Qiskit flipped representation of qubits in CNOT gate?

2

1

Cnot on right and Qiskit representation of cnot on left[1]

The conventional CNOT gate is shown on the right, and the Qiskit version is on the left. Since Qiskit defines it has a flipped representation kindly explain what is happening to the 11 position?

shashanka300

Posted 2020-04-26T13:14:02.593

Reputation: 51

Answers

3

It depends on which qubit is the control qubit. Depending on this we will have different matrix representations. Here is a code that may help to identify the ordering in Qiskit:

from qiskit import *
simulator = Aer.get_backend('qasm_simulator')

circuit_cx_01 = QuantumCircuit(2, 2)
circuit_cx_10 = QuantumCircuit(2, 2)

circuit_cx_01.x(0)
circuit_cx_01.x(1)
circuit_cx_01.cx(0, 1)
circuit_cx_01.measure(0, 0)
circuit_cx_01.measure(1, 1)

circuit_cx_10.x(0)
circuit_cx_10.x(1)
circuit_cx_10.cx(1, 0)
circuit_cx_10.measure(0, 0)
circuit_cx_10.measure(1, 1)

result_cx_01 = execute(circuit_cx_01, simulator, shots=1024).result().get_counts()
result_cx_10 = execute(circuit_cx_10, simulator, shots=1024).result().get_counts()

print("After cx(0, 1) applied on '11' state \n{}".format(result_cx_01))
print("\nAfter cx(1, 0) applied on '11' state \n{}".format(result_cx_10))

The output is:

After cx(0, 1) applied on '11' state 
{'01': 1024}

After cx(1, 0) applied on '11' state 
{'10': 1024}

where in the cx(i, j) the i qubit is the control qubit and j is the target qubit. From this, one can see that in Qiskit 0 index corresponds to the rightmost qubit.

Davit Khachatryan

Posted 2020-04-26T13:14:02.593

Reputation: 3 583