4
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Why is the order reversed on measurement?

from qiskit import(
  QuantumCircuit,
  execute,
  Aer)
from qiskit.visualization import plot_histogram

# Use Aer's qasm_simulator
simulator = Aer.get_backend('qasm_simulator')

# Create a Quantum Circuit acting on the q register
circuit = QuantumCircuit(3, 3)

# Add a X gate on qubit 0
circuit.x(0)

# Add a CX (CNOT) gate on control qubit 0 and target qubit 1
circuit.cx(0, 1)

circuit.barrier()
# Map the quantum measurement to the classical bits
circuit.measure([0,1,2], [0,1,2])

# Execute the circuit on the qasm simulator
job = execute(circuit, simulator, shots=1000)

# Grab results from the job
result = job.result()

# Returns counts
counts = result.get_counts(circuit)
print("\nTotal count:",counts)

# Draw the circuit
circuit.draw()

Got result:

Total count for 00 and 11 are: {'011': 1000}

But I'm expecting '110'.

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I still run into this issue too. If you consider $|q0\rangle$ to be the most significant bit (MSB) you have to map it to the most significant classical bit as well, which is in your case a bit no. 2. Or you can flip your quatnum circuit upside down and then $|q0\rangle$ become the least significant bit (LSB) and the measurement will meet your expectation.

A code

circuit.measure([0,1,2], [0,1,2])

is valid in case $|q0\rangle$ is LSB and

circuit.measure([0,1,2], [2,1,0])

in case $|q0\rangle$ is MSB.

I think that the reason for this arrangement is simply a convention, so you can choose whether $|q0\rangle$ is MSB or LSB and set the measurement procedure accordingly.

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  • $\begingroup$ Just note, the same problem occurs in QASM. $\endgroup$ – Martin Vesely Nov 21 at 22:40

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