While working on circuit construction for Hamiltonian simulation using this answer as reference, I'm unable to see how the following equation is true:

$$ e^{i \sigma_z \otimes \sigma_z t} = \text{CNOT}(I \otimes e^{i \sigma_zt})\text{CNOT} $$

I tried doing the following to the right side:

$$ \begin{align} \text{CNOT}(I \otimes e^{i \sigma_zt})\text{CNOT} &= \text{CNOT}\big[ \cos(t) I\otimes I + i \sin(t) I \otimes \sigma_z \big]\big[|0\rangle \langle 0 | \otimes I + |1\rangle \langle 1 | \otimes \sigma_x \big] \\ &=\text{CNOT}\big[ \cos(t)|0\rangle\langle0|\otimes I+\cos(t)|1\rangle\langle1| \otimes \sigma_x + i\sin(t)|0\rangle\langle0|\otimes\sigma_z+ i\sin(t)|1\rangle\langle1|\otimes\sigma_z\sigma_x \big] \\ &= \cos(t)|0\rangle\langle0|\otimes I + \cos(t)|1\rangle\langle1|\otimes I + i\sin(t)|0\rangle\langle0|\otimes\sigma_z+i\sin(t)|1\rangle\langle1|\otimes\sigma_x\sigma_z\sigma_x \\ &= \cos(t) I+i \sin(t)\big[|0\rangle\langle0|\otimes\sigma_z-|1\rangle\langle1|\otimes\sigma_z\big] \qquad (\text{since }\sigma_x\sigma_z\sigma_x=-\sigma_z) \end{align} $$

And we already know that the left hand side is:

$$ e^{i \sigma_z \otimes \sigma_z t}=\cos(t) I + i \sin(t) \sigma_z \otimes \sigma_z $$

So the only thing I'm missing is showing that $|0\rangle\langle0|\otimes\sigma_z-|1\rangle\langle1|\otimes\sigma_z = \sigma_z\otimes\sigma_z$. Using NumPy I was able to see that effectively both are equal to $\text{diag}\{1, -1, -1, 1\}$.

However, I was wondering if anyone knows a nicer way of showing these two are equal without actually calculating their matrix?

  • $\begingroup$ I'll post the general mechanism behind this equation in addition to the correct answer: If you conjugate a Pauli rotation $e^{i Pt}$ by a Clifford unitary $U$ you will get another Pauli rotation $e^{iQt}$ with $Q=UPU^\dagger$. That is straightforward to prove using the Euler identity. Since $\mathrm{CNOT}( I \otimes Z )\mathrm{CNOT} = Z\otimes Z$, you get the desired equation. $\endgroup$ Sep 21, 2021 at 7:20

1 Answer 1


If your question is only regarding why $| 0 \rangle \langle0 | \otimes \sigma_z - | 1 \rangle \langle 1 | \otimes \sigma_z$ ; you can simply factor it given that trivially: $\sigma_z = | 0 \rangle \langle0 | - | 1 \rangle \langle 1 | $

$$| 0 \rangle \langle0 | \otimes \sigma_z - | 1 \rangle \langle 1 | \otimes \sigma_z$$ $$=\big( | 0 \rangle \langle0 | - | 1 \rangle \langle 1 | \big) \otimes \sigma_z$$ $$=\sigma_z \otimes \sigma_z$$

If you are however looking for a different way to show the equivalence between the operators without using Euler's Identity and whatnot, you can simply write both sides in terms of its matrix elements; since both exponentiations are of diagonal matrices (in the computational basis) their exponentiation is trivial, and really all you have to do is perform 2 simple matrix multiplications.

  • 1
    $\begingroup$ Thanks for the headups, I have no idea how I missed this simplification $\endgroup$
    – epelaaez
    Sep 20, 2021 at 17:43
  • 2
    $\begingroup$ Note, Euler's identity does not immediately apply to operators of the form $\exp(i A\otimes B)$, but fortunately it works in this particular case as in the quoted answer because $(A\otimes B)^2=\mathbb{I}$. $\endgroup$ Sep 20, 2021 at 19:06

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.