# Confusion with the number of CNOTs in a circuit

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):

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

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):

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).

• I don't quite understand the question. The middle part of Figure (a) here seems to be implementing the operation $I\otimes e^{i(Z\otimes Z\otimes Z)\theta_2}$.
– glS
Jan 3 at 8:57
• 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. Jan 3 at 11:18

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 $$Z$$.