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Given a unitary matrix of a unknown gate, can we write a program in Qiskit to output the name of the corresponding gate if that is a standard gate? So suppose I have $U = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$, I want Qiskit to output the Pauli $X$ gate name.

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  • $\begingroup$ Using Simone's methodology, you could also search for specific multiples, fractions, or powers of gates, including the square roots. $\endgroup$
    – rhundt
    Feb 7, 2023 at 22:27

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I don't think Qiskit itself has some functionality to directly get what you want but you could implement it by yourself using something like the following:

from inspect import getmembers, isclass
import numpy as np
from qiskit.circuit.library import standard_gates


def get_qiskit_gate(u):
    for name, gate in getmembers(standard_gates, isclass):
        try:
            unitary = gate().__array__()
        except:
            continue
        if unitary.shape == u.shape and np.allclose(unitary, u):
            return name

However, this will work only for standard gates (defined in the qiskit.circuit.library.standard_gates module) that implement the __array__ attribute. If the given unitary u does not correspond to any of these gates, the function will simply returns None.

Here is an example for the Pauli $X$ gate:

unitary = np.array([[0, 1],
                    [1, 0]])

print(get_qiskit_gate(u=unitary))

XGate

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  • $\begingroup$ Also can you please tell me if I have a standard gate, how can I get its unitary matrix from qiskit $\endgroup$
    – wizzywizzy
    Feb 8, 2023 at 20:26
  • $\begingroup$ Please open a new question and add some more details about what you need to do $\endgroup$ Feb 8, 2023 at 23:46
  • $\begingroup$ unitary = np.array([[0.707, 0.707], [0.707,- 0.707]]). Why isn't the code outputting hadamard for the above unitary matrix $\endgroup$
    – wizzywizzy
    Feb 9, 2023 at 4:19
  • $\begingroup$ That happens simply because of the np.allclose finite precision when comparing the array unitary with the array u. If you put 0.7071, it works fine; unitary = np.array([[1, 1], [1, -1]]) / np.sqrt(2) would be even better $\endgroup$ Feb 9, 2023 at 8:58

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