# Propagation rules for the cross-resonance gate of IBM ($R_{zx}$)

The $$R_{zx}$$ is the fundamental two-qubit gate supported by IBM processors. I'd like to see how Pauli operators propagate over such a gate.

Other well-known cases are shown in the picture below.

For all unitaries $$U$$ and $$V$$, we have $$UV=V'U\iff V'=UVU^\dagger.\tag1$$ Thus, in order to find $$V'$$ into which $$V$$ turns as we propagate $$U$$ over it all we need to do is conjugate $$V$$ by $$U$$. For example, calculation shows that $$\text{CNOT}\circ(X\otimes I)\circ\text{CNOT}^\dagger = X\otimes X\tag2$$ in agreement with the first circuit equality.
In general $$R_{ZX}(\theta)=\exp\left(-i\frac{\theta}{2}Z\otimes X\right)$$ is an infinite gate family, so almost all of them are non-Cliffords. Consequently, $$V'$$ won't generally be a Pauli operator even if $$V$$ is.
However, if $$\theta\in\{0,\frac{\pi}{2},\pi\}$$ then $$R_{ZX}(\theta)$$ is Clifford. The cases $$\theta=0$$ and $$\theta=\pi$$ are trivial. If $$\theta=\frac{\pi}{2}$$ there are two possibilities. If Pauli operator $$P$$ commutes with $$Z\otimes X$$ then \begin{align} P'&=R_{ZX}\left(\frac{\pi}{2}\right)P R_{ZX}\left(\frac{\pi}{2}\right)^\dagger\\ &=PR_{ZX}\left(\frac{\pi}{2}\right)R_{ZX}\left(\frac{\pi}{2}\right)^\dagger=P.\tag3 \end{align} Otherwise, \begin{align} P'&=R_{ZX}\left(\frac{\pi}{2}\right)P R_{ZX}\left(\frac{\pi}{2}\right)^\dagger\\ &=PR_{ZX}\left(\frac{\pi}{2}\right)^\dagger R_{ZX}\left(\frac{\pi}{2}\right)^\dagger\equiv P\circ(Z\otimes X)\tag4 \end{align} where $$\equiv$$ denotes equality up to unobservable global phase.
• Yes, sorry about that, I implicitly refered to the maximally entangling case: $\theta = \pi/2$. I edit the question. Commented Jan 9, 2023 at 18:56
• As a rule, you shouldn't edit questions in a way that changes their meaning after an answer is posted. In this case, your original question is actually more general and includes the edit as a simple special case, so I see no benefit to the edit and took the liberty to roll it back. Hope you don't mind. The present question is genuinely better. I also edited my answer to discuss the case $\theta=\frac{\pi}{2}$ explicitly. Commented Jan 9, 2023 at 19:52