Suppose I have a state $$\vert\psi\rangle = a\left(\frac{1}{\sqrt{2}}\vert\Phi^+\rangle + \frac{1}{\sqrt{2}}\vert\Phi^-\rangle\right) + b\left(\frac{1}{\sqrt{2}}\vert\Psi^+\rangle + \frac{1}{\sqrt{2}}\vert\Psi^-\rangle\right),$$

where I have used the Bell basis states. Suppose I measure the operator $XX$ and obtain $+1$. Is my post measurement state now

$$a\vert\Phi^+\rangle + b\vert\Psi^+\rangle$$

i.e. does such a measurement preserve coherence? Similarly for the $-1$ outcome, is the state

$$a\vert\Phi^-\rangle + b\vert\Psi^-\rangle$$

From the answer, it seems like it depends on how the measurement is implemented. How should one implement the measurement to obtain this result (preserving the coherence)?


1 Answer 1


TLDR: yes.

That might depend on exactly how you implement the measurement, but if you do it properly, using a non-destructive measurement, then yes. Projective measurements project onto their eigenspaces. If that space is not rank 1, then the superposition within the subspace is preserved.

The circuit that you would use is: enter image description here

This is just the standard circuit for measuring an observable $\sigma=\sigma_1\otimes\sigma_2\otimes\ldots$ where $\sigma^2=I$. enter image description here

  • $\begingroup$ Could you add what the Kraus operators of my measurement should be (i.e. how to implement the measurement) to obtain this result? This part is unclear to me. $\endgroup$
    – Vladimir
    Commented May 13 at 15:10
  • $\begingroup$ Well, maybe add that to your question ;) $\endgroup$
    – DaftWullie
    Commented May 13 at 15:12
  • $\begingroup$ Sorry for that, yes I have added it now. $\endgroup$
    – Vladimir
    Commented May 13 at 15:15

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