# Applying Clifford Gate before a Pauli Measurment

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.

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.
• 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 Dec 6, 2022 at 13:27
• 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). Dec 6, 2022 at 18:45