What does it mean for a gate to "commute with a measurement"?

Question

I think I understand intuitively what this means. For example, I can apply an $X$ gate and then perform a measurement in the $X$ basis and I will get the same post-measurement states as if I measured in the $X$ basis and then applied an $X$ gate. This is not the case for a $Z$ gate and $X$ measurement.

Mathematically, what is the precise way of stating this property? Does it mean the gate commutes with all the Kraus operators of the measurement channel? This seems like a sufficient condition but is it also a necessary condition?

Answer 1

I think that mathematically you have an observable and this observable is described by an operator. The gate commute with measurement if the commutator between the gate and the operator associated to your observable is 0.

As you said the measure onto X axis and NOT gate commutes, in fact, the gate X represents the operator associated to the observable "projection onto the X axis". So [X,X]=0

Answer 2

I agree with @Simona99's answer. If you want to formulate it in terms of a quantum channel, consider the quantum channel $\mathcal{E}[\rho] = \sum_i \Pi_i \rho \Pi_i$ that corresponds to a measurement of an observable of the form $O = \sum_i \lambda_i \Pi_i$ with orthogonal projectors $\Pi_i$ and real, distinct eigenvalues $\lambda_i \neq \lambda_j$. Further consider the unitary channel $\mathcal{F}[\rho] = G \rho G^\dagger$ that applies the unitary gate $G$. It is easy to see that, whenever $[G, \Pi_i] = 0 \, \forall i$, then also the two channels commute.

Because the Kraus operators of the measurement channel are orthogonal projectors, it is also a necessary condition. If we start with $[G, O] = 0$, there exists a common eigenbasis of $G$ and $O$. As the $\Pi_i$ project onto a certain eigensubspace of $O$, they also decompose into common eigenvectors, i.e. $[G, \Pi_i] = 0 \, \forall i$.

What the channel does not capture is whether the measurement outcome is also conserved, but this can be straight forwardly checked: ${\rm Tr}(G\rho G^\dagger O) = {\rm Tr}(\rho O)$, if $[G,O]=0$.