Why is measuring $M$ equivalent to measuring $UMU^{-1}$ after applying $U$?

Measurement in quantum mechanics consists of two rules. The first one - called the Born rule - determines the possible measurement outcomes and their probabilities and the second determines the post-measurement state associated with each outcome. Consequently, there are different ways to define equivalence between measurements. In a weaker form, one declares two measurements equivalent if they have the same outcome probability distribution. In a stronger form, one in addition demands that for each outcome the associated post-measurement states are the same.

The authors of the paper have the weaker form of equivalence in mind as can be seen easily by considering a qubit and setting $M=Z$ (the so-called computational basis measurement) and $U=H$ (the Hadamard gate). Measurement of $M$ leaves the qubit either in $|0rangle$ or in $|1rangle$ state. Whereas measurement of $UMU^dagger = HZH = X$ leaves the qubit either in $|+rangle$ or in $|-rangle$ state.

---

Let us derive the weak equivalence in the case of projective measurement. This type of measurement is associated with a Hermitian operator

$$ M = sum_{m} lambda_m P_m $$

where $m$ ranges over eigenvalues $lambda_m$ of $M$ which are the possible measurement outcomes and $P_m$ is the projector on the eigenspace of $M$ associated with eigenvalue $lambda_m$.

Probability $p(m|M,psi)$ of obtaining outcome $lambda_m$ when measuring $M$ on state $|psirangle$ is by the Born rule

$$ p(m|M,psi) = langlepsi|P_m|psirangle. $$

In order to verify the claim that application of $U$ followed by measurement of $M'=UMU^dagger$ is equivalent - in the weaker sense - to measurement of $M$ on the original state we need to check that both yield the same outcomes $lambda_m$ and that the probabilities associated with each outcome are the same.

First, note that $M$ and $UMU^dagger$ are similar so they have the same eigenvalues and so the set of possible outcomes of these two measurements are the same.

Second, suppose we begin in state $|psirangle$, apply $U$ to get $|psi'rangle = U|psirangle$ and measure $M'=UMU^dagger$. Note that

$$ M' = UMU^dagger = U left(sum_{m} lambda_m P_mright)U^dagger = sum_{m} lambda_m UP_mU^dagger = sum_{m} lambda_m P_m' $$

where we defined $P_m' = UP_mU^dagger$. The probability of measuring $lambda_m$ is

$$ p(m|M', psi') = langlepsi'|P_m'|psi'rangle = langlepsi|U^dagger UP_mU^dagger U|psirangle = langlepsi|P_m|psi rangle = p(m|M, psi). $$

Therefore the two measurements are equivalent in the weaker sense.

---

Now, let us see that they are not equivalent in the stronger sense. Post-measurement state when outcome $lambda_m$ was obtained when measuring $M$ on the state $|psirangle$ is

$$ |psi_{m,M}rangle = frac{P_m|psirangle}{sqrt{p(m)}}. $$

Suppose now that we begin in state $|psirangle$, apply $U$ to get $|psi'rangle = U|psirangle$ and measure $M'=UMU^dagger$ obtaining $lambda_m$. This time the post-measurement state is

$$ |psi_{m,M'}'rangle = frac{P_m'|psi'rangle}{sqrt{p(m)}} = frac{UP_mU^dagger U|psirangle}{sqrt{p(m)}} = U|psi_{m,M}rangle. $$

Clearly this is not in general the same state as $|psi_{m,M}rangle$.

Note that strong version of equivalence is in fact operator equality. This justifies the omission of the adjective "weak".