Applying Clifford Gate before a Pauli Measurment

Question

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.

Answer 1

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.