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?


1 Answer 1


They perform a Hadamard Test:

Hadamard-Test circuit


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


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

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.

  • $\begingroup$ @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, 2021 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, 2021 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, 2021 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.