When are the following equivalences correct?

I can't figure out how the equivalences in the picture hold.

The picture comes from this recent publication on PRA.

EDIT: I think I might have been mislead by the gate represenation. In fact, the gate is more common to express the $$CZ$$ gate and not a $$Z\otimes Z$$ gate. What's more, I'm not really used to Ising operators and I can't see why there can be found implementation (both logical and physical) for a gate which can be expressed with single-qubit gates.

• In your post, could you please provide some more details on how you've attempted to answer your question? What aspects confuse you? Jun 9, 2022 at 15:05
• I checked by using the channel-state duality by using Quirk tool and I get different density matrices. I used the $Z$ gate instead of the $R_z$, as they should be equivalent, up to global phase. Jun 10, 2022 at 10:55
• I think that I misunderstood the kind of operator they represent with the red gate. To me that's a controlled-$Z$ not a $Z\otimes Z$. What's more, I don't get the general importance of such kind of complexity for an operator which admits implementation by only using single qubit gates. Jun 10, 2022 at 11:20
– glS
Jun 10, 2022 at 14:21
• Done; hoping now my real doubts comes out! Jun 10, 2022 at 14:37

Start with the first equivalence: ZZ $$\equiv$$ CNOT$$^{(0,1)}$$ $$R_Z^{(1)}(\theta)$$ CNOT$$^{(0,1)}$$ as given, though you can compute this in terms of exponentiated Pauli operators. From your question, I'm guessing you aren't asking about that though.
To get the 2nd equivalence, we have CNOT$$^{(0,1)}$$ ZZ $$=$$ CNOT$$^{(0,1)}$$ CNOT$$^{(0,1)}$$ $$R_Z^{(1)}(\theta)$$ CNOT$$^{(0,1)}$$. Notice the two CNOT gates next to each other. Those cancel each other out becoming the identity CNOT$$^{(0,1)}$$ CNOT$$^{(0,1)}$$ = $$\mathbb{1}$$. So you get CNOT$$^{(0,1)}$$ ZZ $$\equiv$$ $$R_Z^{(1)}(\theta)$$ CNOT$$^{(0,1)}$$.
The 3rd equivalence is the same trick but we also cancel the $$R_Z$$ gate: $$R_Z^{(1)}(-\theta)$$ CNOT$$^{(0,1)}$$ ZZ $$=$$ $$R_Z^{(1)}(-\theta)$$ CNOT$$^{(0,1)}$$ CNOT$$^{(0,1)}$$ $$R_Z^{(1)}(\theta)$$ CNOT$$^{(0,1)}$$. We reduce the CNOTs just like before: $$R_Z^{(1)}(-\theta)$$ CNOT$$^{(0,1)}$$ ZZ $$=$$ $$R_Z^{(1)}(-\theta)$$ $$R_Z^{(1)}(\theta)$$ CNOT$$^{(0,1)}$$. Now we have two single qubit gates that also cancel each other out $$R_Z^{(1)}(-\theta)$$ $$R_Z^{(1)}(\theta) = \mathbb{1}$$. So we get $$R_Z^{(1)}(-\theta)$$ CNOT$$^{(0,1)}$$ ZZ $$=$$ CNOT$$^{(0,1)}$$
• Perhaps it was just a problem of nomenclature. That gate to me is the $CZ$ gate, not the gate $Z\otimes Z$. Jun 10, 2022 at 11:12