= Symmetric z-x-z beam splitter factorization
{title2=$U=R_z(\alpha)H R_z(\beta)H R_z(\alpha)$}
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 gives
$$
U=\begin{pmatrix}e^{-i\alpha}\cos(\beta/2)&-i\sin(\beta/2)\\-i\sin(\beta/2)&e^{i\alpha}\cos(\beta/2)\end{pmatrix}.
$$
The 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.
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