= Solution
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> is
$$
R_z(\alpha)H R_z(\beta)H R_z(\alpha)
=R_z(\alpha)R_x(\beta)R_z(\alpha)
=\begin{pmatrix}
e^{-i\alpha}\cos(\beta/2)&-i\sin(\beta/2)\\
-i\sin(\beta/2)&e^{i\alpha}\cos(\beta/2)
\end{pmatrix}.
$$
The rightmost operator acts first. Each factor is <unitary>, which also proves the given matrix is unitary. Choosing
$$
\boxed{\alpha=\frac\pi2,\qquad\beta=\frac\pi2}
$$
gives $U_{\rm BS}=-iH$, 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 $U_{\rm BS}=R_x(\beta)$, and direct multiplication gives
$$
U_{\rm BS}P(\phi)U_{\rm BS}|0\rangle
=(c^2-e^{i\phi}s^2)|0\rangle-ics(1+e^{i\phi})|1\rangle.
$$
Thus the upper-port probability is
$$
\boxed{P_0=c^4+s^4-2c^2s^2\cos\phi
=1-\sin^2\beta\cos^2\frac\phi2,}
$$
with $P_1=\sin^2\beta\cos^2(\phi/2)$. These <identical beam splitters in a two-path interferometer> give a nonconstant <interference fringe> whenever $\sin\beta\ne0$. With this matrix and port convention, two identical balanced splitters swap the bright and dark ports relative to the two-Hadamard model: at $\phi=0$, $R_x(\pi/2)^2=-iX$. 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> is
$$
\mathcal V=\frac{P_{\max}-P_{\min}}{P_{\max}+P_{\min}}
=\frac{\sin^2\beta}{1+\cos^2\beta}.
$$
Both measures of contrast are maximal precisely for
$$
\boxed{\beta=\frac\pi2+k\pi,\qquad k\in\mathbb Z.}
$$
These 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.
\Image[/past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2007/5-beam-splitter-fringes.png]
{title=Interferometer fringes with explicit beam-splitter phases and the balance condition for full visibility}
{description=Original ideal-model calculations, not measured data. The first panel compares the Hadamard convention with two identical $R_x$ splitters; the output-port phase shift remains visible even at 50:50 balance. The second panel shows the normalized upper-port visibility.}
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