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I am still confused when I am trying to use the IBM Q composer to reproduce some quantum circuits I found in different papers, and I am wondering if it is because there are two ways to represent a multi-qubit state in a quantum circuit.

The multi-qubit state $|xy\rangle$ can be represented in a quantum circuit by

  1. putting the $|x\rangle$ qubit (left) on top, and the $|y\rangle$ one below, or by
  2. putting the $|y\rangle$ qubit (right) on top, and the $|x\rangle$ one below.

Is that correct?

  1. The convention of putting on top of the circuit the qubit that is the furthest to the left on the state, is called big-endian, while
  2. The convention of putting on top of the circuit the qubit that is the furthest to the right on the state, is called little-endian.

Is that correct?

IBM Q composer is putting the on top of the circuit the qubit that is the furthest to the right on the state, meaning it is using the little-endian representation of multi-qubit states.

Is that correct?

Sanchayan Dutta
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Patrice
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1 Answers1

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IBM Q Experience and Qiskit both use the little-endian format.

The multi-qubit state $|xy\rangle$ can be represented in a quantum circuit by

  1. putting the $|x\rangle$ qubit (left) on top, and the $|y\rangle$ one below, or by
  2. putting the $|y\rangle$ qubit (right) on top, and the $|x\rangle$ one below.

Is that correct?

Yes. The IBM Q Composer uses convention 1.

  1. The convention of putting on top of the circuit the qubit that is the furthest to the left on the state, is called big-endian, while
  2. The convention of putting on top of the circuit the qubit that is the furthest to the right on the state, is called little-endian.

Is that correct?

Yes. The point being that the qubits are numbered starting from the top (i.e., q[0], q[1], q[2],...).

IBM Q composer is putting the on top of the circuit the qubit that is the furthest to the right on the state, meaning it is using the little-endian representation of multi-qubit states.

Is that correct?

Correct.

Sanchayan Dutta
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