 Solution

Represent the single-photon state by its two path probability amplitudes. The first Hadamard gate sends $|0⟩$ to $(|0⟩+|1⟩)/\sqrt{2}$. With $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$ for the lower-arm phase shifter, the complete Mach-Zehnder interferometer givesThe global phase has no effect on detection. By the Born rule,They sum to one. At $\varphi =0$, port 1 is a dark port; at $\varphi =\pi$, the roles reverse. The result uses coherent addition of the path amplitudes, rather than addition of separate path probabilities.