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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?

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  • $\begingroup$ 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 at 7:20
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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.

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    $\begingroup$ Thanks for the headups, I have no idea how I missed this simplification $\endgroup$
    – epelaaez
    Sep 20 at 17:43
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    $\begingroup$ 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 at 19:06

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