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I am reading the VQLS paper and equation C2 on page 10 they have:

$ \delta_{ll'}^j = \beta_{ll'} + \langle0|V^\dagger A_{l'}^\dagger U(Z_j \otimes I_\bar{j}) U^\dagger A_lV|0\rangle $

Here they define the $I_\bar{j}$ as "the identity on all qubits except qubit $j$" (see the text below equation 7).

They then go on to say that equation C2 can be calculated from the circuit in Fig.9c (reproduced below).

enter image description here I understand how the circuit is made except for the $CZ$ gate. Is it the case that $(Z_j \otimes I_\bar{j}) = CZ$ and if so, can you demonstrate how this is the case?

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They perform a Hadamard Test:

Hadamard-Test circuit

with

$$ \vert \psi \rangle = V(\alpha)\vert 0 \rangle $$

and:

$$ M = A_{l'}^\dagger U (Z_j\otimes I_{\overline{j}}) U^\dagger A_l $$

By replacing the expression of $M$ in the circuit, you end up with the circuit you provided except that the two $U$ are controlled. "Removing" the controls here is a simplification that can be performed because $U$ and $U^\dagger$ simplify to the identity. To convince yourself about this:

  1. If the control qubit is $\vert 0 \rangle$ then the $M$ is not applied. By removing the controls from the $U$ and $U^\dagger$, they are both applied (because they are no more controlled) but they "cancel" each other so the final operation is the identity, as expected.
  2. If the control qubit is $\vert 1 \rangle$ then $U$ and $U^\dagger$ are supposed to be applied too, which is the case.

So the controlled-$Z$ comes from the fact that the whole $M$ gate is controlled, which means that each of the gates that compose $M$ will be controlled too, $Z$ being one of them.

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  • $\begingroup$ @thespaceman, Obviously, I am new to this. Could you give me some intuition why local cost function has Z and can replace 0><0 from the global cost? $\endgroup$ Jun 24 '21 at 17:25
  • $\begingroup$ @JohnParker, The Z gate comes from the identity $|0_j\rangle\langle0_j|=(I_j+Z_j)/2$. This is discussed in the VQLS paper just above equation (C2) in the appendix. Unfortunately, I cannot provide any intuition as to why the local cost function has a Z gate as I am new to the field myself. The only discussion that I have found in this paper is the text surrounding equation (6), but they really just state the Hamiltonian used without explaining it. $\endgroup$ Jun 24 '21 at 22:01
  • $\begingroup$ The talk of Patrick Coles at QHack2021 might be a good start to understand. I'm linking a specific timestamp where he starts talking about global VS local cost, but the whole video is worth looking at: youtu.be/bwmLfxelwUA?t=665 $\endgroup$ Jun 25 '21 at 9:32

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