I have read an article Unified derivations of measurement-based schemes for quantum computation. In page 3, Section A it says that The classical outcome j corresponds to the measurement of $(U^\dagger X U)\otimes (V^\dagger ZV)$, because $(H \otimes I)\Lambda(Z)(U \otimes V)$ maps the $\pm 1$ eigenspace of $(U^\dagger X U)\otimes (V^\dagger ZV)$ onto the $\pm 1$ eigenspace of $Z \otimes I$. (where $\Lambda(Z)$ denote contolled-Z gate)

My question is:

I think I know that eigenspace of Z is $\{|+ \rangle ,|-\rangle\}$ , but what is eigenspace of $I$?

Why use the controlled-Z gate in the measurement of $(U^\dagger X U)\otimes (V^\dagger ZV)$? What does it do?

  • 5
    $\begingroup$ You could improve your question by defining the various operators you are referring to. And by the way if $I$ is referring to the identity operator then note that $I |\psi \rangle = |\psi\rangle$ holds for all vectors $|\psi\rangle$ and so the identity operator has eigenvalues $\{1\}$ and the corresponding eigenspace is just the whole space. $\endgroup$
    – Rammus
    Aug 9, 2021 at 13:20
  • 4
    $\begingroup$ The eigenspace of $Z = \{|0\rangle,|1\rangle\}$ if $Z$ is meant to be the Pauli-Z operator. $\{|+\rangle,|-\rangle\}$ is the eigenspace of the Pauli-X operator. $\endgroup$ Aug 9, 2021 at 14:36

1 Answer 1


That sentence of the paper is referring to the following quantum circuit. Figure in Eq. (17) of the paper.

(Note that the notation for the CZ gate in the middle is not standard anymore.)

The gates in this circuit are exactly the operation $(H \otimes V^\dagger)\Lambda(Z)(U \otimes V)$. The authors are stating the circuits of this type are going to appear several times in the paper. Then, they note that you can think of this as a measurement of $(U^\dagger X U)\otimes (V^\dagger ZV)$. You don't need a CZ gate to measure this operator; the logic is going in the opposite direction. It's kind of like saying, "this circuit looks like a complicated multi-qubit operation, but it produce the same measurement outcomes as a much simpler operation."

Note that $(Z\otimes I)(H \otimes V^\dagger)\Lambda(Z)(U \otimes V) \neq (U^\dagger X U)\otimes (V^\dagger ZV)$. So, the post-measurement states are not the same. However, they have the same eigenvalues and eigenspaces. The exact equivalent circuit is shown in Eq. (18). However, if all you want to do is verify the claim, you can do the following somewhat tedious calculation.

First note that it is easier to work with projectors than eigenspaces. The projector onto the $+1$ eigenspace of $Z\otimes I$ is $P_+ = |0\rangle\!\langle 0|\otimes I$ and similarly for the $-1$ eigenspace. The projector onto the $+1$ eigenspace of $X\otimes Z$ is $Q_+ = |+\rangle\!\langle +|\otimes |0\rangle\!\langle 0|+ |-\rangle\!\langle -|\otimes |1\rangle\!\langle 1|$. The projector of the $+1$ eigenspace of $(U^\dagger X U)\otimes (V^\dagger ZV)$ is then $(U^\dagger \otimes V^\dagger) Q_+ (U\otimes V)$. Call the circuit above $C$ and all you have to do is verify $C P_+ C^\dagger = (U^\dagger \otimes V^\dagger) Q_+ (U\otimes V)$.

(You can do this with a half-page calculation using the fact that $\Lambda(Z) = P_+ + |1\rangle\!\langle 1|\otimes Z$).

  • $\begingroup$ Thanks very much, but I have another question, What exactly does $M_{Z \otimes Z}$ mean in Eq.(32). How does it measure on two qubits and outcome one classical result $j$. I can't understand what does $M_{Z \otimes Z}$ do. $\endgroup$
    – chris
    Aug 12, 2021 at 13:08
  • $\begingroup$ $M_O$ is a "measurement" of the operator $O$, which means the outcomes are one of the eigenvalues of $O$ with probability given by the Born rule. You can analyze the circuit by considering a separate case for each outcome. Two circuits are equivalent if they give the same outcomes with the same probabilities. $\endgroup$ Aug 13, 2021 at 1:17

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