I am looking for a way to decompose any given operator U into ZYZ rotations. And then plug the values back into ei∗α∗Rz(θ0)∗Ry(θ1)∗Rz(θ2), to 'recompose' the gate and check if the gate decomposed correctly. Can anyone help me with the Julia codes for this? If not, then Python codes are also welcome


1 Answer 1


welcome to QCSE.

I have understood your question, but next time make sure that you use MathJax notation for math equations, it is much cleaner and understoodable.

I don't know about Julia, but I have wrote the following code in Python-Qiskit and I think that's what you are looking for:

import numpy as np
import math

from qiskit import QuantumCircuit, quantum_info
from qiskit.visualization import array_to_latex

#U = np.array([[1/math.sqrt(2),1/math.sqrt(2)],[1/math.sqrt(2),-1/math.sqrt(2)]])
U = np.array([[0,1],[1,0]])
display(array_to_latex(U, prefix = "U = "))

decomposer = quantum_info.OneQubitEulerDecomposer(basis = "ZYZ")
data = decomposer.angles_and_phase(U)
theta = data[0]
phi = data[1]
lamda = data[2]
phase = data[3]

print("\033[1m ZYZ decomposition of U: \033[0m")
print(" theta = {0}, phi = {1}, lambda = {2}, phase = {3}".format(theta,phi,lamda,phase))

qc = QuantumCircuit(1)
qc.rz(lamda, 0)
qc.ry(theta, 0)
qc.rz(phi, 0)

phase_exp = math.e ** (1j * phase)
U_recomposed = np.array(phase_exp * quantum_info.Operator(qc))
display(array_to_latex(U_recomposed, prefix = "U recomposed = "))

And the output is:

enter image description here

I have used built-in class of qiskit quantum_info.OneQubitEulerDecomposer and its methods in order to easily compute the angles and phase of the ZYZ decomposition. Take a look at the documentation of the OneQubitEulerDecomposer class and its method angles_and_phase() Note that I gave identical names to the angles just as in the documentation, to avoid confusions.

Finally, I have used the qunatum_info.Operator class to obtain the unitary matrix for the circuit created, and after multiplying the result by the complex exponent of the phase we have got the original U operator.

  • $\begingroup$ Alright, thanks so much for this $\endgroup$
    – Hamzah
    Aug 12, 2022 at 0:04

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