= Solution
Without the upper-arm rotation, the <photon polarization> still labels the path. Before BS2 the <quantum state> is
$$
\alpha|\ell\rangle|H\rangle+i\beta e^{i\theta}|u\rangle|V\rangle.
$$
After the balanced <beam splitter>, the polarization-valued output amplitudes are
$$
|\chi_1\rangle=\frac{i}{\sqrt2}\bigl(\alpha|H\rangle+\beta e^{i\theta}|V\rangle\bigr),\qquad |\chi_0\rangle=\frac1{\sqrt2}\bigl(\alpha|H\rangle-\beta e^{i\theta}|V\rangle\bigr).
$$
The detectors do not resolve <photon polarization>. Since $\langle H|V\rangle=0$, the cross terms vanish in the <Born rule>:
$$
\boxed{P(D_1)=\langle\chi_1|\chi_1\rangle=\frac{|\alpha|^2+|\beta|^2}{2}=\frac12,\qquad P(D_0)=\frac12.}
$$
These <probabilities> are independent of the tunable <quantum phase> and the input <photon polarization>, so there is no usable <wave interference> fringe. Equivalently, taking the <partial trace> over <photon polarization> removes the <quantum coherence> between the two path <quantum states>. The <which-path information> is retained in orthogonal <photon polarizations> even without anybody measuring it. Rotating the two occupied <photon polarization> <quantum states> to the same <quantum state> removes that distinction and allows the cross term in part (ii).
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