enter image description here

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.

  • 2
    $\begingroup$ In your post, could you please provide some more details on how you've attempted to answer your question? What aspects confuse you? $\endgroup$
    – Jacob
    Jun 9, 2022 at 15:05
  • $\begingroup$ 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. $\endgroup$ Jun 10, 2022 at 10:55
  • $\begingroup$ 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. $\endgroup$ Jun 10, 2022 at 11:20
  • $\begingroup$ @DanieleCuomo please add any relevant detail by editing the post, not in comments $\endgroup$
    – glS
    Jun 10, 2022 at 14:21
  • $\begingroup$ Done; hoping now my real doubts comes out! $\endgroup$ Jun 10, 2022 at 14:37

1 Answer 1


These are indeed correct. See this paper as well starting at Eq. (31).

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$^{(0,1)}$

  • 1
    $\begingroup$ Perhaps it was just a problem of nomenclature. That gate to me is the $CZ$ gate, not the gate $Z\otimes Z$. $\endgroup$ Jun 10, 2022 at 11:12

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.