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Write $R_z(\theta )=e^{-i\theta Z/2}$ and $R_x(\theta )=e^{-i\theta X/2}$, using the Pauli operators. The Hadamard gate obeys $HZH=X$, so unitary conjugation of the matrix exponential gives $H{R}_z(\beta )H=R_x(\beta )$. Therefore the symmetric z-x-z beam splitter factorization isThe rightmost operator acts first. Each factor is unitary, which also proves the given matrix is unitary. Choosinggives $U_{\mathrm{BS}}=-iH$, so it is exactly a Hadamard gate up to global phase.

For the specified identical splitters with $\alpha =0$, put $c=\cos (\beta /2)$ and $s=\sin (\beta /2)$. Here $U_{\mathrm{BS}}=R_x(\beta )$, and direct multiplication givesThus the upper-port probability iswith $P_1={\sin }^2\beta {\cos }^2(\varphi /2)$. These identical beam splitters in a two-path interferometer give a nonconstant interference fringe whenever $\sin \beta \ne 0$. With this matrix and port convention, two identical balanced splitters swap the bright and dark ports relative to the two-Hadamard model: at $\varphi =0$, $R_x(\pi /2{)}^2=-iX$. The reflection phases must remain in the calculation.

The upper-port maximum and minimum are $1$ and ${\cos }^2\beta$. Hence the peak-to-peak modulation is ${\sin }^2\beta$ and its normalized interferometric visibility isBoth measures of contrast are maximal precisely forThese are balanced beam splitters, with equal single-input transmission and reflection probabilities. When $\beta \in \pi \mathbb{Z}$, only one route contributes to the output and the fringes disappear.