5
$\begingroup$

I’ll define my question more specifically:

Lets suppose we have an $n$ bit quantum computer, with qubits labelled $q_{1},…,q_{n}$, and let $b$ be any bit string of length $n$ (let’s assume we’ve labelled each qubits’ states $0$ and $1$ so that we obtain an $n$ bit bit string when we measure the qubits).

The question is, is the probability that we obtain $b$ when we measure each qubits’ state dependent on the order in which we measure the qubits? I have a feeling the probability would be independent of the order the qubits are measured, but am struggling to prove this.

$\endgroup$
  • $\begingroup$ Consider the case where we take the qubits one light year apart, measure them in close succession, and then bring them back together $\endgroup$ – John Dvorak Dec 7 '16 at 23:13
  • $\begingroup$ @JanDvorak How could measuring the qubits when they're 1 light year apart and then bringing them back together alter the probabilities (of obtaining each measurement outcome), from the case where qubits are close together (as they would be in a quantum computer)? $\endgroup$ – Alex Michael Dec 7 '16 at 23:32
  • 2
    $\begingroup$ It's an axiom of quantum mechanics that space-like separated measurements commute (otherwise you'd be getting different results depending on your frame of reference). If your measurements can be space-like separated, they commute. $\endgroup$ – John Dvorak Dec 7 '16 at 23:34
  • $\begingroup$ @JanDvorak This is a little above my understanding of Quantum Physics (I'm a Computer Scientist studying Quantum Computing :) ) $\endgroup$ – Alex Michael Dec 8 '16 at 1:26
  • $\begingroup$ Have you tried proving this using the definitions? That's always a good start. $\endgroup$ – Yuval Filmus Dec 8 '16 at 7:16
8
$\begingroup$

No, the order doesn't matter.

Proofs

  • Algebra. Take an input state $\sum_k c_k |k_0 k_1 k_2 ...\rangle$. Apply the definition of measurement from your textbook to it. Compute the expression for the probabilities and outputs of each case when measuring qubit 0 then qubit 1. Do the same for measuring qubit 1 then qubit 0. Notice that the two expressions are equal. Generalize.

  • Circuit moves. Measurement is equivalent to a CNOT gate from the target qubit onto an ancilla qubit that you simply don't use for anything else:

    measure as cnot

    So if you think you can change the outcome statistics by reordering measurements, you should think you can do the same without involving measurements at all (until a simultaneous measurement of all qubits at the end of the circuit). All you should need is independent CNOTs. But clearly independent CNOTs can be re-ordered.

  • Play. For me, the fact that ordering doesn't matter wasn't proven so much as experienced. I dragged gates around in Quirk and noticed that measurement is easily the most boring operation. All measurement does is throw away off-diagonal density matrix elements. Even for the target qubit, measurement doesn't change the computational basis probabilities.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.