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Does the order in which qubits are measured matter in quantum computing?
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qubitstheorderaremeasureddoeswhichcomputingmatterquantum
Problem
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.
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.
Solution
No, the order doesn't matter.
Proofs
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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.
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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:
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.
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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.
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:
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.
Context
StackExchange Computer Science Q#67082, answer score: 9
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