# How does this expression from the VQLS paper result in a CZ-gate?

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).

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?

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.

• @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? Jun 24 '21 at 17:25
• @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. Jun 24 '21 at 22:01
• 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 Jun 25 '21 at 9:32