I am a bit puzzled on the following circuit. According to this Quantum Computing SE thread it holds that $$ e^{i(Z\otimes Z)t} = {\rm CNOT} (I\otimes e^{iZt}){\rm CNOT} \qquad (1) $$

As a result we have the following circuit (C1):enter image description here

Furthermore, for $e^{iZ\otimes Z\otimes Z t}$ we obtain the following circuit (C2):

enter image description here

I was reading arxiv:2003.13599 and in Figure 3 we see three unitaries corresponding to $I\otimes Z \otimes I$, $I \otimes Z \otimes Z$ and $Z \otimes Z \otimes I$ seen below (C3): enter image description here

Which is very confusing. Specifically, let's focus for example on the middle part of Diagram (a) corresponding to $I\otimes Z \otimes Z$. The $I$ term (in the first qubit) can be ignored. Then, if we ignore the first wire we essentially should have the term $Z\otimes Z$ from Eq. (1) but nevertheless we see two pairs of CNOT gates just like in (C2) that corresponds to $e^{iZ\otimes Z \otimes Z t}$.

Why is this the case? I suspect this is somehow related to the extra wire we ( C2 has three wires but for some reason C3 has four).

  • $\begingroup$ I don't quite understand the question. The middle part of Figure (a) here seems to be implementing the operation $I\otimes e^{i(Z\otimes Z\otimes Z)\theta_2}$. $\endgroup$
    – glS
    Jan 3, 2022 at 8:57
  • $\begingroup$ I agree. But in the same diagram it is claimed this is $e^{i(I\otimes Z \otimes Z)\theta_2}$ that is two $Z$ gates and not 3. $\endgroup$
    – user39726
    Jan 3, 2022 at 11:18

1 Answer 1


I think the key fact you're missing is that $Z_2 \otimes Z_3 \otimes Z_4 = Z_2 \otimes Z_3$ when you know qubit 4 is in the $|0\rangle$ state; in the +1 eigenstate of $Z$.

I'm not sure why that paper is using six CNOT gates instead of four. The ancilla isn't helping. Maybe it's just supposed to be an example for a more general case where it is helpful.

enter image description here

  • $\begingroup$ Yeah, I don't understand this precisely. Why use an ancilla state? 4 CNOTs should be enough. $\endgroup$
    – user39726
    Jan 3, 2022 at 11:20

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