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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.

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    $\begingroup$ 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

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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$^{(0,1)}$

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    $\begingroup$ 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

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