How can we represent a CNOT gate using a cascade of rotation gates?

Question

I have been working using qiskit to implement the CNOT decomposition into a cascade of rotation gates from this source. After computing the unitary matrix, the resultant matrix is not the same as the matrix for a regular CNOT.

$$ CNOT=\begin{bmatrix} 1 & 0 & 0& 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \end{bmatrix} $$

$$ \text{qiskit output}=\begin{bmatrix} 0.71+0.71j& 0 & 0 & 0 \\ 0 & 0 & 0 & 0.71+0.71j \\ 0 & 0 & 0.71+0.71j & 0\\0 & 0.71+0.71j & 0 & 0 \end{bmatrix} $$

This is the code for reproduction.

from qiskit import Aer, QuantumCircuit, execute
import numpy as np

if __name__ == '__main__':
    # create a qiskit circuit that implements a cnot gate using rotation gates
    n = 2
    alpha = 1
    circuit = QuantumCircuit(n)

    circuit.ry(alpha*np.pi/2, 0)
    circuit.rxx(alpha*np.pi/2, 0, 1)
    circuit.ry(-alpha*np.pi/2, 0)
    circuit.rx(-np.pi/2, 1)
    circuit.rz(-np.pi/2, 0)
    
    circuit.ry(np.pi/2, 0)
    circuit.ry(np.pi/2, 1)

    backend = Aer.get_backend('unitary_simulator')
    job = execute(circuit, backend, shots=8192)
    result = job.result()

    print(result.get_unitary(circuit, 2))

Answer 1

The matrix you get is actually correct. In ordet to see it, you have to recall two things:

1. Qiskit uses little-endian ordering.
2. Global phases are irrelevant.

Concerning the first one, you can find it in the Qiskit documentation. By using Qiskit and considering the first wire as the one which will give the first bit once measured, you'll end up getting things in reverse.

In Qiskit, the first wire is associated to the last bit. Thus, performing a CNOT controlled on the first wire actually performs the following transformations: $$\begin{align}|00\rangle\to|00\rangle\\|01\rangle\to|11\rangle\\|10\rangle\to|10\rangle\\|11\rangle\to|01\rangle\end{align}$$ which corresponds to the following matrix: $$\begin{pmatrix}1&0&0&0\\0&0&0&1\\0&0&1&0\\0&1&0&0\end{pmatrix}$$

Concerning the second points, note that the matrix you get is exactly the matrix you seek up to a factor $0.71+0.71\mathrm{i}\approx\frac{1+\mathrm{i}}{\sqrt{2}}=\mathrm{e}^{\mathrm{i}\frac\pi4}$. This factor is a global phase and is irrelevant in implementing a quantum gate, as long as you don't build a controlled-gate out of it. Thus, you can definitely use the gate you've created as a CNOT gate :)