raman active modes, 2 IR active modes, 8 IR/Raman active, and 2 modes that are neither Raman or IR active. This gives a total of 21 vibrational modes which is consistent with the total degrees of freedom (27=3x9) minus 6 for translation and rotation.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.12%3A_Cyclopropane.txt
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The following table shows the vibrational frequencies of CH4. Assuming CH4 belongs to the Td point group, fill in the gaps in the table. Use S for stretch and B for bend to designate type of vibration.
$\begin{pmatrix} \text{Frequency} & 2917 cm^{-1} & 1534 cm^{-1} & 3019 cm^{-1} & 1306 cm^{-1} \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \frac{ \text{Activity}}{ \text{IR or Raman}} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
The following table shows the vibrational frequencies of CD4. Assuming CD4 belongs to the Td point group, fill in the gaps in the table. Use S for stretch and B for bend to designate type of vibration.
$\begin{pmatrix} \text{Frequency} & 2109 cm^{-1} & 1092 cm^{-1} & 2259 cm^{-1} & 966 cm^{-1} \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \frac{ \text{Activity}}{ \text{IR or Raman}} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
The following table summarizes the infrared activity CH2D2. Assuming it belongs to the C2v point group, complete the table. Use S for stretch and B for bend to designate the type of vibration.
$\begin{pmatrix} \text{Frequency} & 2974 & 2202 & 1436 & 1033 & 1333 & 3013 & 1090 & 2234 & 1155 \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \frac{ \text{Activity}}{ \text{IR or Raman}} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
The following table summarizes the infrared activity of CH3D. Assuming CH3D belongs to the C3v point group, fill in the gaps in the table. Use S for stretch and B for bend to designate type of vibration.
$\begin{pmatrix} \text{Frequency} & 2945 cm^{-1} & 2200 cm^{-1} & 1300 cm^{-1} & 3017 cm^{-1} & 1471 cm^{-1} & 1155 cm^{-1} \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \frac{ \text{Activity}}{ \text{IR or Raman}} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
Formaldehde has vibrational frequencies at 2843, 2776, 1501, 1251, 1746, and 1167 cm-1. How many are IR active and how many are Raman active. Separate them into bends and stretches.
$\begin{pmatrix} \text{Frequency} & 2843 & 2776 & 1501 & 1251 & 1746 & 1167 \ \text{Stretch or Bend} & A_1 & A_1 & A_1 & B_1 & B_2 & B_2 \ \text{Symmetry} & \text{Stretch} & \text{Stretch} & \text{Bend} & \text{Bend} & \text{Stretch} & \text{Bend} \ \frac{ \text{Activity}}{ \text{IR or Raman}} & IR,R & IR,R & IR,R & IR,R & IR,R & IR,R \end{pmatrix} \nonumber$
PH3 (C3v) has IR and Raman active vibrations at 2421, 2327, 1121, and 991 cm-1. PD3 has IR and Raman active vibrations at 1698, 1694, 806 and 730 cm-1. Make assignments in terms of stretches and bends.
The infrared spectrum of BCl3 shows vibrational bands at 995, 480, and 244 cm-1, while Raman bands appear at 995, 471, and 244 cm-1. Is the geometry of the molecule trigonal pyramid (C3v) or trigonal planar (D3h)? Is your answer to this question consistent with chemical bonding principles? Assign symmetry labels to the vibrational bands and identify the stretches and bends.
$\begin{pmatrix} \text{Frequency} & 995 cm^{-1} & 480 cm^{-1} & 471 cm^{-1} & 224 cm^{-1} \ \frac{ \text{Activity}}{ \text{IR or Raman}} & IR,R & IR & R & IR,R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
CH3CN has 12 vibrational degrees of freedom, but 8 fundamental vibrational frequencies appear in the infrared at 2999, 2942, 2249, 1440, 1376, 1124, 918, and 380 cm-1. Explain.
The infrared spectrum of methane shows two absorptions: a bend at 1306 cm-1 and a stretch at 3019 cm-1. Demonstrate that a symmetry analysis assuming tetrahedral symmetry for methane is consistent with this spectroscopic data. Also predict how many Raman active modes methane should have.
The infrared spectrum of XeF4 has absorptions at 161, 291, and 586 cm-1 (two bends, one stretch), while the Raman spectrum has peaks at 218, 524, and 554 cm-1 (one bend, two stretches). Is its molecular structure tetrahedral or square planar? References: J. Am. Chem. Soc. 1963, 85, 1927; J. Phys. Chem. 1971, 54, 5247.
Cubane, C8H8, has 42 vibrational degrees of freedom, but only three IR active modes. Cubane belongs to the octahedral point group. Show that group theory predicts three IR active modes. Determine how many vibrational modes will be Raman active. Will there be any coincidences between the IR and Raman active modes? The synthesis and characterization of cubane was reported in 1964 by Philip Eaton and Thomas Cole in JACS 1964, 86, 3157-3158. They reported three IR bands at 3000, 1231, and 851 cm-1.
When Paquette's group synthesized dodecahedrane, C20H20, they measured its infrared and Raman spectra (JACS 1983, 105, 5446-5450). They found three IR active bands at 2945, 1298, and 728 cm-1 and eight Raman frequencies at 2924, 2938, 1324, 1164, 1092, 840, 676, and 480 cm-1. Use group theory to show that these data are consistent with the fact that dodecahedrane has icosahedral symmetry.
$\begin{pmatrix} \text{Frequency cm} & 2938 & 2924 & 1324 & 1164 & 1092 & 840 & 676 & 480 & 2945 & 1298 & 728 \ \text{Symmetry} & A_g & A_g & H_g & H_g & H_g & H_g & H_g & H_g & T_{1u} & T_{1u} & T_{1u} \ \frac{ \text{IR, R}}{ \text{Activity}} & R & R & R & R & R & R & R & R & IR & IR & IR \ \text{Stretch or Bend} & \text{stretch} & \text{stretch} & \text{stretch} & \text{stretch} & \text{stretch} & \text{bend} & \text{bend} & \text{bend} & \text{stretch} & \text{stretch} & \text{bend} \end{pmatrix} \nonumber$
Sulfur tetrafluoride represents a difficult case which can't be resolved to complete satisfaction on the basis of IR and Raman data alone. The experimental spectra show eight (five certain and three likely) IR bands and eight (five certain and three likely) Raman bands, and eight (five certain and three likely) coincidences between the two. This information plus nmr spectra lead to the conclusion that the symmetry is C2v or the see-saw structure predicted by VSEPR.
The March 28, 2003 issue of Science reported the synthesis and characterization of Al2H6, the aluminum analog of diborane (therefore, dialane). The researchers reported the following experimental IR frequencies in cm-1: B1u (1932, 836); B2u (1268, 632); B3u (1915, 1408,702). Do a symmetry analysis of Al2H6 which belongs to the D2h point group. How many vibrational modes are there? Which ones are IR active and what are their symmetry designations. The researchers reported that one of the expected low frequency vibrations (~200 cm-1) was not observed. What is the symmetry of this vibrational mode?
Buckminsterfullerene (C60) has four IR active vibrational modes (528, 577, 1180, 1430 cm-1) and ten Raman active modes (273, 436, 496, 710, 773, 110, 1250, 1435, 1470, 1570 cm-1). Demonstrate that the assumption of icosahedral symmetry for C60 is consistent with this data.
Do a symmetry analysis of the 60π orbitals of C60 and show that the results are in agreement with a Huckel calculation.
The following Raman and IR frequencies have been observed for the tetrahedral P4 molecule. Is the assignment of tetrahedral geometry to this molecule in agreement with the spectroscopic data? Explain.
$\begin{pmatrix} R & R & R,IR \ \frac{614}{cm} & \frac{372}{cm} & \frac{466}{cm} \end{pmatrix} \nonumber$
Isotopic subsitution for one of the 31P atoms with a 32P atom reduces the symmetry to C3v. Redo the symmetry analysis and predict the number of IR and Raman active vibrational modes.
Tetrahedrane, C4H4, belongs to the Td point group. Use group theory to predict the number of IR and Raman active vibrational modes it has. Predict also the number of stretches and bends will appear in each type of spectroscopy. To date tetrahedrane has not been synthesized.
Do symmetry analyses on cis-difluoroethene (C2v) and trans-difluoroethene (C2h). Can spectroscopic methods (IR and Raman) be used to distinquish between these isomers. Explain. What about 1,1 difluoroethene? What point group does it belong to? Can spectroscopic methods distinguish it from the cis and trans isomers examined above? Cis-MA2B2 has C2v symmetry and trans-MA2B2 has D2h symmetry. Determine the IR and Raman active modes for each molecule and discuss how such spectroscopic evidence can be used to distinguish the two isomers.
CH3Cl has IR and Raman active modes at 3017, 2937, 1452, 1355, 1017, and 732 cm-1. Is this data consistent with C3v symmetry assignment for chloromethane?
Diborane, D2h, has 18 vibrational degrees of freedom. Nine modes are Raman active and nine are IR active. There are no coincidences. Do a symmetry analysis of diborane to confirm the assignments made in the table below. Identify stretches and bends. The xy plane is the plane of the paper. The four terminal H atoms of diborane lie in the xz plane and the two bridging atoms lie in the xy plane.
$\begin{pmatrix} \text{D}_{2h} & A_g & A_g & A_g & A_g & B_{1g} & B_{1g} & B_{2g} & B_{2g} & B_{3g} \ \frac{ \text{Raman}}{ \text{cm}} & 2524 & 2104 & 1180 & 794 & 1768 & 1035 & 2591 & 920 & 1012 \ \text{Stretch or Bend} & S & S & B & B & S & B & S & B & B \ \text{D}_{2h} & A_u & B_{1u} & B_{1u} & B_{1u} & B_{2u} & B_{2u} & B_{3u} & B_{3u} & B_{3u} \ \frac{ \text{IR}}{ \text{cm}} & ia & 2612& 950 & 368 & 1915 & 973 & 2525 & 1606 & 1177 \ \text{Stretch or Bend} & B & S & B & B & S & B & S & S & B \end{pmatrix} \nonumber$
The following IR and Raman spectroscopic data is available for cyclopropane, C3H6. Demonstrate that this data is consistent with a D3h symmetry assignment for cyclopropane. In addition complete the table below.
$\begin{pmatrix} \frac{ \text{Frequency}}{ \text{cm}^{-1}} & 3038 & 1479 & 1188 & 3025 & 1438 & 1029 & 866 & 3103 & 854 & 3082 & 188 & 734 \ \text{Activity} & IR & IR & IR & R,IR & R,IR & R,IR & R,IR & IR & IR & R & R & R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
$\begin{pmatrix} \frac{ \text{Frequency}}{ \text{cm}^{-1}} & 3038 & 1479 & 1188 & 3025 & 1438 & 1029 & 866 & 3103 & 854 & 3082 & 188 & 734 \ \text{Activity} & IR & IR & IR & R,IR & R,IR & R,IR & R,IR & IR & IR & R & R & R \ \text{Symmetry} & S & S & B & S & S & B & B & \S & B & S & B & B \ \text{Stretch or Bend} & A1' & A1' & A1' & E' & E' & E' & E' & A2" & A2" & E" & E" & E" \end{pmatrix} \nonumber$
PX5 has trigonal bipyramidal geometry and therefore belongs to the D3h point group. Use the model provided to do a symmetry analysis of PX5 by determining Γuma and Γbonds.
The following spectroscopic information is available.
$\begin{pmatrix} \text{Frequency} & 816 cm^{-1} & 648 cm^{-1} & 947 cm^{-1} & 525 cm^{-1} & 1024 cm^{-1} & 533 cm^{-1} & 174 cm^{-1} & 520 cm^{-1} \ \text{Activity} & R & R & IR & IR & R,IR & R, IR & R, IR & R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
Is the symmetry analysis consistent with the spectroscopic data? Explain in detail.
$\begin{pmatrix} \text{Frequency} & 816 cm^{-1} & 648 cm^{-1} & 947 cm^{-1} & 525 cm^{-1} & 1024 cm^{-1} & 533 cm^{-1} & 174 cm^{-1} & 520 cm^{-1} \ \text{Activity} & R & R & IR & IR & R,IR & R, IR & R, IR & R \ \text{Symmetry} & A_1 & A_1 & A_2 & A_2 & E & E & E & E \ \text{Stretch or Bend} & S & S & S & B & S & B & B & B \end{pmatrix} \nonumber$
The following IR and Raman vibrational data is available for tetrahedral methane. Complete the table. Also show that vibrational data is not consistent with a square planar (D4h) or square pyramid (C4v) geometry.
$\begin{pmatrix} \text{Frequency} & 3019 cm^{-1} & 2717 cm^{-1} & 1534 cm^{-1} & 1306 cm^{-1} \ \text{Activity} & IR,R & R & R & IR, R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
The following IR and Raman spectroscopic data is available for ethene, C2H4. Demonstrate that this data is consistent with a D2h symmetry assignment for ethene. In addition complete the table below.
$\begin{pmatrix} \text{Frequency} & 3018 & 3106 & 3019 & 2990 & 1623 & 1444 & 1342 & 1236 & 949 & 943 & 810 \ \text{Activity} & R & IR & R & IR & R & IR & R & R & R & IR & IR \ \text{Stretch or Bend} & S & S & S & S & S & B & B & B & B & B & B \ \text{Symmetry} & Ag & B2u & Ag & B3u & B1g & B1u & Ag & B1g & B2g & B2u & B3u \end{pmatrix} \nonumber$
IR and Raman data forXeOF4, which has C4v symmetry. Establish that the symmetry assignment is correct.
$\begin{pmatrix} \text{Frequency cm} & 926 & 576 & 286 & 232 & 220 & 527 & 609 & 364 & 161 \ \text{Activity} & R,IR & R,IR & R,IR & R & R & R & R,IR & R,IR & R,IR \ \text{Symmetry} & A_1 & A_1 & A_1 & \frac{B_1}{B_2} & \frac{B_1}{B_2} & B_1 & E & E & E \ \text{Stretch or Bend} & S & S & B & B & B & S & S & B & B \end{pmatrix} \nonumber$
IR and Raman data for [PtCl4]2- which has D4h symmetry. Establish that the symmetry assignment is correct.
$\begin{pmatrix} \text{Frequency} & 332 cm^{-1} & 320 cm^{-1} & 314 cm^{-1} & 183 cm^{-1} & 170 cm^{-1} & 93 cm^{-1} \ \text{Activity} & R & IR & R & IR & R & IR \ \text{Symmetry} & A_{1g} & E_u & B_{1g} & E_u & B_{2g} & A_{2u} \ \text{Stretch or Bend} & S & S & S & B & B & B \end{pmatrix} \nonumber$
The vibrational modes of the ions [BrF2]- and [BrF2]+ are given below. Identify which one is A and which one is B.
$\begin{bmatrix} A & B \ 596 (IR) & 715 (IR,R) \ 442(R) & 706 (IR,R) \ 198 (IR) & 366(IR,R) \end{bmatrix} \nonumber$
The following spectroscopic data for [ClO4]- is available. Show that it is consistent with Td symmetry.
$\begin{pmatrix} \text{Frequency} & 1102 cm^{-1} & 935 cm^{-1} & 628 cm^{-1} & 462 cm^{-1} \ \text{Activity} & IR,R & R & IR,R & R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
MoF5 has trigonal bipyramidal geometry and therefore belongs to the D3h point group. Use the model provided to do a symmetry analysis of MoF5 by determining Γuma and Γbonds. The following spectroscopic information is available.
$\begin{pmatrix} \text{Frequency} & 747 cm^{-1} & 732 cm^{-1} & 703 cm^{-1} & 685 cm^{-1} & 500 cm^{-1} & 440 cm^{-1} & 239 cm^{-1} & 203 cm^{-1} \ \text{Activity} & R & IR,R & R & IR & IR & R & IR,R & IR,R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
The Raman and IR spectra of a sample of N2F2 are measured and the results are shown below. Is the sample cis-difluordiazine (C2v) or trans-difluordiazine (C2h)?
$\begin{pmatrix} \text{Frequency} & 1636 cm^{-1} & 1010 cm^{-1} & 989 cm^{-1} & 592 cm^{-1} & 412 cm^{-1} & 360 cm^{-1} \ \text{Activity} & R & R & IR & R & IR & IR \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
Benzene has IR active modes at 675, 1035, 1479, and 3036 cm-1. Demonstrate that this is consistent with D6h symmetry. Separate the vibrations into stretches and bends.
Do a symmetry analysis of the π orbitals of benzene and show that it is consistent with the results of a Huckel calculation on benzene.
Allene, C3H4, belongs to the D2d point group.
$\begin{pmatrix} \frac{ \text{Frequency}}{cm^{-1}} & 3015 & 1443 & 1073 & 865 & 3007 & 1957 & 1398 & 3086 & 999 & 841 & 355 \ \text{Symmetry} & A_1 & A_1 & A_1 & B_1 & B_2 & B_2 & B_2 & E & E & E & E \ \text{Bend or Stretch} & S & S & B & B & S & S & B & S & B & B & B \end{pmatrix} \nonumber$
The following data is available for ethene and deuterated ethene which have D2h symmetry. To each of the vibrational modes determine whether it is a stretch or a bend and its symmetry.
$\begin{pmatrix} C_2H_4 & C_2D_4 & \text{Activity} & \frac{ \text{Stretch}}{ \text{Bend}} & \text{Symmetry} \ 3108 & 2304 & R & \blacksquare & \blacksquare \ 3106 & 2345 & IR & \blacksquare & \blacksquare \ 3018 & 2251 & R & \blacksquare & \blacksquare \ 2990 & 2200 & IR & \blacksquare & \blacksquare \ 1623 & 1515 & R & \blacksquare & \blacksquare \ 1444 & 1078 & IR & \blacksquare & \blacksquare \ 1342 & 981 & R & \blacksquare & \blacksquare \ 1236 & 1009 & R & \blacksquare & \blacksquare \ 1007 & 726 & IA & \blacksquare & \blacksquare \ 949 & 721 & R & \blacksquare & \blacksquare \ 943 & 780 & IR & \blacksquare & \blacksquare \ 810 & 586 & IR & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
BH3 has D3h symmetry. Determine the reducible representation for the hydrogen 1s orbitals (collectively). What linear combination of boron valence orbitals has the same symmetry.
CH4 has Td symmetry. Determine the reducible representation for the hydrogen 1s orbitals (collectively). What linear combination of carbon valence orbitals has the same symmetry.
The following vibrational frequencies are available for cis-dichloroethene. Complete the table assuming that the molecule has C2v symmetry.
$\begin{pmatrix} \text{Frequency cm} & 3077 & 1587 & 1179 & 711 & 173 & 876 & 406 & 697 & 3072 & 1303 & 857 & 571 \ \text{Symmetry} & A_1 & A_1 & A_1 & A_1 & A_1 & A_2 & A_2 & B_1 & B_2 & B_2 & B_2 & B_2 \ \frac{ \text{IR}}{ \text{Activity}} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{no} & \text{no} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} \ \frac{ \text{Raman}}{ \text{Activity}} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} \ \text{Type} & \text{stretch} & \text{stretch} & \text{stretch} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{stretch} & \text{bend} \end{pmatrix} \nonumber$
The following vibrational frequencies are available for trans-dichloroethene. Complete the table assuming that the molecule has C2h symmetry.
$\begin{pmatrix} \text{Frequency cm} & 3073 & 1578 & 1274 & 846 & 350 & 763 & 900 & 227 & 3090 & 1200 & 828 & 250 \ \text{Symmetry} & A_g & A_g & A_g & A_g & A_g & B_g & A_u & A_u & B_u & B_u & B_u & B_u \ \frac{ \text{IR}}{ \text{Activity}} & \text{no} & \text{no} & \text{no} & \text{no} & \text{no} & \text{no} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} \ \frac{ \text{Raman}}{ \text{Activity}} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{no} & \text{no} & \text{no} & \text{no} & \text{no} & \text{no} \ \text{Type} & \text{stretch} & \text{stretch} & \text{stretch} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{stretch} & \text{bend} \end{pmatrix} \nonumber$
The following vibrational frequencies are available for 1,1-dichloroethene. Complete the table assuming that the molecule has C2v symmetry.
$\begin{pmatrix} \text{Frequency cm} & 3035 & 1627 & 1400 & 603 & 299 & 686 & 875 & 460 & 3130 & 1095 & 800 & 372 \ \text{Symmetry} & A_1 & A_1 & A_1 & A_1 & A_1 & A_2 & A_2 & B_1 & B_2 & B_2 & B_2 & B_2 \ \frac{ \text{IR}}{ \text{Activity}} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{no} & \text{no} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} \ \frac{ \text{Raman}}{ \text{Activity}} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} & \text{yes} \ \text{Type} & \text{stretch} & \text{stretch} & \text{stretch} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{bend} & \text{stretch} & \text{bend} \end{pmatrix} \nonumber$
Pyrazine has D2h symmetry. Confirm the entries in the following table.
$\begin{pmatrix} \text{Symmetry} & A_g & A_g & A_g & A_g & A_g & B_{1g} & B_{2g} & B_{2g} & B_{3g} & B_{3g} & B_{3g} & B_{3g} \ \text{Frequency cm} & 3054 & 1578 & 1230 & 1015 & 596 & 757 & 919 & 703 & 3041 & 1524 & 1118 & 641 \ \text{Type} & \text{Stretch} & \text{Stretch} & \text{Stretch} & \text{Bend} & \text{Bend} & \text{Bend} & \text{Bend} & \text{Stretch} & \text{Stretch} & \text{Bend} & \text{Bend} \ \frac{ \text{Ir, R}}{ \text{Activity}} R & R & R & R & R & R & R & R & R & R & R & R \ \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ & \circ \ \text{Symmetry} & A_u & A_u & B_{1u} & B_{1u} & B_{1u} & B_{1u} & B_{2u} & B_{2u} & B_{2u} & B_{2u} & B_{3u} & B_{3u} \ \text{Frequency cm} & na & 363 & 3066 & 1484 & 1135 & 1021 & 3066 & 1418 & 1346 & 1063 & 804 & 416 \ \text{Type} & \text{Bend} & \text{Bend} & \text{Stretch} & \text{Stretch} & \text{Bend} & \text{Bend} & \text{Stretch} & \text{Stretch} & \text{Stretch} & \text{Bend} & \text{Bend} & \text{Bend} \ \frac{ \text{IR,R}}{ \text{Activity}} & IR & IR & IR & IR & IR & IR & IR & IR & IR & IR & IR & IR \end{pmatrix} \nonumber$
Is the following spectroscopic data consistent with the assignment of D4h symmetry to tetrachloroplatinate, PtCl42-?
$\begin{pmatrix} \text{Frequency} & 332 cm^{-1} & 314 cm^{-1} & 170 cm^{-1} & 320 cm^{-1} & 183 cm^{-1} & 93 cm^{-1} \ \text{Activity} & R & R & R & IR & IR & IR \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
$\begin{pmatrix} \text{Frequency} & 332 cm^{-1} & 314 cm^{-1} & 170 cm^{-1} & 320 cm^{-1} & 183 cm^{-1} & 93 cm^{-1} \ \text{Activity} & R & R & R & IR & IR & IR \ \text{Symmetry} & A_{1g} & B_{1g} & B_{2g} & E_u & A_{2u} & B_{2u} \ \text{Stretch or Bend} & S & S & B & S & B & B \end{pmatrix} \nonumber$
Given the following spectroscopic data determine the whether InCl52- has C4v or D3h symmetry.
$\begin{pmatrix} \text{Frequency cm} & 294 & 287 & 283 & 274 & 193 & 165 & 143 & 140 & 108 \ \text{Activity} & IR,R & R & IR,R & IR,R & R & R & IR,R & IR,R & IR,R \ \text{Symmetry} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \ \text{Stretch or Bend} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare \end{pmatrix} \nonumber$
Symmetry analysis supports C4v symmetry as indicated below.
$\begin{pmatrix} \text{Activity} & IR & R & \text{Coincidences} \ \text{Exp} & 6 & 9 & 6 \ C_{4v} & 6 & 9 & 6 \ D_{3h} & 5 & 6 & 3 \end{pmatrix} \nonumber$
Do symmetry analyses on the three C10H8 isomers: azulene (C2v), fulvalene (D2h) and napthalene (D2h). Determine whether IR and Raman spectroscopy can be used to distinguish between the isomers. Completing the table below should facilitate answering the question.
$\begin{pmatrix} \text{Molecule} & \frac{ \text{IR}}{ \text{Stretches}} & \frac{ \text{Raman}}{ \text{Stretches}} & \text{Coincidences} & \frac{ \text{IR}}{ \text{bends}} & \frac{ \text{Raman}}{ \text{Bends}} & \text{Coincidences} & \frac{ \text{InActive}}{ \text{ Modes}} \ \text{Azulene} & 19 & 19 & 19 & 23 & 29 & 23 & 0 \ \text{Fulvalene} & 9 & 10 & 0 & 11 & 14 & 0 & 4 \ \text{Napthalene} & 9 & 10 & 0 & 11 & 14 & 0 & 4 \end{pmatrix} \nonumber$
Propellane has, as shown below, has D3h symmetry. To date it hasn't been synthesized, but theoreticians debate whether or not it has a bridging carbon-carbon bond as shown in the figure. Do a symmetry analysis with and without the bridging bond to determine whether vibrational spectroscopy could decide the issue if the molecule ever became available.
$\begin{pmatrix} \text{Propellane} & \frac{ \text{IR}}{ \text{Stretches}} & \frac{ \text{Raman}}{ \text{Stretches}} & \frac{ \text{IR}}{ \text{Bends}} & \frac{ \text{Raman}}{ \text{Bends}} & \frac{ \text{InActive}}{ \text{Modes}} \ \text{Bond} & 4 & 9 & 4 & 6 & 3 \ \text{No Bond} & 4 & 8 & 4 & 7 & 3 \end{pmatrix} \nonumber$
Raman spectroscopy would be required to answer this question, because it indicates one less stretch and one more bend in the vibrational spectroscopy. This makes sense since the number of vibrational degrees of freedom must be conserved. If there is one less bond, there must be one less stretch and therefore one more bend. In this case, the change is observable (hypothetically) in Raman spectroscopy.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/06%3A_Group_Theory_with_Mathcad/6.13%3A_An_Extensive_Set_of_Group_Theory_Problems_for_Chemists.txt
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Thumbnail: United States Air Force laser experiment. (Public Domain).
07: Quantum Optics
The schematic diagram below shows a Mach-Zehnder interferometer for photons. When the experiment is run so that there is only one photon in the apparatus at any time, the photon is always detected at $D_2$ and never at $D_1$.(1,2,3) The qualitative explanation is that there are two paths to each detector and, therefore, the probability amplitudes for these paths may interfere constructively or destructively. For detector $D_2$ the probability amplitudes for the two paths interfere constructively, while for detector $D_1$ they interfere destructively.
A quantitative quantum mechanical analysis of this striking phenomenon is outlined below. The photon leaves the source, S, traveling in the y-direction. Whether the photon takes the upper or lower path it interacts with a beam splitter, a mirror, and another beam splitter before reaching the detectors.
Orthonormal basis states: (1 x 2 vectors)
Photon moving in the x-direction: $| x > = \begin{vmatrix} 1\ 0 \end{vmatrix}~~~< x | = (1~~~0)~~~< x | x > = 1$
Photon moving in the y-direction: $| y > = \begin{vmatrix} 0 \ 1 \end{vmatrix}~~~< y | = (0~~~1) ~~~< y | y > = 1$
< y | x > = < x | y > = 0
Operators: (2 x 2 matrices)
Operator for photon interaction with the mirror:
$\hat{M} = $\begin{vmatrix} 0 & 1\ 1 & 0 \end{vmatrix} \nonumber$ Operator for photon interaction with the beam splitter: $\hat{BS} = \(\begin{vmatrix} T & iR\ iR & T \end{vmatrix} \nonumber$ \(T$ and $R$ are the transmission and reflection amplitudes. For the half-silvered mirrors used in this example they are:
$T = R = \left( \dfrac{1}{2} \right) ^{ \frac{1}{2}} = 0.707 \nonumber$
Operations:
After interacting with a beam splitter, a photon is in a linear superposition of |x> and |y> in which the components are 90 degrees out of phase.
$\hat{BS}|x \rangle = \dfrac{[ |x + i|y]}{2}^{ \frac{1}{2}} \nonumber$
$\hat{BS}|y \rangle = \dfrac{[ i|x + i|y]}{2}^{ \frac{1}{2}} \nonumber$
BS M BS|y > = i|y>
Interaction with the mirror merely changes the direction of the photon.
$\hat{M} | x \rangle = | y \rangle \nonumber$
$\hat{M} | y \rangle = | x \rangle \nonumber$
Matrix elements:
< x | M | x > = 0 < y | M | x > = 1 < x | M | y > = 1 < y | M | y > = 0
< x | BS | x > = < y | BS | y > = $\frac{1}{2}^{ \frac{1}{2}}$ < y | BS | x > = < x | BS | y > = $\frac{i}{2}^{ \frac{1}{2}}$
Dirac brackets are read from right to left. In Dirac's notation < x | M | y > is the amplitude that a photon initially moving in the y-direction will be moving in the x-direction after interacting with the mirror. |< x | M | y >|2 is the probability that a photon initially moving in the y-direction will be moving in the x-direction after interacting with the mirror. |< y | BS | y >|2 is the probability that a photon initially moving in the y-direction will be found moving in the y-direction after interacting with the beam splitter.
(A) For the photon to be detected at D1 it must be in the state |x> after interacting with two beam splitters and a mirror in the configuration shown above. The probability that a photon will be detected at D1:
< x | BS M BS | y > = 0 thus |< x | BS M BS |y>|2 = 0
(B) For the photon to be detected at D2 it must be in the state |y> after interacting with two beam splitters and a mirror in the configuration shown above. The probability that a photon will be detected at D2:
< y |BS M BS| y > = i thus |< y |BS M BS| y >|2 = 1
It is also instructive to use Dirac's notation to examine upper and lower paths.
(A')
\begin{align} \langle D_1| y \rangle &= \langle D_1 | y \rangle_{upper} + \langle D_1 | y \rangle_{lower} \[4pt] &= \langle x | \textbf{BS} | x \rangle\langle x | \textbf{M} | y \rangle\langle y | \textbf{BS} | y \rangle + \langle x | \textbf{BS} | y \rangle\langle y | \textbf{M} | x \rangle\langle x | \textbf{BS} | y > \[4pt] &= \frac{i}{2}^{ \frac{1}{2}} \times 1 \times \frac{i}{2}^{ \frac{1}{2}} + \frac{i}{2}^{ \frac{1}{2}} \times 1 \times \frac{i}{2}^{ \frac{1}{2}} \[4pt] &= \frac{i}{2} - \frac{i}{2} = i \end{align} \nonumber
This shows that upper and lower paths have the photon arriving 180 degrees out of phase. Thus the photon suffers destructive interference at D1.
(B') < D2 | y > = < D2 | y >upper + < D2 | Y >lower
= < y | BS | x >< x | M | y >< y | BS | y > + < y | BS | y >< y | M | x >< x | BS | y >
= $\frac{i}{2}^{ \frac{1}{2}} \times 1 \times \frac{i}{2}^{ \frac{1}{2}}$ + $\frac{i}{2}^{ \frac{1}{2}} \times 1 \times \frac{i}{2}^{ \frac{1}{2}}$
= $\frac{i}{2} - \frac{i}{2} = i$ i
Thus, |< D2 | y >|2 = 1
This calculation shows that the upper and lower paths have the photon arriving in phase at D2.
If either path (upper or lower) is blocked the interference no longer occurs and the photon reaches D1 25% of the time and D2 25%. Of course, 50% of the time it is absorbed by the blocker.
Lower path blocked:
Probability photon reaches D1: |< x | BS | x >< x | M | y >< y | BS | y >|2 = $\frac{1}{4}$
Probability photon reaches D2: |< y | BS | x >< x | M | y >< y | BS | y >|2 = $\frac{1}{4}$
Upper path blocked:
Probability photon reaches D1: |< x | BS | y >< y | M | x >< x | BS | y >|2 = $\frac{1}{4}$
Probability photon reaches D2: |< y | BS | y >< y | M | x >< x | BS | y >|2 = $\frac{1}{4}$
References:
1. P. Grangier, G. Roger, and A. Aspect, "Experimental Evidence for Photon Anticorrelation Effects on a Beam Splitter: A New Light on Single Photon Interferences," Europhys. Lett. 1, 173-179 (1986).
2. V. Scarani and A. Suarez, "Introducing Quantum Mechanics: One-particle Interferences," Am. J. Phys. 66, 718-721 (1998).
3. Kwiat, P, Weinfurter, H., and Zeilinger, A, "Quantum Seeing in the Dark," Sci. Amer. Nov. 1996, pp 72-78.
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Using Dirac Notation to Analyze Single Particle Interference
The schematic diagram below shows a Mach-Zehnder interferometer for photons. When the experiment is run so that there is only one photon in the apparatus at any time, the photon is always detected at D2 and never at D1. (1,2,3)
The quantum mechanical analysis of this striking phenomenon is outlined below. The photon leaves the source, S, and whether it takes the upper or lower path it interacts with a beam splitter, a mirror, and another beam splitter before reaching the detectors. At the beam splitters there is a 50% chance that the photon will be transmitted and a 50% chance that it will be reflected.
After the first beam splitter the photon is in an even linear superposition of being transmitted and reflected. Reflection involves a 90o (π/2) phase change which is represented by exp(iπ/2) = i, where i = (-1)1/2. (See the appendix for a simple justification of the 90o phase difference between transmission and reflection.) Thus the state after the first beam is given by equation 257.1.
$| \psi = \left( \frac{[ | T> + i|R ]}{2} \right) ^{ \frac{1}{2}} \nonumber$
Now |T> and |R> will be written in terms of |D1> and |D2> the states they evolve to at detection. |T> reaches |D1> by transmission and |D2> by reflection.
$|T> = \left( \frac{[|D1> + i|D2>]}{2} \right) ^{ \frac{1}{2}} \nonumber$
|R> reaches |D1> by reflection and |D2> by transmission.
$|R> = \left( \frac{[i|D1> + |D2>]}{2} \right) ^{ \frac{1}{2}} \nonumber$
Equations 257.2 and 257.3 are substituted into equation 257.1.
$| \psi > = \frac{[|D1> + i|D2> + i2|D1> + i|D2>]}{2} \nonumber$
It is clear (i2 = -1) that the first and third terms cancel (the amplitudes are 180o out of phase), so that we end up with a final state given by equation 257.5.
$| \psi> = i|D_2> \nonumber$ The probability of an event is the square of the absolute magnitude of the probability amplitude.
$P(D_2) = |i|^2 = 1 \nonumber$
Thus this analysis is in agreement with the experimental outcome that no photons are ever detected at D1.
Appendix:
Suppose there is no phase difference between transmission and reflection. Then equations 257.1, 257.2, and 257.3 become
$| \psi = \left( \frac{[|T> + |R>]}{2} \right) ^{ \frac{1}{2}} \nonumber$
$| T> = \left( \frac{[|D1> + |D2>]}{2} \right) ^{ \frac{1}{2}} \nonumber$
$| \psi = \left( \frac{[|D_1> + |D_2> |D_1> + D_2>]}{2} \right) ^{ \frac{1}{2}} \nonumber$
Substitution of equations 257.8 and 257.9 into equation 257.7 yields
$| \psi > = |D_1> + |D_2> \nonumber$
Thus, the detection probabilities at the two detectors are:
$P(D_1) = 1 and P(D_2) = 1 \nonumber$
This result violates the principle of conservation of energy because the original photon has a probability of 1 of being detected at D1 and also a probability of 1 of being detected at D2. In other words, the number of photons has doubled. Thus, there must be a phase difference between transmission and reflection, and a 90o phase difference, as shown above, conserves energy.
References:
1. P. Grangier, G. Roger, and A. Aspect, "Experimental Evidence for Photon Anticorrelation Effects on a Beam Splitter: A New Light on Single Interferences," Europhys. Lett. 1, 173-179 (1986).
2. V. Scarani and A. Suarez, "Introducing Quantum Mechanics: One-particle Interferences," Am. J. Phys. 66, 718-721 (1998).
3. Kwiat, P, Weinfurter, H., and Zeilinger, A, "Quantum Seeing in the Dark, Sci. Amer. Nov. 1996, pp 72-78.
7.03: Single-photon Interference - Third Version
The schematic diagram below shows a Mach-Zehnder interferometer for photons. When the experiment is run so that there is only one photon in the apparatus at any time, the photon is always detected at D2 and never at D1.(1,2,3) The quantum mechanical analysis of this striking phenomenon is outlined below. There are two paths (upper and lower) to each detector, and they both contain a beam splitter, a mirror, and another beam splitter before the detectors are reached. At the beam splitters the the probability amplitude for transmission is $\left( \frac{1}{2} \right)^{ \frac{1}{2}}$, while for reflection it is $\left( \frac{i}{2} \right)^{ \frac{1}{2}}$. The origin of the 90° phase difference between transmission and reflection is found in the principle of energy conservation.
Because there are two paths to each detector the probability amplitudes for these paths may interfere constructively or destructively when added. For detector D2 the probability amplitudes for the two paths interfere constructively, while for detector D1 they interfere destructively.
For example, the probability for the photon being detected at $D_2$ is calculated as follows:
$P(D_2) = |< D_2 | S >|^2 = |< D_2 | T >< T | S > + < D_2 | R >< R | S >|^2 = \bigg| \left( \frac{i}{2} \right)^{ \frac{1}{2}} \left( \frac{1}{2} \right)^{ \frac{1}{2}} + \left( \frac{1}{2} \right)^{ \frac{1}{2}} \left( \frac{i}{2} \right)^{ \frac{1}{2}} \bigg|^2 = 1 \label{258.1}$
The probability that the photon will be detected at $D_1$ is:
$P(D_1) = |< D_1 | S >|^2 = |< D_1 | T >< T | S > + < D_1 | R >< R | S >|^2 = \left| \left( \frac{1}{2} \right)^{ \frac{1}{2}} \left( \frac{1}{2} \right)^{ \frac{1}{2}} + \left( \frac{i}{2} \right)^{ \frac{1}{2}} \left( \frac{i}{2} \right)^{ \frac{1}{2}} \right|^2 = 0 \label{258.2}$
energy conservation
Suppose there is no phase difference between transmission and reflection. Then the probability amplitudes for transmission and reflection are both $\left( \frac{1}{2} \right)^{ \frac{1}{2}}$. Under these circumstances Equations \ref{258.1} and \ref{258.2} become
$P(D_2) = \left| \left( \frac{1}{2} \right)^{ \frac{1}{2}} \left( \frac{1}{2} \right)^{ \frac{1}{2}} + \left( \frac{1}{2} \right)^{ \frac{1}{2}} \left( \frac{1}{2} \right)^{ \frac{1}{2}} \right|^2 = 1 \nonumber$
$P(D_1) = \left| \left( \frac{1}{2} \right)^{ \frac{1}{2}} \left( \frac{1}{2} \right)^{ \frac{1}{2}} + \left( \frac{1}{2} \right)^{ \frac{1}{2}} \left( \frac{1}{2} \right)^{ \frac{1}{2}} \right|^2 = 1 \nonumber$
This result violates the principle of conservation of energy because the original photon has a probability of 1 of being detected at D1 and also a probability of 1 of being detected at $D_2$. In other words, the number of photons has doubled. Thus, there must be a phase difference between transmission and reflection, and a 90o phase difference, as shown above, conserves energy.
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This analysis of the operation of a Mach-Zehnder Interferometer (MZI) will use tensor algebra and the creation and annihilation operators.
An interferometer arm can be occupied or unoccupied. These states are represented by the following vectors.
$Unoccupied: | 0 \rangle = \begin{pmatrix} 1 \ 0 \end{pmatrix} \nonumber$
$Occupied: | 1 \rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix} \nonumber$
After the first beam splitter the photon is in an even superposition of being in both arms of the interferometer. By convention a 90 degree phase shift is assigned to arm b to preserve probability. In terms of the concept of occupation, the superposition takes the following form in tensor algebra.
$|S \rangle \xrightarrow{i} \frac{1}{ \sqrt{2}} [|1 \rangle_a | 0 \rangle_b + i | 0 \rangle_a 1 | \rangle_b ] = \frac{1}{ \sqrt{2}} \bigg[ \begin{pmatrix} 0\ 1 \end{pmatrix}_a \otimes \begin{pmatrix} 1\ 0 \end{pmatrix}_b + i \begin{pmatrix} 1\ 0 \end{pmatrix}_a \otimes \begin{pmatrix} 0\ 1 \end{pmatrix}_b \bigg] = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0\ i\ 1\ 0 \end{pmatrix} \nonumber$
$\Psi = \frac{1}{ \sqrt{2}} \begin{pmatrix} 0\ i\ 1\ 0 \end{pmatrix} \nonumber$
The matrix operators required for this analysis are as follows.
Creation:
$C = \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix} \nonumber$
Annihilation:
$A = \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix} \nonumber$
Number:
$N = \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix} \nonumber$
Identity:
$I = \begin{pmatrix} 0 & 0 \ 1 & 0 \end{pmatrix} \nonumber$
The effect of the creation, annihilation and number operators on |0> and |1>:
$C \begin{pmatrix} 1\ 0 \end{pmatrix} = \begin{pmatrix} 0\ 1 \end{pmatrix} \nonumber$
$A \begin{pmatrix} 0\ 1 \end{pmatrix} = \begin{pmatrix} 1\ 0 \end{pmatrix} \nonumber$
$N \begin{pmatrix} 1\ 0 \end{pmatrix} = \begin{pmatrix} 0\ 0 \end{pmatrix} \nonumber$
$N \begin{pmatrix} 0\ 1 \end{pmatrix} = \begin{pmatrix} 0\ 1 \end{pmatrix} \nonumber$
The creation operator is the Hermitian adjoint of the annihilation operator and the annihilation operator is the Hermitian adjoint of the creation operator.
$\overline{(A^T)} = \begin{pmatrix} 0 & 0\ 1 & 0 \end{pmatrix} \nonumber$
$\overline{(C^T)} = \begin{pmatrix} 0 & 1\ 0 & 0 \end{pmatrix} \nonumber$
The number operator is the product of the creation and annihilation operators.
$CA = \begin{pmatrix} 0 & 0\ 0 & 1 \end{pmatrix} \nonumber$
$\overline{(A^T)} A = \begin{pmatrix} 0 & 0\ 0 & 1 \end{pmatrix} \nonumber$
$\overline{(C^T)} C = \begin{pmatrix} 0 & 0\ 0 & 1 \end{pmatrix} \nonumber$
The eigenvectors of the number operator are |0> and |1> with eigenvalues 0 and 1, respectively:
$eigenvecs(N) = \begin{pmatrix} 1 & 0\ 0 & 1 \end{pmatrix} \nonumber$
$eigenvals(N) = \begin{pmatrix} 0\ 1 \end{pmatrix} \nonumber$
There are two paths to each detector. This provides the opportunity for constructive and destructive interference. To arrive at D1 the a-arm photon state is reflected (90 degree phase shift) at BS2 and the b-arm photon state is transmitted at BS2. Therefore, photon detection requires the annihilation of the superposition of these paths to D1. The annihilation is achieved with the following operator.
$\frac{iA_a + A_b}{ \sqrt{2}} \nonumber$
The product of this operator with its Hermitian conjugate (see above) creates the number operator for photon detection at D1.
$N_{D1} = \frac{iC_a + C_b}{ \sqrt{2}} + \frac{iA_a + A_b}{ \sqrt{2}} \nonumber$
The D1 number operator is formed using Mathcad's kronecker command as follows:
ND1 = $\frac{1}{2}$ (-i kronecker (C, I) + kronecker (I, C)) (i kronecker (A,I) + kronecker (I, A))
To arrive at D2 the a-arm photon state is transmitted at BS2 and the b-arm photon state is reflected (90 degree phase shift) at BS2. Photon detection at D2 requires the annihilation of the superposition of these paths to the detector. The annihilation is represented the following operator.
$\frac{iA_a + A_b}{ \sqrt{2}} \nonumber$
Therefore, the number operator for photon detection at D2 is:
$N_{D2} = \frac{C_a + iC_b}{ \sqrt{2}} \frac{A_a + iA_b}{ \sqrt{2}} \nonumber$
The D2 number operator is formed using Mathcad's kronecker command as follows.
ND2 = $\frac{1}{2}$ (kronecker (C, I) - i kronecker (I, C)) (kronecker (A,I) + i kronecker (I, A))
We now show that the photon always arrives at D1 and never at D2 for an equal arm MZI.
Expectation value for photon detection at D1:
$\overline{ \Psi ^T} N_{D1} \Psi = 1 \nonumber$
Expectation value for photon detection at D2:
$\overline{ \Psi ^T} N_{D2} \Psi = 1 \nonumber$
Equivalent results can be obtained algebraically. Operating on $\Psi$ with the D1 number operator yields $\Psi$. In other words, $\Psi$ is an eigenfunction of ND1 with eigenvalue 1.
$\bigg[ \frac{-C_a + C_b}{ \sqrt{2}} \frac{iA_a + A_b}{\sqrt{2}} \bigg] \frac{1}{ \sqrt{2}} \big[ |1 \rangle_a |0 \rangle_b + i|0 \rangle_a |1 \rangle_b \big] = \frac{1}{ \sqrt{2}} \big[ |1 \rangle_a |0 \rangle_b + i|0_a |1 \rangle_b \big] \nonumber$
Operating on $\Psi$ with the D2 number operator yields 0.
$\bigg[ \frac{C_a - iC_b}{ \sqrt{2}} \frac{A_a + iA_b}{\sqrt{2}} \bigg] \frac{1}{ \sqrt{2}} \big[ |1 \rangle_a |0 \rangle_b + i|0 \rangle_a |1 \rangle_b \big] = 0 \nonumber$
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The schematic diagram below shows a Mach-Zehnder interferometer for photons. When the experiment is run so that there is only one photon in the apparatus at any time, the photon is always detected at D1 and never at D2.
This surprising phenomenon will be analyzed using matrix mechanics. State vectors for photon motion in the x- and y-direction, plus matrix operators for beam splitters and mirrors are defined below. For background and references to the primary literature see: V. Scarani and A. Suarez, "Introducing Quantum Mechanics: One-particle Interferences," Am. J. Phys. 66, 718-721 (1998).
Orthonormal basis states:
Photon moving in x-direction: $x = \begin{bmatrix} 1\ 0 \end{bmatrix} ~~~x^T x = 1 \nonumber$
Photon moving in y-direction: $y = \begin{bmatrix} 0\ 1 \end{bmatrix} ~~~y^T y = 1~~~ x^T y = 0 ~~~ y^T x = 0 \nonumber$
Operators:
Operator for interaction with the mirror:
$M = \begin{bmatrix} 0 & 1\ 1 & 0 \end{bmatrix} \nonumber$
Operator for interaction with a 50/50 beam splitter:
$BS = \frac{1}{ \sqrt{2}} \begin{bmatrix} 1 & i\ i & 1 \end{bmatrix} \nonumber$
Operations:
$M (x) = \begin{bmatrix} 0\ 1 \end{bmatrix} ~~~ M (y) = \begin{bmatrix} 1\ 0 \end{bmatrix} ~~~ BS (x) = \begin{bmatrix} 0.707\ 0.707i \end{bmatrix} ~~~ BS (y) = \begin{bmatrix} 0.707i\ 0.707 \end{bmatrix} ~~~ BS (M) BS (x) = \begin{bmatrix} i\ 0 \end{bmatrix} \nonumber$
Quantum Mechanical Calculation of Experimental Results:
To be detected at D1 the photon must be moving in the x-direction (photon state = |x>). To be detected at D2 the photon must be moving in the y-direction (photon state = |y>). It is shown below that the probability the photon is moving in the x-direction is 1 and the probability it is moving in the y-direction is 0. In its course from source to detector the photon encounters a beam splitter, a mirror, and another beam splitter. Thus,
Probability photon will arrive at detector D1:
$(| x^T BS(M)BS(x)|)^2 = 1 \nonumber$
Probability photon will arrive at detector D2:
$(| x^T BS(M)BS(x)|)^2 = 1 \nonumber$
Further analysis:
The photon leaves the source traveling in the x-direction. Interaction with the beam splitter puts the photon in an even linear superposition of traveling in the x- and y-directions with a 90o phase shift ( $\frac{ \pi}{2}$ or i) assigned by convention to motion in the y-direction (see note below).
$BS (x) = \begin{bmatrix} 0.707\ 0.707i \end{bmatrix} ~~~ \frac{x+iy}{ \sqrt{2}} = \begin{bmatrix} 0.707\ 0.707i \end{bmatrix} \nonumber$
The interaction of this state with the mirrors transfers the 90o phase shift to motion in the x-direction.
$(M) BS (x) = \begin{bmatrix} 0.707i\ 0.707 \end{bmatrix} = \frac{ix+y}{ \sqrt{2}} = \begin{bmatrix} 0.707i\ 0.707 \end{bmatrix} \nonumber$
Finally the interaction of this state with the second beam splitter yields a 90o phase-shifted photon travelling in the x-direction.
$BS(M)BS(x) = \begin{bmatrix} i\ 0 \end{bmatrix} ~~~ ix = \begin{bmatrix} i\ 0 \end{bmatrix} \nonumber$
Thus, the probability amplitude that it will be detected at D1 is i and the probability amplitude that it will be detected at D2 is 0.
$x^T BS(M)BS(x) = (i) ~~~ y^T BS(M)BS(x) = (0) \nonumber$
The probability for an event is the square of the absolute magnitude of the probability amplitude, so the probability that the photon will be detected at D1 is 1.
Note: The justification for the 90 ($\frac{ \pi}{2}$ or i) phase shift between transmission and reflection at the beam splitter is conservation of energy. Assuming there is no phase shift requires that the BS operator be defined as:
$BS = \frac{1}{ \sqrt{2}} \begin{bmatrix} 1 & 1\ 1 & 1 \end{bmatrix} \nonumber$
This leads to the following calculation for the probability of the detection events:
Probability photon will arrive at detector D1: $(| x^T BS(M)BS(x)|)^2 = 1$
Probability photon will arrive at detector D2: $(| y^T BS(M)BS(x)|)^2 = 1$
According to this analysis, the single photon leaving the source has arrived at both detectors. One photon has become two, an obvious violation of the energy conservation principle.
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A polarizing beam splitter (PBS) and PBS interferometer (PBSI) can be used to illustrate the superposition principle. In this analysis the quantum math explaining the operation of a PBSI is presented.
A PBS transmits vertically polarized photons, $|v\rangle$, and reflects horizontally polarized photons, |h>. As the diagrams below show, the photons propagate in the x- or y-directions, $|x\rangle$ and $|y\rangle$. The interaction of $|v\rangle$ and $|h\rangle$ photons emitted by the source in the x-direction, $|x\rangle$, with the PBS is summarized in the two equations below. Note that we need to keep track of two degrees of freedom, direction of propagation and polarization state. This information is stored in the product of two Dirac kets, | >| >.
$S \rightarrow |x \rangle |v \rangle \xrightarrow{PBS} |x \rangle |v \rangle \nonumber$
$S \rightarrow |x \rangle |h \rangle \xrightarrow{PBS} |y \rangle |h \rangle \nonumber$
We will proceed with our analysis assuming that the source emits a diagonally polarized photon, |x>|d>. A diagonally polarized photon is an even superposition of being vertically and horizontally polarized. Thus the photon state incident on the PBS is,
$S \rightarrow |x \rangle |d \rangle = |x \rangle \bigg[ \frac{1}{ \sqrt{2}} |v \rangle + \frac{1}{ \sqrt{2}} |h \rangle \bigg] = \frac{1}{ \sqrt{2}} |x \rangle |v \rangle + \frac{1}{ \sqrt{2}} |x \rangle |h \rangle \nonumber$
After the interaction with the PBS and the mirrors the photon state is,
$\frac{1}{ \sqrt{2}} |x \rangle |v \rangle + \frac{1}{ \sqrt{2}} |x \rangle |h \rangle \xrightarrow{PBS} \frac{1}{ \sqrt{2}} |x \rangle |v \rangle + \frac{1}{ \sqrt{2}} |y \rangle |h \rangle \xrightarrow{M} \frac{1}{ \sqrt{2}} |y \rangle |v \rangle + \frac{1}{ \sqrt{2}} |x \rangle |h \rangle \nonumber$
The photon is in an even superposition of moving in the y-direction with vertical polarization and moving in the x-direction with horizontal polarization. For a statistically meaningful number of experiments, this analysis predicts that in 50% of the experiments the photon will appear at Dx with horizontal polarization and in 50% it will appear at Dy with vertical polarization. This is confirmed by experiment.
Now, at the photon path intersection just before the detectors a second PBS is inserted as shown below. This second PBS recombines the photon paths and creates a PBSI.
The interaction of the photon with the second PBS yields the following state.
$\frac{1}{ \sqrt{2}} |y \rangle |v \rangle + \frac{1}{ \sqrt{2}} |x \rangle |h \rangle \xrightarrow{PBS} \frac{1}{ \sqrt{2}} |y \rangle |v \rangle + \frac{1}{ \sqrt{2}} |y \rangle |h \rangle = | y \rangle \bigg[ \frac{1}{ \sqrt{2}} |v \rangle + \frac{1}{ \sqrt{2}} |h \rangle \bigg] = |y \rangle |d \rangle \nonumber$
The photon is always detected at Dy with diagonal polarization.
In summary, the source emits a photon in the x-direction in a v-h superposition state. Inside the interferometer the photon is in a superposition of both direction of propagation and polarization state. It emerges from the interferometer in the original polarization state |d>, but moving in the y-direction.
$|x \rangle | d \rangle = |x \rangle \bigg[ \frac{1}{ \sqrt{2}} |v \rangle + \frac{1}{ \sqrt{2}} |h \rangle \bigg] = \frac{1}{ \sqrt{2}} |x \rangle |v \rangle + \frac{1}{ \sqrt{2}} |x \rangle |h \rangle \nonumber$
PBS
$\frac{1}{ \sqrt{2}} |x \rangle |v \rangle + \frac{1}{ \sqrt{2}} |y \rangle |h \rangle \nonumber$
M
$\frac{1}{ \sqrt{2}} |y \rangle |v \rangle + \frac{1}{ \sqrt{2}} |x \rangle |h \rangle \nonumber$
PBS
$\frac{1}{ \sqrt{2}} |y \rangle |v \rangle + \frac{1}{ \sqrt{2}} |y \rangle |h \rangle = |y \rangle \bigg[ \frac{1}{ \sqrt{2}} |v \rangle + \frac{1}{ \sqrt{2}} |h \rangle \bigg] = |y \rangle |d \rangle \nonumber$
For an arbitrary polarization state |θ> we have,
$|x \rangle | \theta \rangle = |x \rangle \big[ \cos \theta |v \rangle + \sin \theta |h \rangle \big] = |x \rangle \cos \theta | v \rangle + |x \rangle \sin \theta |h \rangle \nonumber$
PBS
$|x \rangle \cos \theta | v \rangle + |y \rangle \sin \theta |h \rangle \nonumber$
M
$|x \rangle \cos \theta | v \rangle + |x \rangle \sin \theta |h \rangle \nonumber$
PBS
$|y \rangle \cos \theta |v \rangle + |y \rangle \sin \theta |h \rangle = |y \rangle \big[ \cos \theta |v \rangle + \sin \theta |h \rangle \big] = |y \rangle \theta \rangle \nonumber$
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The purpose of this tutorial is to analyze a Mach-Zehnder (MZ) interferometer with polarizing beam splitters (PBS) using tensor algebra. First we will review the traditional MZ with non-polarizing beam splitters using matrix algebra.
The source emits photons in the x-direction illuminating a 50-50 beam splitter which splits the beam into a superposition of motion in the x- and y-directions. The reflected beam collects a $\frac{ \pi}{2}$ (i) phase shift relative to the transmitted beam. Mirrors redirect the two beams to a second 50-50 BS. For the ideal case of an equal arm interferometer the photons are always registered at Dx and never at Dy. One way to explain this is shown on the figure above. Each photon has two paths to each of the detectors. At Dx the photon's paths add in phase each being shifted by $\frac{ \pi}{2}$, resulting in constructive interference. At Dy the photon's paths are 180 degrees out of phase causing destructive interference.
The matrix analysis requires vectors to represent photon states, direction of propagation and polarization, and matrices to represent the devices that the photon interacts with, such as beam splitters and mirrors. The vectors representing direction of propagation are,
$x = \begin{pmatrix} 1\ 0 \end{pmatrix} ~~~ y = \begin{pmatrix} 0\ 1 \end{pmatrix} \nonumber$
The beam splitters and mirrors are represented by the following matrices which operate on the motional states.
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i\ i & 1 \end{pmatrix} ~~~ M = \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \nonumber$
The probability that the photon will be moving in the x-direction after the second BS and therefore detected at Dx is 1. The probability it will be registered at Dy is 0.
$\big( | x^T BS(M)BS(x) | \big) ^2 = 1 ~~~ \big( | y^T BS(M)BS(x) | \big) ^2 = 0 \nonumber$
In a polarization MZ interferometer the traditional beam splitters are replaced with polarizing beam splitters, which in this case transmit horizontal polarization and reflect vertical polarization.
Now in addition the direction of propagation, the photons have a polarization state which is initially restricted to horizontal or vertical polarization.
$h = \begin{pmatrix} 1\ 0 \end{pmatrix} ~~~ v = \begin{pmatrix} 0\ 1 \end{pmatrix} \nonumber$
There are now four photon states: |xh>, |xv>, |yh> and |yv>, which can also be written |x>|h>, |x>|v>, |y>|h> and |y>|v>. These are vector tensor products and they take us from the two-dimensional space of the previous example to a four-dimensional Hilbert space.
$|xh \rangle = |x \rangle |h \rangle = \begin{pmatrix} 1\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1\ 0 \end{pmatrix} = \begin{pmatrix} 1\ 0\ 0\ 0 \end{pmatrix} ~~~ |xv \rangle = |x \rangle |v \rangle = \begin{pmatrix} 1\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0\ 1 \end{pmatrix} = \begin{pmatrix} 0\ 1\ 0\ 0 \end{pmatrix} \nonumber$
$|yh \rangle = |y \rangle |h \rangle = \begin{pmatrix} 0\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1\ 0 \end{pmatrix} = \begin{pmatrix} 0\ 0\ 1\ 0 \end{pmatrix} ~~~ |yv \rangle = |y \rangle |v \rangle = \begin{pmatrix} 0\ 1 \end{pmatrix} \otimes \begin{pmatrix} 0\ 1 \end{pmatrix} = \begin{pmatrix} 0\ 0\ 0\ 1 \end{pmatrix} \nonumber$
$xh = \begin{pmatrix} 1\ 0\ 0\ 0 \end{pmatrix} ~~~ xv = \begin{pmatrix} 0\ 1\ 0\ 0 \end{pmatrix} ~~~ yh = \begin{pmatrix} 0\ 0\ 1\ 0 \end{pmatrix} ~~~ yv = \begin{pmatrix} 0\ 0\ 0\ 1 \end{pmatrix} \nonumber$
As mentioned previously the PBS transmits horizontal polarization and reflects vertical polarization. Unlike the nonpolarizing BS, there is no phase change on reflection. The matrix representing a PBS can be constructed by considering the fate of the individual photon states encountering a PBS.
$\widehat{PBS} = |xh \rangle \langle xh | + |yv \rangle \langle xv | + | yh \rangle \langle yh | + | xv \rangle \langle yv | \nonumber$
$PBS: \begin{pmatrix} 1\ 0\ 0\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0\ 0\ 0\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0\ 0\ 1\ 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0 \end{pmatrix} + \begin{pmatrix} 0\ 1\ 0\ 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 1 \end{pmatrix} \nonumber$
$PBS = \begin{pmatrix} 1 & 0 & 0 & 0\ 0 & 0 & 0 & 1\ 0 & 0 & 1 & 0\ 0 & 1 & 0& 0 \end{pmatrix} \nonumber$
This gives us the PBS operator in the four-dimensional Hilbert space. Next we construct the matrix for the mirror in this space. The matrices representing the various photon interactions must be 4x4 because the photon vector states are 1x4. The mirror only affects the direction of propagation and not the polarization state of the photon. Since the motional state appears first in the kets, we form the 4-D mirror matrix by the following tensor (Kronecker) product of M and the identity matrix.
$\widehat{M}' = \widehat{M} \otimes \widehat{I} = \begin{pmatrix} 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\ 1 & 0 & 0 & 0\ 0 & 1 & 0& 0 \end{pmatrix} \nonumber$
$M' = kronecker \bigg[ M, \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \bigg] ~~~ M' = \begin{pmatrix} 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\ 1 & 0 & 0 & 0\ 0 & 1 & 0& 0 \end{pmatrix} \nonumber$
As constructed this matrix changes the photon's direction of propagation with out affecting its state of polarization.
Now suppose that the source emits a horizontally polarized photon, |xh>. Calculation shows that it arrives at Dy with horizontal polarization,|yh>, 100% of the time.
$\big( | yh^T PBS(M')PBS(xh) | \big) ^2 = 1 \nonumber$
If the source emits a vertically polarized photon, |xv>. Calculation shows that it arrives at Dy with vertical polarization,|yv>, 100% of the time.
$\big( | yv^T PBS(M')PBS(xv) | \big) ^2 = 1 \nonumber$
A photon emitted by the source polarized at an angle of $\theta$ relative to the horizontal has the following state vector.
$|x \theta \rangle = \begin{pmatrix} 1\ 0 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \theta\ \sin \theta\ 0\ 0 \end{pmatrix} \nonumber$
It is easy to show that in the MZ polarization interferometer, the polarization state of the source photon is preserved while the direction of propagation changes.
$|y \theta \rangle = \begin{pmatrix} 0\ 1 \end{pmatrix} \otimes \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\big| \rangle y \theta | \widehat{PBS} \widehat{M'} \widehat{PBS} | x \theta \rangle \big|^2 = 1 \nonumber$
$\Bigg[ \Bigg| \begin{pmatrix} 0 & 0 & \cos \theta & \sin \theta \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\ 0 & 0 & 0 & 1\ 0 & 0 & 1 & 0\ 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\ 1 & 0 & 0 & 0\ 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\ 0 & 0 & 0 & 1\ 0 & 0 & 1 & 0\ 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta\ 0\ 0 \end{pmatrix} \Bigg| \Bigg]^2 simplify \rightarrow 1 \nonumber$
This is an excellent example of the superposition principle. The first PBS creates a x-h, y-v superposition state.
$\begin{pmatrix} 1 & 0 & 0 & 0\ 0 & 0 & 0 & 1\ 0 & 0 & 1 & 0\ 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ \sin \theta\ 0\ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \cos \theta \ 0 \ 0\ \sin \theta \end{pmatrix} \nonumber$
$\begin{pmatrix} 1\ 0 \end{pmatrix} \begin{pmatrix} \cos \theta\ 0 \end{pmatrix} + \begin{pmatrix} 0\ 1 \end{pmatrix} \begin{pmatrix} 0\ \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \theta\ 0\ 0\ \sin \theta \end{pmatrix} \nonumber$
The mirror changes this to a y-h, x-v superposition.
$\begin{pmatrix} 0 & 0 & 1 & 0\ 0 & 0 & 0 & 1\ 1 & 0 & 0 & 0\ 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} \cos \theta \ 0 \ 0\ \sin \theta \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ \sin \theta \ \cos \theta \ 0 \end{pmatrix} \nonumber$
$\begin{pmatrix} 0\ 1 \end{pmatrix} \begin{pmatrix} \cos \theta\ 0 \end{pmatrix} + \begin{pmatrix} 1\ 0 \end{pmatrix} \begin{pmatrix} 0\ \sin \theta \end{pmatrix} = \begin{pmatrix} 0\ \sin \theta\ \cos \theta\ 0 \end{pmatrix} \nonumber$
The second PBS restores the original polarization state, but changes the direction of propagation.
$\begin{pmatrix} 1 & 0 & 0 & 0\ 0 & 0 & 0 & 1\ 0 & 0 & 1 & 0\ 0 & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 \ \sin \theta \ \cos \theta \ 0 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \ 0 \ \cos \theta \ \sin \theta \end{pmatrix} \nonumber$
$\begin{pmatrix} 0\ 1 \end{pmatrix} \begin{pmatrix} \cos \theta\ 0 \end{pmatrix} + \begin{pmatrix} 0\ 1 \end{pmatrix} \begin{pmatrix} 0\ \sin \theta \end{pmatrix} = \begin{pmatrix} 0\ 1 \end{pmatrix} \begin{pmatrix} \cos \theta\ \sin \theta \end{pmatrix} = \begin{pmatrix} 0\ 0\ \cos \theta\ \sin \theta \end{pmatrix} \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.07%3A_Mach-Zehner_Polarization_Interferometer_Analyzed_Using_Tensor_Algebra.txt
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Abstract
Single-photon interference in a Mach-Zehnder interferometer is used to illustrate the superposition principle. Three methods of analysis of an historically important experiment are presented at a level appropriate for an undergraduate course in quantum chemistry or physics. The importance of the superposition principle in chemistry is also discussed.
Introduction
Introducing the double-slit experiment as paradigmatic of the bizarre nature of quantum mechanical behavior, Richard Feynman wrote,
We choose to examine a phenomenon which is impossible, absolutely impossible, to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery. We cannot make the mystery go away by “explaining” how it works. We will just tell you how it works. In telling you how it works we will have told you about the basic peculiarities of all quantum mechanics.(1)
Amplifying the last sentence of this quotation Feynman said the double-slit experiment is so fundamental that if asked a question about quantum mechanics one can always reply, “You remember the case of the experiment with the two holes? It’s the same thing.”(2) The crucial point being made is that the double-slit experiment is the simplest manifestation of the ubiquitous superposition principle and its attendant interference effects. The superposition principle, according to Feynman, permeates quantum mechanics, and is the origin of the strangeness we associate with quantum mechanical phenomena.
Having identified the mystery and conceptual difficulty of quantum mechanics Feynman went on to point out that computationally it was frequently quite simple.
We have come to the conclusion that what are usually called the advanced parts of quantum mechanics are, in fact, quite simple. The mathematics that is involved is particularly simple, involving simple algebraic operations and no differential equations or at most very simple ones. (3)
Single-Photon Interference
Recently Scarani and Suarez (4) illustrated this mathematical simplicity by providing an elementary quantum mechanical analysis of an historically important experiment (5) that is a close relative of the famous double-slit experiment. Fig. 1 shows a schematic diagram of an equal-arm Mach-Zehnder interferometer consisting of two 50-50 beam splitters and two mirrors. The experiment is run at low source intensity such that there is only one photon in the interferometer at a time.
Three idealized experiments are considered for pedagogical purposes. (I) If either path is blocked 50% of the photons get through, and 25% reach D1 and 25% reach D2. (II) In the absence of the second beam splitter (BS2) 50% of the time the photon is detected at D1 and 50% of the time it is detected at D2. (III) However, in the presence of BS2 the photon is always detected at D1 and never at D2. Three methods will be used to analyze the surprising result of experiment III. Of course the key will be that in this experiment each detector can be reached by two paths, whereas in experiments I and II there is only one path to each detector.
First Method
We will begin with a brief review of the analysis given by Scarani and Suarez for experiment III before presenting two alternative approaches. After the first beam splitter the photon is in an even linear superposition of being transmitted $|T \rangle$ and reflected $|R \rangle$.
$|S \rangle \rightarrow \frac{1}{ \sqrt{2}} \big[ |T \rangle + i |R \rangle \big] \nonumber$
By convention a $\frac{ \pi}{2}$ (i) phase shift is assigned to the reflected beam. More will be said about this below.
The mirrors (6) direct the two beams to the second beam splitter where, according to the superposition principle, $|T \rangle$ and $|R \rangle$ evolve as follows,
$|T \rangle \rightarrow \frac{1}{ \sqrt{2}} \big[ i|D_1 \rangle + |D_2 \rangle \big] \nonumber$
$|R \rangle \rightarrow \frac{1}{ \sqrt{2}} \big[ |D_1 \rangle + i|D_2 \rangle \big] \nonumber$
where again the $\frac{ \pi}{2}$ phase shift is assigned to the reflection. Substitution of eqns (2) and (3) into (1) yields
$|S \rangle \rightarrow i |D_1 \rangle \nonumber$
Thus, the probability that the photon will arrive at D1 is
$\big| \langle D_1 |S \rangle \big|^2 = 1 \nonumber$
It is easy to show that if there is no phase difference between reflection and transmission 1 2 (i.e. replace i by 1 in equations (1), (2), and (3)) the probabilities for arrival at D1 and D2 are both 1. This is a clear violation of energy conservation because one photon has become two photons with no mechanism for a reduction of energy of the individual photons. Thus, conservation of energy is a compelling argument for a $\frac{ \pi}{2}$ phase difference between transmission and reflection.
Second Method
The second method uses Dirac notation to enumerate the probability amplitudes for arrival at the two detectors. As in the above analysis, at the beam splitters the probability amplitude for transmission is $\frac{1}{ \sqrt{2}}$, and for reflection it is $\frac{i}{ \sqrt{2}}$. Because the photon path is not observed the probability is calculated as the square of the absolute magnitude of the sum of the probability amplitudes for each path. Thus, the probability amplitudes for the two paths may interfere constructively and destructively. As shown below the probability amplitudes to reach D1 are in phase (TR+RT) and those to reach D2 are 180° out of phase (TT+RR), so the photon is never detected at D2.
$\big| \langle D_1 |S \rangle \big|^2 = \big| \langle D_1 |T \rangle \langle T |S \rangle + \langle D_1 |R \rangle \langle R|S \rangle \big|^2 = \bigg| \frac{i}{ \sqrt{2}} \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2}} \frac{i}{ \sqrt{2}} \bigg|^2 = 1 \nonumber$
$\big| \langle D_2 |S \rangle \big|^2 = \big| \langle D_2 |T \rangle \langle T |S \rangle + \langle D_2 |R \rangle \langle R|S \rangle \big|^2 = \bigg| \frac{1}{ \sqrt{2}} \frac{1}{ \sqrt{2}} + \frac{i}{ \sqrt{2}} \frac{i}{ \sqrt{2}} \bigg|^2 = 0 \nonumber$
While it is customary to refer to this experiment as an example of single-particle interference, Glauber (7) recommends more careful language.
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.
What Glauber is saying here is clearly revealed in this second method of analysis.
Third Method
The third method illustrates how the experiment is analyzed using matrix mechanics.(8) In this approach the photon’s translational states $|x \rangle$ (horizontal) and $|y \rangle$ (vertical) are represented by the orthonormal basis vectors shown below.
$|x \rangle = \begin{pmatrix} 1\ 0 \end{pmatrix} ~~~ |y \rangle = \begin{pmatrix} 0\ 1 \end{pmatrix} ~~~ \langle x | = \begin{pmatrix} 1 & 0 \end{pmatrix} ~~~ \langle y| = \begin{pmatrix} 0 & 1 \end{pmatrix} \nonumber$
The operators for the interaction of the photon with a 50-50 beam splitter and a mirror (6) are represented by the following matrices.
$BS = \frac{1}{ \sqrt{2}} \begin{pmatrix} 1 & i\ i & 1 \end{pmatrix} ~~~ M = \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \nonumber$
To be detected at D1 a photon must be in the translational state $|x \rangle$ after interacting with two beam splitters and a mirror in the configuration shown in Fig. 1. To be detected at D2 it must be in the translational state $|y \rangle$. The probabilities for these outcomes are calculated using equations (8) and (9) as follows:
$\big| \langle x |BS (M) BS | x \rangle \big|^2 = 1 ~~~ \big| \langle y |BS (M) BS | x \rangle \big|^2 = 0 \nonumber$
It is easy to confirm the rest by hand calculation.
Recapitulation
Dirac emphasized the unique role of the superposition principle in his seminal treatise on quantum mechanics by saying, “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.”(9)
The best way to get accustomed to the ubiquitous, but non-classical nature of the quantum mechanical superposition is through the study of those phenomena where it reveals itself most directly. Single-photon interference in a Mach-Zehnder interferometer provides a very simple example of the superposition principle in action, and is therefore suitable for presentation to undergraduates studying quantum chemistry or quantum physics. The same is true, of course, for the more widely known double-slit experiment.
The Superposition Principle in Chemistry
Having introduced the superposition with three analyses of an experiment in quantum optics, we will now briefly review its importance in chemistry. The superposition principle informs the chemists view of atomic and molecular structure, and is especially important in unifying the many facets of chemical bonding (10).
For example, an electronic wave function, whether atomic or molecular, is a weighted linear superposition. In atoms and molecules...
Electrons are characterized by their entire distributions (called wave functions or orbitals) rather than by instantaneous positions and velocities: an electron may be considered always to be (with appropriate probability) at all points of its distribution (which does not vary with time). (11)
Electrons confined in atoms and molecules are not moving in a classical sense; if they were they would radiate energy continuously, and atomic and molecular stability would be raised to the status of a scientific miracle. Atomic and molecular electrons are not here and later there, rather they are, at the same time, here and there. Just as the photon is present simultaneously in both arms of the Mach-Zehnder interferometer, the electron is present at all possible locations (properly weighted) in atoms and molecules. The confined electron is in an atomic or molecular stationary state.
Similar arguments pertain to molecular vibrations (12) and rotations. In the absence of external perturbations, such as electromagnetic radiation, molecules do not really vibrate and rotate about their centers of mass for the same reason electrons do not orbit the nucleus; such behavior would lead to the continuous emission of electromagnetic radiation, and again call into question the stability of matter. In the nanoscopic world of atoms and molecules motion only occurs during transitions from one stationary state to another – during the so called quantum jump.
The superposition principle also provides a viable model for this quantum jump, the transition between quantized stationary states, which is the essential process in all forms of spectroscopy. (13) Under the influence of an electromagnetic perturbation, a molecule or atom in an initial stationary state may move into a time-dependent linear superposition of the initial state and some final state.
$\Psi (r, t) = c_i (t) \Psi _i (r, t) exp \left( - \frac{iE_i t}{ \hbar} \right) + c_f (t) \Psi _f (r, t) exp \left( - \frac{iE_f t}{ \hbar} \right) \nonumber$
If the Bohr frequency condition is met, $v = \frac{E_f - E_i}{h}$, and $| \Psi (r, t)|^2$ exhibits oscillating electric dipole character, the transition may occur, otherwise it is forbidden. In summary, the superposition principle provides an interpretation of quantized stationary states and the transitions between them induced by electromagnetic radiation.
Conclusion
Feynman has identified the superposition principle as the fundamental mystery of quantum mechanics. Accepting this view is good pedagogy, because it helps us teachers to see the coherence of this strange and awesome subject, and hopefully to pass that vision on to our students. It is characteristic of great thinkers, like Feynman, that they see the threads that make the tapestry.
Literature cited
1. Feynman, R. P.; Leighton, R. B.; Sands, M., The Feynman Lectures on Physics, Vol. 3; Addison-Wesley: Reading, 1965, p. 1-1.
2. Feynman, R. P. The Character of Physical Law; MIT Press: Cambridge, 1967; p. 130.
3. Feynman, R. P.; Leighton, R. B.; Sands, M., The Feynman Lectures on Physics, Vol. 3; Addison-Wesley: Reading, 1965, p. 3-1.
4. Scarani, V.; Suarez, A. “Introducing quantum mechanics: One-particle interferences,” Am. J. Phys. 1998, 66, 718-721.
5. Grangier, P.; G. Roger, G.; Aspect, A. “Experimental Evidence for Photon Anticorrelation Effects on a Beam Splitter: A New Light on Single-Photon Interferences,” Europhys. Lett 1986, 1, 173-179.
6. The $\frac{ \pi}{2}$ phase shift accompanying reflection at the mirrors can be ignored because there are mirrors in both arms of the interferometer.
7. Glauber, R. J. “Dirac’s famous dictum,” Am. J. Phys. 1995, 63, 12.
8. This suggestion was made to the author by V. Scarani in a private communication.
9. Dirac, P. A. M. Principles of Quantum Mechanics, 4th ed.; Oxford U. P.: London, 1958, p.12.
10. Weinhold, F. A. “Chemical Bonding as a Superposition Phenomenon,” J. Chem. Educ. 1999, 76, 1141-1146.
11. Harris, F. E. “Molecules,” The Encyclopedia of Physics, 2nd ed.; R. G. Lerner, Ed. New York, VCH Publishers, 1990, p. 763.
12. Baskin, J. S.; Zewail, A. H. “Freezing Atoms in Motion: Principles of Femtochemistry and Demonstration by Laser Stroboscopy,” J. Chem. Educ. 2001, 78, 737-751.
Fig. 1 Schematic diagram of a Mach-Zehnder interferometer. S = source; BS = beam splitter; M = mirror 1 ; R = reflected; T = transmitted; D = detector; TT = transmitted at BS and 2 1 2 transmitted at BS ; TR = transmitted at BS and reflected at BS ; etc.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/07%3A_Quantum_Optics/7.08%3A_Illustrating_the_Superposition_Principle_with_Single_Photon_Interference.txt
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