# Useful additional packages 
import matplotlib.pyplot as plt
%matplotlib inline
import numpy as np
from math import pi

from qiskit import QuantumCircuit, ClassicalRegister, QuantumRegister
from qiskit import available_backends, execute, register, get_backend
from qiskit.tools.visualization import circuit_drawer
from qiskit.tools.qi.qi import state_fidelity
from qiskit import Aer

backend = Aer.get_backend('unitary_simulator')

"""
Explanation: <img src="../../images/qiskit-heading.gif" alt="Note: In order for images to show up in this jupyter notebook you need to select File => Trusted Notebook" width="500 px" align="left">
Summary of Quantum Operations
In this section we will go into the different operations that are available in Qiskit Terra. These are:
- Single-qubit quantum gates
- Multi-qubit quantum gates
- Measurements
- Reset
- Conditionals
- State initialization
We will also show you how to use the three different simulators:
- unitary_simulator
- qasm_simulator
- statevector_simulator
End of explanation
"""


q = QuantumRegister(1)

"""
Explanation: Single Qubit Quantum states
A single qubit quantum state can be written as
$$|\psi\rangle = \alpha|0\rangle + \beta |1\rangle$$
where $\alpha$ and $\beta$ are complex numbers. In a measurement the probability of the bit being in $|0\rangle$ is $|\alpha|^2$ and $|1\rangle$ is $|\beta|^2$. As a vector this is
$$
|\psi\rangle =
\begin{pmatrix}
\alpha \
\beta
\end{pmatrix}.
$$
Note due to conservation probability $|\alpha|^2+ |\beta|^2 = 1$ and since global phase is undetectable $|\psi\rangle := e^{i\delta} |\psi\rangle$ we only requires two real numbers to describe a single qubit quantum state.
A convenient representation is
$$|\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)e^{i\phi}|1\rangle$$
where $0\leq \phi < 2\pi$, and $0\leq \theta \leq \pi$.  From this it is clear that there is a one-to-one correspondence between qubit states ($\mathbb{C}^2$) and the points on the surface of a unit sphere ($\mathbb{R}^3$). This is called the Bloch sphere representation of a qubit state.
Quantum gates/operations are usually represented as matrices. A gate which acts on a qubit is represented by a $2\times 2$ unitary matrix $U$. The action of the quantum gate is found by multiplying the matrix representing the gate with the vector which represents the quantum state.
$$|\psi'\rangle = U|\psi\rangle$$
A general unitary must be able to take the $|0\rangle$ to the above state. That is 
$$
U = \begin{pmatrix}
\cos(\theta/2) & a \
e^{i\phi}\sin(\theta/2) & b 
\end{pmatrix}
$$ 
where $a$ and $b$ are complex numbers constrained such that $U^\dagger U = I$ for all $0\leq\theta\leq\pi$ and $0\leq \phi<2\pi$. This gives 3 constraints and as such $a\rightarrow -e^{i\lambda}\sin(\theta/2)$ and $b\rightarrow e^{i\lambda+i\phi}\cos(\theta/2)$ where $0\leq \lambda<2\pi$ giving 
$$
U = \begin{pmatrix}
\cos(\theta/2) & -e^{i\lambda}\sin(\theta/2) \
e^{i\phi}\sin(\theta/2) & e^{i\lambda+i\phi}\cos(\theta/2) 
\end{pmatrix}.
$$
This is the most general form of a single qubit unitary.
Single-Qubit Gates
The single-qubit gates available are:
- u gates
- Identity gate
- Pauli gates
- Cliffords gates
- $C3$ gates
- Standard rotation gates 
We have provided a backend: unitary_simulator to allow you to calculate the unitary matrices.
End of explanation
"""


qc = QuantumCircuit(q)
qc.u3(pi/2,pi/2,pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: u gates
In Qiskit we give you access to the general unitary using the $u3$ gate
$$
u3(\theta, \phi, \lambda) = U(\theta, \phi, \lambda) 
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.u2(pi/2,pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: The $u2(\phi, \lambda) =u3(\pi/2, \phi, \lambda)$ has the matrix form
$$
u2(\phi, \lambda) = 
\frac{1}{\sqrt{2}} \begin{pmatrix}
1 & -e^{i\lambda} \
e^{i\phi} & e^{i(\phi + \lambda)}
\end{pmatrix}.
$$
This is a useful gate as it allows us to create superpositions
End of explanation
"""


qc = QuantumCircuit(q)
qc.u1(pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: The $u1(\lambda)= u3(0, 0, \lambda)$ gate has the matrix form
$$
u1(\lambda) = 
\begin{pmatrix}
1 & 0 \
0 & e^{i \lambda}
\end{pmatrix},
$$
which is a useful as it allows us to apply a quantum phase.
End of explanation
"""


qc = QuantumCircuit(q)
qc.u0(pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: The $u0(\delta)= u3(0, 0, 0)$ gate is the identity matrix. It has the matrix form
$$
u0(\delta) = 
\begin{pmatrix}
1 & 0 \
0 & 1
\end{pmatrix}.
$$
The identity gate does nothing (but can add noise in the real device for a period of time equal to fractions of the single qubit gate time)
End of explanation
"""


qc = QuantumCircuit(q)
qc.iden(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Identity gate
The identity gate is $Id = u0(1)$.
End of explanation
"""


qc = QuantumCircuit(q)
qc.x(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Pauli gates
$X$: bit-flip gate
The bit-flip gate $X$ is defined as:
$$
X   =
\begin{pmatrix}
0 & 1\
1 & 0
\end{pmatrix}= u3(\pi,0,\pi)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.y(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: $Y$: bit- and phase-flip gate
The $Y$ gate is defined as:
$$
Y  = 
\begin{pmatrix}
0 & -i\
i & 0
\end{pmatrix}=u3(\pi,\pi/2,\pi/2)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.z(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: $Z$: phase-flip gate
The phase flip gate $Z$ is defined as:
$$
Z = 
\begin{pmatrix}
1 & 0\
0 & -1
\end{pmatrix}=u1(\pi)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.h(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Clifford gates
Hadamard gate
$$
H = 
\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & 1\
1 & -1
\end{pmatrix}= u2(0,\pi)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.s(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: $S$ (or, $\sqrt{Z}$ phase) gate
$$
S = 
\begin{pmatrix}
1 & 0\
0 & i
\end{pmatrix}= u1(\pi/2)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.sdg(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: $S^{\dagger}$ (or, conjugate of $\sqrt{Z}$ phase) gate
$$
S^{\dagger} = 
\begin{pmatrix}
1 & 0\
0 & -i
\end{pmatrix}= u1(-\pi/2)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.t(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: $C3$ gates
$T$ (or, $\sqrt{S}$ phase) gate
$$
T = 
\begin{pmatrix}
1 & 0\
0 & e^{i \pi/4}
\end{pmatrix}= u1(\pi/4) 
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.tdg(q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: $T^{\dagger}$ (or, conjugate of $\sqrt{S}$ phase) gate
$$
T^{\dagger} =
\begin{pmatrix}
1 & 0\
0 & e^{-i \pi/4}
\end{pmatrix}= u1(-pi/4)
$$
They can be added as below.
End of explanation
"""


qc = QuantumCircuit(q)
qc.rx(pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Standard Rotations
The standard rotation gates are those that define rotations around the Paulis $P={X,Y,Z}$. They are defined as 
$$ R_P(\theta) = \exp(-i \theta P/2) = \cos(\theta/2)I -i \sin(\theta/2)P$$
Rotation around X-axis
$$
R_x(\theta) = 
\begin{pmatrix}
\cos(\theta/2) & -i\sin(\theta/2)\
-i\sin(\theta/2) & \cos(\theta/2)
\end{pmatrix} = u3(\theta, -\pi/2,\pi/2)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.ry(pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Rotation around Y-axis
$$
R_y(\theta) =
\begin{pmatrix}
\cos(\theta/2) & - \sin(\theta/2)\
\sin(\theta/2) & \cos(\theta/2).
\end{pmatrix} =u3(\theta,0,0)
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.rz(pi/2,q)
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Rotation around Z-axis
$$
R_z(\phi) = 
\begin{pmatrix}
e^{-i \phi/2} & 0 \
0 & e^{i \phi/2}
\end{pmatrix}\equiv u1(\phi)
$$
Note here we have used an equivalent as is different to u1 by global phase $e^{-i \phi/2}$.
End of explanation
"""


q = QuantumRegister(2)

"""
Explanation: Note this is different due only to a global phase
Multi-Qubit Gates
Mathematical Preliminaries
The space of quantum computer grows exponential with the number of qubits. For $n$ qubits the complex vector space has dimensions $d=2^n$. To describe states of a multi-qubit system, the tensor product is used to "glue together" operators and basis vectors.
Let's start by considering a 2-qubit system. Given two operators $A$ and $B$ that each act on one qubit, the joint operator $A \otimes B$ acting on two qubits is
$$\begin{equation}
    A\otimes B = 
    \begin{pmatrix} 
        A_{00} \begin{pmatrix} 
            B_{00} & B_{01} \
            B_{10} & B_{11}
        \end{pmatrix} & A_{01}  \begin{pmatrix} 
                B_{00} & B_{01} \
                B_{10} & B_{11}
            \end{pmatrix} \
        A_{10}  \begin{pmatrix} 
                    B_{00} & B_{01} \
                    B_{10} & B_{11}
                \end{pmatrix} & A_{11}  \begin{pmatrix} 
                            B_{00} & B_{01} \
                            B_{10} & B_{11}
                        \end{pmatrix}
    \end{pmatrix},                    
\end{equation}$$
where $A_{jk}$ and $B_{lm}$ are the matrix elements of $A$ and $B$, respectively.
Analogously, the basis vectors for the 2-qubit system are formed using the tensor product of basis vectors for a single qubit:
$$\begin{equation}\begin{split}
    |{00}\rangle &= \begin{pmatrix} 
        1 \begin{pmatrix} 
            1  \
            0
        \end{pmatrix} \
        0 \begin{pmatrix} 
            1  \
            0 
        \end{pmatrix}
    \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \0 \end{pmatrix}~~~|{01}\rangle = \begin{pmatrix} 
    1 \begin{pmatrix} 
    0 \
    1
    \end{pmatrix} \
    0 \begin{pmatrix} 
    0  \
    1 
    \end{pmatrix}
    \end{pmatrix} = \begin{pmatrix}0 \ 1 \ 0 \ 0 \end{pmatrix}\end{split}
\end{equation}$$
$$\begin{equation}\begin{split}|{10}\rangle = \begin{pmatrix} 
    0\begin{pmatrix} 
    1  \
    0
    \end{pmatrix} \
    1\begin{pmatrix} 
    1 \
    0 
    \end{pmatrix}
    \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \end{pmatrix}~~~   |{11}\rangle = \begin{pmatrix} 
    0 \begin{pmatrix} 
    0  \
    1
    \end{pmatrix} \
    1\begin{pmatrix} 
    0  \
    1 
    \end{pmatrix}
    \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \1 \end{pmatrix}\end{split}
\end{equation}.$$
Note we've introduced a shorthand for the tensor product of basis vectors, wherein $|0\rangle \otimes |0\rangle$ is written as $|00\rangle$. The state of an $n$-qubit system can described using the $n$-fold tensor product of single-qubit basis vectors. Notice that the basis vectors for a 2-qubit system are 4-dimensional; in general, the basis vectors of an $n$-qubit sytsem are $2^{n}$-dimensional, as noted earlier.
Basis vector ordering in Qiskit
Within the physics community, the qubits of a multi-qubit systems are typically ordered with the first qubit on the left-most side of the tensor product and the last qubit on the right-most side. For instance, if the first qubit is in state $|0\rangle$ and second is in state $|1\rangle$, their joint state would be $|01\rangle$. Qiskit uses a slightly different ordering of the qubits, in which the qubits are represented from the most significant bit (MSB) on the left to the least significant bit (LSB) on the right (big-endian). This is similar to bitstring representation on classical computers, and enables easy conversion from bitstrings to integers after measurements are performed. For the example just given, the joint state would be represented as $|10\rangle$. Importantly, this change in the representation of multi-qubit states affects the way multi-qubit gates are represented in Qiskit, as discussed below.
The representation used in Qiskit enumerates the basis vectors in increasing order of the integers they represent. For instance, the basis vectors for a 2-qubit system would be ordered as $|00\rangle$, $|01\rangle$, $|10\rangle$, and $|11\rangle$. Thinking of the basis vectors as bit strings, they encode the integers 0,1,2 and 3, respectively.
Controlled operations on qubits
A common multi-qubit gate involves the application of a gate to one qubit, conditioned on the state of another qubit. For instance, we might want to flip the state of the second qubit when the first qubit is in $|0\rangle$. Such gates are known as controlled gates. The standard multi-qubit gates consist of two-qubit gates and three-qubit gates. The two-qubit gates are:
- controlled Pauli gates
- controlled Hadamard gate
- controlled rotation gates
- controlled phase gate
- controlled u3 gate
- swap gate
The three-qubit gates are: 
- Toffoli gate 
- Fredkin gate
Two-qubit gates
Most of the two-gates are of the controlled type (the SWAP gate being the exception). In general, a controlled two-qubit gate $C_{U}$ acts to apply the single-qubit unitary $U$ to the second qubit when the state of the first qubit is in $|1\rangle$. Suppose $U$ has a matrix representation
$$U = \begin{pmatrix} u_{00} & u_{01} \ u_{10} & u_{11}\end{pmatrix}.$$
We can work out the action of $C_{U}$ as follows. Recall that the basis vectors for a two-qubit system are ordered as $|00\rangle, |01\rangle, |10\rangle, |11\rangle$. Suppose the control qubit is qubit 0 (which, according to Qiskit's convention, is one the right-hand side of the tensor product). If the control qubit is in $|1\rangle$, $U$ should be applied to the target (qubit 1, on the left-hand side of the tensor product). Therefore, under the action of $C_{U}$, the basis vectors are transformed according to
$$\begin{align}
C_{U}: \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle} &\rightarrow \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle}\
C_{U}: \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle} &\rightarrow \underset{\text{qubit}~1}{U|0\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle}\
C_{U}: \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle} &\rightarrow \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle}\
C_{U}: \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle} &\rightarrow \underset{\text{qubit}~1}{U|1\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle}\
\end{align}.$$
In matrix form, the action of $C_{U}$ is
$$\begin{equation}
    C_U = \begin{pmatrix}
    1 & 0 & 0 & 0 \
    0 & u_{00} & 0 & u_{01} \
    0 & 0 & 1 & 0 \
    0 & u_{10} &0 & u_{11}
        \end{pmatrix}.
\end{equation}$$
To work out these matrix elements, let
$$C_{(jk), (lm)} = \left(\underset{\text{qubit}~1}{\langle j |} \otimes \underset{\text{qubit}~0}{\langle k |}\right) C_{U} \left(\underset{\text{qubit}~1}{| l \rangle} \otimes \underset{\text{qubit}~0}{| k \rangle}\right),$$
compute the action of $C_{U}$ (given above), and compute the inner products.
As shown in the examples below, this operation is implemented in Qiskit as cU(q[0],q[1]).
If qubit 1 is the control and qubit 0 is the target, then the basis vectors are transformed according to
$$\begin{align}
C_{U}: \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle} &\rightarrow \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle}\
C_{U}: \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle} &\rightarrow \underset{\text{qubit}~1}{|0\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle}\
C_{U}: \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{|0\rangle} &\rightarrow \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{U|0\rangle}\
C_{U}: \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{|1\rangle} &\rightarrow \underset{\text{qubit}~1}{|1\rangle}\otimes \underset{\text{qubit}~0}{U|1\rangle}\
\end{align},$$
which implies the matrix form of $C_{U}$ is
$$\begin{equation}
    C_U = \begin{pmatrix}
    1 & 0 & 0  & 0 \
    0 & 1 & 0 & 0 \
    0 & 0 & u_{00} & u_{01} \
    0 & 0 & u_{10} & u_{11}
        \end{pmatrix}.
\end{equation}$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.cx(q[0],q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Controlled Pauli Gates
Controlled-X (or, controlled-NOT) gate
The controlled-not gate flips the target qubit when the control qubit is in the state $|1\rangle$. If we take the MSB as the control qubit (e.g. cx(q[1],q[0])), then the matrix would look like
$$
C_X = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 1 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & 1 & 0
\end{pmatrix}. 
$$
However, when the LSB is the control qubit, (e.g. cx(q[0],q[1])), this gate is equivalent to the following matrix:
$$
C_X = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & 1 & 0\
0 & 1 & 0 & 0
\end{pmatrix}. 
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.cy(q[0],q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Controlled $Y$ gate
Apply the $Y$ gate to the target qubit if the control qubit is the MSB
$$
C_Y = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 1 & 0 & 0\
0 & 0 & 0 & -i\
0 & 0 & i & 0
\end{pmatrix},
$$
or when the LSB is the control
$$
C_Y = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 0 & 0 & -i\
0 & 0 & 1 & 0\
0 & i & 0 & 0
\end{pmatrix}.
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.cz(q[0],q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Controlled $Z$ (or, controlled Phase) gate
Similarly, the controlled Z gate flips the phase of the target qubit if the control qubit is $1$. The matrix looks the same regardless of whether the MSB or LSB is the control qubit:
$$
C_Z = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 1 & 0 & 0\
0 & 0 & 1 & 0\
0 & 0 & 0 & -1
\end{pmatrix}
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.ch(q[0],q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary']/(0.707+0.707j), 3)

"""
Explanation: Controlled Hadamard gate
Apply $H$ gate to the target qubit if the control qubit is $|1\rangle$. Below is the case where the control is the LSB qubit.
$$
C_H = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\
0 & 0 & 1 & 0\
0 & \frac{1}{\sqrt{2}}  & 0& -\frac{1}{\sqrt{2}}
\end{pmatrix}
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.crz(pi/2,q[0],q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Controlled rotation gates
Controlled rotation around Z-axis
Perform rotation around Z-axis on the target qubit if the control qubit (here LSB) is $|1\rangle$.
$$
C_{Rz}(\lambda) = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & e^{-i\lambda/2} & 0 & 0\
0 & 0 & 1 & 0\
0 & 0 & 0 & e^{i\lambda/2}
\end{pmatrix}
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.cu1(pi/2,q[0], q[1])
circuit_drawer(qc)

"""
Explanation: Controlled phase rotation
Perform a phase rotation if both qubits are in the $|11\rangle$ state. The matrix looks the same regardless of whether the MSB or LSB is the control qubit.
$$
C_{u1}(\lambda) = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 1 & 0 & 0\
0 & 0 & 1 & 0\
0 & 0 & 0 & e^{i\lambda}
\end{pmatrix}
$$
End of explanation
"""


job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: I THINK SHOULD BE CALLED $C_\mathrm{PHASE}(\lambda)$
End of explanation
"""


qc = QuantumCircuit(q)
qc.cu3(pi/2, pi/2, pi/2, q[0], q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Controlled $u3$ rotation
Perform controlled-$u3$ rotation on the target qubit if the control qubit (here LSB) is $|1\rangle$. 
$$
C_{u3}(\theta, \phi, \lambda) \equiv 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & e^{-i(\phi+\lambda)/2}\cos(\theta/2) & 0 & -e^{-i(\phi-\lambda)/2}\sin(\theta/2)\
0 & 0 & 1 & 0\
0 & e^{i(\phi-\lambda)/2}\sin(\theta/2) & 0 & e^{i(\phi+\lambda)/2}\cos(\theta/2)
\end{pmatrix}.
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.swap(q[0], q[1])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: NOTE I NEED TO FIX THIS AND DECIDE ON CONVENTION - I ACTUALLY THINK WE WANT A FOUR PARAMETER GATE AND JUST CALL IT CU AND TO REMOVE THIS GATE.
SWAP gate
The SWAP gate exchanges the two qubits. It transforms the basis vectors as
$$|00\rangle \rightarrow |00\rangle~,~|01\rangle \rightarrow |10\rangle~,~|10\rangle \rightarrow |01\rangle~,~|11\rangle \rightarrow |11\rangle,$$
which gives a matrix representation of the form
$$
\mathrm{SWAP} = 
\begin{pmatrix}
1 & 0 & 0 & 0\
0 & 0 & 1 & 0\
0 & 1 & 0 & 0\
0 & 0 & 0 & 1
\end{pmatrix}.
$$
End of explanation
"""


q = QuantumRegister(3)

qc = QuantumCircuit(q)
qc.ccx(q[0], q[1], q[2])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Three-qubit gates
There are two commonly-used three-qubit gates. For three qubits, the basis vectors are ordered as
$$|000\rangle, |001\rangle, |010\rangle, |011\rangle, |100\rangle, |101\rangle, |110\rangle, |111\rangle,$$
which, as bitstrings, represent the integers $0,1,2,\cdots, 7$. Again, Qiskit uses a representation in which the first qubit is on the right-most side of the tensor product and the third qubit is on the left-most side:
$$|abc\rangle : \underset{\text{qubit 2}}{|a\rangle}\otimes \underset{\text{qubit 1}}{|b\rangle}\otimes \underset{\text{qubit 0}}{|c\rangle}.$$
Toffoli gate ($ccx$ gate)
The Toffoli gate flips the third qubit if the first two qubits (LSB) are both $|1\rangle$:
$$|abc\rangle \rightarrow |bc\oplus a\rangle \otimes |b\rangle \otimes c \rangle.$$
In matrix form, the Toffoli gate is
$$
C_{CX} = 
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0
\end{pmatrix}.
$$
End of explanation
"""


qc = QuantumCircuit(q)
qc.cswap(q[0], q[1], q[2])
circuit_drawer(qc)

job = execute(qc, backend)
np.round(job.result().get_data(qc)['unitary'], 3)

"""
Explanation: Controlled swap gate (Fredkin Gate)
The Fredkin gate, or the controlled swap gate, exchanges the second and third qubits if the first qubit (LSB) is $|1\rangle$:
$$ |abc\rangle \rightarrow \begin{cases} |bac\rangle~~\text{if}~c=1 \cr |abc\rangle~~\text{if}~c=0 \end{cases}.$$
In matrix form, the Fredkin gate is
$$
C_{\mathrm{SWAP}} = 
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}.
$$
End of explanation
"""


q = QuantumRegister(1)
c = ClassicalRegister(1)

"""
Explanation: Non unitary operations
Now we have gone through all the unitary operations in quantum circuits we also have access to non-unitary operations. These include measurements, reset of qubits, and classical conditional operations.
End of explanation
"""


qc = QuantumCircuit(q, c)
qc.measure(q, c)
circuit_drawer(qc)

backend = Aer.get_backend('qasm_simulator')
job = execute(qc, backend, shots=1024)
job.result().get_counts(qc)

"""
Explanation: Measurements
We don't have access to all the information when we make a measurement in a quantum computer. The quantum state is projected onto the standard basis. Below are two examples showing a circuit that is prepared in a basis state and the quantum computer prepared in a superposition state.
End of explanation
"""


qc = QuantumCircuit(q, c)
qc.h(q)
qc.measure(q, c)
circuit_drawer(qc)

job = execute(qc, backend, shots=1024)
job.result().get_counts(qc)

"""
Explanation: The simulator predicts that 100 percent of the time the classical register returns 0.
End of explanation
"""


qc = QuantumCircuit(q, c)
qc.reset(q[0])
qc.measure(q, c)
circuit_drawer(qc)

job = execute(qc, backend, shots=1024)
job.result().get_counts(qc)

qc = QuantumCircuit(q, c)
qc.h(q)
qc.reset(q[0])
qc.measure(q, c)
circuit_drawer(qc)

job = execute(qc, backend, shots=1024)
job.result().get_counts(qc)

"""
Explanation: The simulator predicts that 50 percent of the time the classical register returns 0 or 1. 
Reset
It is also possible to reset qubits to the $|0\rangle$ state in the middle of computation. Note that reset is not a Gate operation, since it is irreversible.
End of explanation
"""


qc = QuantumCircuit(q, c)
qc.x(q[0]).c_if(c, 0)
qc.measure(q,c)
circuit_drawer(qc)

job = execute(qc, backend, shots=1024)
job.result().get_counts(qc)

"""
Explanation: Here we see that for both of these circuits the simulator always predicts that the output is 100 percent in the 0 state.
Conditional operations
It is also possible to do operations conditioned on the state of the classical register
End of explanation
"""


qc = QuantumCircuit(q, c)
qc.h(q)
qc.measure(q,c)
qc.x(q[0]).c_if(c, 0)
qc.measure(q,c)
circuit_drawer(qc)

job = execute(qc, backend, shots=1024)
job.result().get_counts(qc)

"""
Explanation: Here the classical bit always takes the value 0 so the qubit state is always flipped.
End of explanation
"""


# Initializing a three-qubit quantum state
import math
desired_vector = [
    1 / math.sqrt(16) * complex(0, 1),
    1 / math.sqrt(8) * complex(1, 0),
    1 / math.sqrt(16) * complex(1, 1),
    0,
    0,
    1 / math.sqrt(8) * complex(1, 2),
    1 / math.sqrt(16) * complex(1, 0),
    0]


q = QuantumRegister(3)

qc = QuantumCircuit(q)

qc.initialize(desired_vector, [q[0],q[1],q[2]])

backend = Aer.get_backend('statevector_simulator')
job = execute(qc, backend)
qc_state = job.result().get_statevector(qc)
qc_state 

"""
Explanation: Here the classical bit by the first measurement is random but the conditional operation results in the qubit being deterministically put into $|1\rangle$.
Arbitrary initialization
What if we want to initialize a qubit register to an arbitrary state? An arbitrary state for $n$ qubits may be specified by a vector of $2^n$ amplitudes, where the sum of amplitude-norms-squared equals 1. For example, the following three-qubit state can be prepared:
$$|\psi\rangle = \frac{i}{4}|000\rangle + \frac{1}{\sqrt{8}}|001\rangle + \frac{1+i}{4}|010\rangle + \frac{1+2i}{\sqrt{8}}|101\rangle + \frac{1}{4}|110\rangle$$
End of explanation
"""


state_fidelity(desired_vector,qc_state)

"""
Explanation: Fidelity is useful to check whether two states are same or not.
For quantum (pure) states $\left|\psi_1\right\rangle$ and $\left|\psi_2\right\rangle$, the fidelity is
$$
F\left(\left|\psi_1\right\rangle,\left|\psi_2\right\rangle\right) = \left|\left\langle\psi_1\middle|\psi_2\right\rangle\right|^2.
$$
The fidelity is equal to $1$ if and only if two states are same.
End of explanation
"""


