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 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 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 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.