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

I read a paper talking about Bell basis measurement circuit that simultaneously measures $$XX, YY,$$ and $$ZZ$$.

It said: after applying a quantum gate $$U$$, a target measurement of $$M$$ on the original state has become equivalent to a measurement of $$UMU^{-1}$$. I was wondering how comes the $$UMU^{-1}$$. I think the measurement of $$M$$ should be something like $$\langle\phi|M|\phi\rangle$$.

• measuring $M$ after evolving a state $|\psi\rangle$ through $U$ means measuring $M$ on the state $U|\psi\rangle$, i.e. $\langle \psi|U^\dagger M U |\psi\rangle$. I don't understand the connection between this and the second question
– glS
Dec 8, 2020 at 21:28
• How does $UMU^{-1}$ become $U^{\dagger}MU$? Actually, the second question is not related to the first one. The second question is about the Bell basis simultaneous measurement. Dec 9, 2020 at 8:39
• posts should ask about a single, laser-focused question. You can ask separate questions on separate posts. Reading the sentence in the paper, I think they mean that measuring $M$ on $|\psi\rangle$, which gives $\langle \psi|M|\psi\rangle$, is equivalent to measuring $UMU^\dagger$ on $U|\psi\rangle$, which gives you $\langle \psi|U^\dagger UMU^\dagger U|\psi\rangle=\langle\psi|M|\psi\rangle$
– glS
Dec 9, 2020 at 8:46
• I see! Thanks! I will post another question for the second one. Dec 9, 2020 at 10:40
• You are quoting the expectation values as a measurement. This is not consistent with QM. Measurement will yield a specific eigenvalue of the operator and the state will be the corresponding eigenstate.
– user196418
Dec 11, 2020 at 12:58

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 $$|0\rangle$$ or in $$|1\rangle$$ 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 $$|\psi\rangle$$ is by the Born rule

$$p(m|M,\psi) = \langle\psi|P_m|\psi\rangle.$$

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 $$|\psi\rangle$$, apply $$U$$ to get $$|\psi'\rangle = U|\psi\rangle$$ and measure $$M'=UMU^\dagger$$. Note that

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

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

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

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 $$|\psi\rangle$$ is

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

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

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

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".

• This answers measurement equivalence part. Left the part about $XX, YY, ZZ$ to the other question to be posted (see scope discussion under the question). Dec 20, 2020 at 23:54