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.


1 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.