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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.

Alex Michael
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  • Consider the case where we take the qubits one light year apart, measure them in close succession, and then bring them back together – John Dvorak Dec 07 '16 at 23:13
  • @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)? – Alex Michael Dec 07 '16 at 23:32
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    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. – John Dvorak Dec 07 '16 at 23:34
  • @JanDvorak This is a little above my understanding of Quantum Physics (I'm a Computer Scientist studying Quantum Computing :) ) – Alex Michael Dec 08 '16 at 01:26
  • Have you tried proving this using the definitions? That's always a good start. – Yuval Filmus Dec 08 '16 at 07:16
  • @Yuval Filmus I can see for the case where the qubits are not entangled that the order wouldn't matter, but for the case when they are entangled I can't as I'm not sure on the following issue: Lets suppose qubits $q_{1}$ and $q_{2}$ are entangled. Then I know that if we measure $q_{1}$, that $q_{2}$'s superposition state will collapse, but I'm not sure on what the probability that $q_{2}$ will take $0$ or $1$ is - does it depend on the state $q_{1}$ took at all when measured, or is it just dependent on $q_{2}$'s superposition state at the time $q_{1}$ was measured – Alex Michael Dec 08 '16 at 10:17
  • Try using formulas for the probability to compute the probability (or rather, density) of getting a certain outcome when you measure the bits in the order $q_1,q_2$. You should get something symmetric in the indices. – Yuval Filmus Dec 08 '16 at 14:27

1 Answers1

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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.

Craig Gidney
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