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I am reading a paper where they describe different randomised measurement protocols. I am confused about a simple case of this protocol that they discuss.

The link of the paper is here , and I am talking about the subsection "post-processing protocol" under the section "Experimental recipe and post-processing of the measurements"

First, we prepare a state $ \rho $ of $ n $ qubits. Secondly, we apply a local random operation $ U=\otimes_{n=1}^N U_n$ once to these qubits. In the context of my question, I assume that each of the $ U_n $ are sampled from the single-qubit Clifford group. Finally, we perform a projective measurement in the computational basis.

We repeat this protocol $ K $ times.

Here is where I am confused. They claim that repeating this random measurement over and over again is equivalent to measuring a random string of Pauli observables $ \otimes_{n=1}^N U_n^{\dagger} Z U_n =:\otimes_{n=1}^N W_n $, where $ W_n \in \{X,Y,Z\} $ a total of $ K $ times.

I was wondering if anyone could help me see explicitly why this is the case. I think it is related to how we perform a $ Z $-basis measruement at the end, but I wasn't sure.

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Pauli measurements are done by rotating the state into the eigenbasis of the corresponding operator. For example, the eigenbasis of $X$ is $\{ |+\rangle, |-\rangle \}$, and applying the Hadamard gate $H$ performs the aforementioned rotation:

$$ \langle \psi | X | \psi \rangle = \langle \psi | H^\dagger ZH| \psi \rangle = \langle \psi | HZH| \psi \rangle $$

Likewise, $HS^\dagger$ rotates into the $Y$-basis, and $Z$ is already in the computational basis so we don't need any additional rotations.

By the definition of the Clifford group, conjugation by any Clifford operator maps $Z$ to a Pauli operator; consequently, applying a random Clifford gate and measuring $Z$ is equivalent to measuring a random Pauli.

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  • $\begingroup$ Thank you for your clear answer. I am still confused about one thing, however. It says that they perform a measurement in the computational $Z$-basis. So, it is a fixed Pauli measurement that is done. For your example with the $X$ operator for example, you wouldn't need $H$ I would have thought because you want to perform a $Z$-measurement at the end. However, maybe you were using that just to illustrate your point because we wouldn't be applying the $X$ gate since I don't think it is not in the Clifford group $\endgroup$ Dec 6, 2022 at 13:27
  • $\begingroup$ Oh, all they're saying is that they're making a Pauli measurement precisely by adding a rotation followed by a $Z$ measurement. To measure $X$, you apply the $H$ and then measure $Z$ (you don't apply an $X$ gate at any point). $\endgroup$
    – Cody Wang
    Dec 6, 2022 at 18:45

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