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In qiskit you can get a unitary matrix from a circuit (circuit to unitary matrix example). Is the opposite direction possible? Can you input a unitary matrix and have qiskit come up with a circuit? If it helps you can restrict the matrices to be clifford + multi-qubit controlled pauli strings.

Here's an example that goes from circuit -> unitary and then attempts to get the original circuit back based on the answer below :

import numpy as np
np.set_printoptions(threshold=np.inf)

import qiskit

backend=qiskit.Aer.get_backend('unitary_simulator')

qr=qiskit.QuantumRegister(4,name="qr")

CirA=qiskit.QuantumCircuit(qr);
CirA.cx(3,2)
CirA.h(0)
CirA.cx(0,2)
CirA.h(1)
CirA.cx(1,3)
print(CirA)

job=qiskit.execute(CirA,backend,shots=1)
result=job.result()
MatA=result.get_unitary(CirA,3)

CirB=qiskit.QuantumCircuit(qr);
CirB.unitary(MatA,[ 0, 1, 2, 3 ],label='CirB')
print(CirA)

unroller = qiskit.transpiler.passes.Unroller(basis=['u', 'cx'])

uCirA = qiskit.converters.dag_to_circuit(unroller.run(qiskit.converters.circuit_to_dag(CirA)))
print(uCirA)

uCirB = qiskit.converters.dag_to_circuit(unroller.run(qiskit.converters.circuit_to_dag(CirB)))
print(uCirB)
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1 Answer 1

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You can easily create a quantum circuit which implements a unitary by appending a UnitaryGate to the circuit, or using QuantumCircuit.unitary() method:

# Get some random unitary:
from qiskit.quantum_info import random_unitary
num_qubits = 4
U = random_unitary(2 ** num_qubits)

# Create the quantum circuit:
qr = QuantumRegister(num_qubits, 'q')
circ = QuantumCircuit(qr)
circ.unitary(U, qr)

For more information about how Qiskit constructs the circuit from the unitary matrix see here.

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  • $\begingroup$ thanks for the answer. I can define a circuit using a unitary as you describe but what I'm really after is the actual synthesis into a gate netlist. In your linked answer you refer to qiskit.quantum_info.synthesis.qsd which is probably what I need to look into. I can't find examples on how to use it. Do you have any? is it restricted to clifford gates? $\endgroup$
    – unknown
    Oct 6, 2022 at 14:38
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    $\begingroup$ Show the answer here: quantumcomputing.stackexchange.com/a/23547/9474 $\endgroup$
    – Egretta.Thula
    Oct 6, 2022 at 16:14
  • $\begingroup$ I edited the question to experiment with the circuit->unitary->circuit conversions. In principle CirA=CirB, but the unrolled versions (uCirA, uCirB) are different. I would expect them to give the same expansion. Ideally both would have recovered the original circuit which uses only h,cx gates...maybe with a better choice of expansion basis? $\endgroup$
    – unknown
    Oct 6, 2022 at 17:56

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