I have a few questions about measurement in Bell-state basis. In particular, if $Z = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$ is for a measurement on the computational basis, then what is the representative matrix for a measurement in Bell-state basis.

I know that such a matrix can be constructed using spectral decomposition, but my Professor says the eigenvalues corresponding to 4 Bell-states remain unknown, so basically there is currently no physical quantity that helps on this kind of measurement.

However, Nielsen and Chuang (p.27) give a circuit for teleportation (basically Bell-basis measurement)

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I wondered if $U^\dagger (Z\otimes Z)U$, where $U = (H\otimes I)CNOT$, is the needed matrix. It turns out that its eigenvectors are not Bell states. Can someone explain where I'm wrong here?


1 Answer 1


I think your reasoning is just fine and I checked that the Bell states are indeed eigenvectors of $$M=U^\dagger(Z\otimes Z)U,$$ as $$\begin{align} M|\phi^+\rangle =& \phantom{{}-{}}|\phi^+\rangle,\\ M|\phi^-\rangle =& -|\phi^-\rangle,\\ M|\psi^+\rangle =& -|\psi^+\rangle,\\ \text{and } M|\psi^-\rangle =& \phantom{{}-{}}|\psi^-\rangle \end{align}.$$ I like to see it this way: The matrix that you assign to a measurement depends on what measurement results you assign to a certain outcome state. For the measurement in the computational basis $\{|0\rangle, |1\rangle\}$ you are assigning $1$ to the $|0\rangle$ outcome and $-1$ to the $|1\rangle$ outcome. That's why you say that, using the spectral decomposition, $$1 |0\rangle \langle 0| + (-1) |1\rangle \langle 1| = Z$$ is the corresponding matrix.

Now you can do the same with the Bell basis, e.g. $$1 |\phi^+\rangle\langle\phi^+| + 1 |\psi^+\rangle\langle\psi^+| + (-1) |\phi^-\rangle\langle\phi^-| + (-1) |\psi^-\rangle\langle\psi^-|.$$ This is quite arbitrary, of course. You are free to choose the values that you assign to each outcome. Typically they would be $\pm 1$, but really you can assign any number (in the broadest sense) to it.

  • $\begingroup$ Oh I see I miscalculated the eigenvectors. So, according to this, we can measure Bell-basis measurement with $CNOT$ and $H$. Then why my Prof told that measurement does not exist experimentally in this time. Do you have any idea? $\endgroup$ Commented Jul 8, 2020 at 17:00
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    $\begingroup$ Experiments are a completely different story. Probably your Prof was referring to implementing the measurement on photonic qubits with unit success probability and using only linear optics. This is indeed not possible as far as I know. $\endgroup$
    – M. Stern
    Commented Jul 8, 2020 at 17:03
  • $\begingroup$ +1 This is a good answer, but there is a caveat. You're right that we can choose to assign any eigenvalues to the eigenvectors as we construct the measurement operators. However, the choice of repeated eigenvalues is not great since it fails to keep the eigenspaces distinct. For example, any linear combination of $|\phi^+\rangle$ and $|\psi^+\rangle$ is now also an eigenvector. In fact, the operator is diagonal in the product basis $|{++}\rangle$, $|{+-}\rangle$, $|{-+}\rangle$, $|{--}\rangle$ which probably defeats the intention of it being an entangling measurement! $\endgroup$ Commented Mar 16, 2021 at 21:47

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