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I have a quantum circuit with $N$ qubits represented by the unitary $U$. The initial state is $| 00...0\rangle$ and $\psi=U|00...0\rangle$.

Given $\langle\psi| Z_1...Z_{N} |\psi\rangle$ and $\langle \psi|Z_i |\psi\rangle$ $\forall i$, is there a way to find the expectation value $\langle \psi|Z_1...Z_{N-1} |\psi\rangle$?

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Not in general, no.

Consider the state $$ |\psi\rangle=\frac{1}{\sqrt{2}}\left(|0x\rangle+|1\bar x\rangle\right)|+\rangle $$ for $x\in\{0,1\}$. We have $\langle Z_i\rangle=0$, and $\langle Z_1Z_2Z_3\rangle=0$ (to see this most trivially, look at the third qubit). However, the first two qubits are an eigenstate of $Z_1Z_2$ of eigenvalue $(-1)^x$. From the values you have (which do not depend on $x$) you cannot get the result for the two-qubit observable because it depends on $x$.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – stopper
    Jan 5 at 13:50
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2 assumptions for my answer (because I did not really understood your setup):

  • Let's assume a pure state like |0010110> (not superposition)
  • Assume you are doing the same experiment again and again (because in each time you collapse and project your state)

The N Zs operator is actually counting if there is an odd or even number of ones in the state.

Using all Zi, you can also know exactly which of them was 0 and which is 1. (and you don't need the

So sure you can know the (N-1) Zs expectation value. (even without the N Zs measure)

In the case of superposition, the expectation value is probabilistic. So you can't say you know it.

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  • $\begingroup$ Yes I can't say I know it in a superposition state. I supposed that I'm in a perfect classical simulator, so I can actually know every expectation value I need. I was not clear in the question, sorry. Thank you for the reply, @DaftWullie told me what i wanted to know! $\endgroup$
    – stopper
    Jan 5 at 14:00

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