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I want to understand better what it means by any commuting set of operators can be measured simultaneously.

Suppose I have an $n$-qubit arbitrary pure state $\rho = \lvert \psi \rangle \langle \psi \rvert$. Is it always true that all $X_1 = X\otimes \cdots \otimes I$, ..., $X_n = I \otimes \cdots \otimes X $ can be measured simultaneously? If so, then is it always true that measuring all $X_i, Y_i, Z_i$ $\forall i \in [n]$ can be done by using $3$ copies of quantum states? If not, then what does it mean by "any commuting set of operators can be measured simultaneously"? What are conditions?

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  • $\begingroup$ My you expand your claim or put the reference? $\endgroup$ Dec 23, 2022 at 14:17

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A set of commuting operators can be measured without one measurement effecting another regardless of the purity of the state. In other words, when measurements commute, then the observables may all be measured in any order or at the same time without one measurement distrubing another. There is nothing preventing someone from trying to measure non-commuting observables simultaneously, for example this paper, but then measurement outcomes will disturb one another.

For your second question, if you can simultaneously measure different observables (not all hardware can), then yes you could measure all $X_i$ then all $Y_i$ then all $Z_i$ on 3 different copies without one outcome effecting another. Theoretically, you can do other things like measure X on all even qubits and Z on all odd qubits since these commute. These outcomes are still random variables the depend on the underlying state but not on other measurement outcomes.

Finally, a little example to illustrate this. Say you have a 2-qubit state $|\psi \rangle = |00 \rangle$. If you measure $Z_1, X_2$, then you'd always get a 0 for the result of $Z_1$ and you'd get a 0 or 1 with equal probability for the result of $X_2$ regardless of the order in which you measure. Now if you measure $X_2$ first, and get outcome +1, then the resulting state becomes $|0\rangle |+\rangle$. If you then measure $Z_2$ then you'd get 0 half the time and 1 half the time. Compare this to measuring $Z_2$ first then $X_2$. In that case, you'd always get 0 when you measure $Z_2$ first then you'd get a 50-50 outcome for $X_2$. Since $X_2, Z_2$ don't commute, they disturb each other and the order that the measurement is done changes the statistics of the outcomes. Measuring $Z_2, X_2$ simultaneously is then difficult to think about: which one happens first?

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  • $\begingroup$ So in principle (without taking account of any hardware implication), is it correct to say that any set of mutually commuting observable can be measured simultaneously? $\endgroup$
    – Jon Megan
    Dec 24, 2022 at 0:43

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