 Solution

Use the symmetric lossless beam splitter convention with transmission amplitude $t=1/\sqrt{2}$ and reflection amplitude $r=i/\sqrt{2}$. The two mirrors contribute the same fixed quantum phase, which can be suppressed. At the rightward port $D_1$, the upper-arm route has one reflection at BS1 and one transmission at BS2, while the lower-arm route has one transmission at BS1 and one reflection at BS2. Their probability amplitudes therefore add asBy the Born rule,For the upward port, the two amplitudes are $r^2{e}^{i\theta }$ and $t^2$, giving $a_0=(1-e^{i\theta })/2$ and $P(D_0)={\sin }^2(\theta /2)$. They sum to one, as required for an ideal lossless Mach-Zehnder interferometer. Thus $D_1$ is bright at zero relative quantum phase and dark at quantum phase $\pi$. A fixed unequal reflection quantum phase or a swapped port labeling would shift the fringe; the stated convention fixes the port and quantum phase calibration used below.