= Solution
Represent the <single-photon state> by its two path <probability amplitudes>. The first <Hadamard gate> sends $|0\rangle$ to $(|0\rangle+|1\rangle)/\sqrt2$. With $P(\phi)=\operatorname{diag}(1,e^{i\phi})$ for the lower-arm <phase shifter>, the complete <Mach-Zehnder interferometer> gives
$$
H P(\phi)H|0\rangle
=\frac{1+e^{i\phi}}2|0\rangle+\frac{1-e^{i\phi}}2|1\rangle
=e^{i\phi/2}\left(\cos\frac\phi2|0\rangle-i\sin\frac\phi2|1\rangle\right).
$$
The <global phase> has no effect on detection. By the <Born rule>,
$$
\boxed{P_0(\phi)=\cos^2\frac\phi2,\qquad P_1(\phi)=\sin^2\frac\phi2.}
$$
They sum to one. At $\phi=0$, port 1 is a <dark port>; at $\phi=\pi$, the roles reverse. The result uses coherent addition of the path amplitudes, rather than addition of separate path <probabilities>.
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