 Symmetric z-x-z beam splitter factorization

Since $HZH=X$, unitary conjugation gives $H{e}^{-i\beta Z/2}H=e^{-i\beta X/2}$. Multiplying by the two outer $R_z(\alpha )$ rotations givesThe case $\alpha =\beta =\pi /2$ equals $-iH$. Equal output probabilities for one input do not by themselves identify an optical component with the Hadamard gate; relative wave phases also matter.