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Let $|\psi\rangle$ be a fixed state and $M$ and $N$ two commuting operators corresponding to projective measurements. Consider the probability distribution $p$ obtained by measuring $N$ on $|\psi\rangle$. Is this probability distribution exactly the same as the probability distribution $q$ that we would get if we measured first $M$ on $|\psi\rangle$, discarded the result, then measured $N$ on the post-measurement state?

I managed to prove it using the basic law of total probability and the existence of a common orthogonal basis, but I want to be sure since I have never seen it.

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Yes.

Let's give a decomposition of both operators in terms of projectors onto their eigenspaces $$ N=\sum_{q}n_{q}Q_{q},\qquad M=\sum_{p}m_{p}P_{p}. $$ The probability of getting measurement result $q$ on measuring $N$ is $$ p_q=\text{Tr}(Q_q|\psi\rangle\langle\psi|). $$

Instead, imagine we first measure $M$ and discard the result. After that measurement, we have $$ \rho=\sum_pP_p|\psi\rangle\langle\psi|P_p. $$ If we now measure $N$, the probability of getting result $q$ is $$ \tilde{p}_q=\text{Tr}(Q_q\rho). $$ Using the fact that $[Q_q,P_p]=0$, this can be written as $$ \tilde{p}_q=\text{Tr}(\sum_pP_pQ_q|\psi\rangle\langle\psi|P_p). $$ Now use the cyclic property of trace, and the fact that $P_p^2=P_p$ to give $$ \tilde p_q=\text{Tr}\left(\left(\sum_pP_p\right)Q_q|\psi\rangle\langle\psi|\right). $$ Since $\sum_pP_p=I$, this shows that $p_q=\tilde p_q$, as required.

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