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

Question

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.

Answer 1

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.