# How to find $\langle Z_1...Z_{N-1} \rangle$ knowing $\langle Z_1...Z_{N} \rangle$ and $\langle Z_i \rangle$

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 $$\langle \psi|Z_1...Z_{N-1} |\psi\rangle$$?

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 $$x$$.

• Thank you very much! Jan 5 at 13:50

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.

• 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! Jan 5 at 14:00