Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 58 / 1

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.

Write the unknown pure photon polarization qubit asThe polarizing beam splitter sends the $H$ component down the lower path and the $V$ component up the upper path. After rotating the upper photon polarization to $H$, both paths have the same internal quantum state and can interfere at BS2. Absorb fixed apparatus phases into a calibrated zero for the tunable quantum phase. With the preceding reflection convention, the quantum state before BS2 is $(\alpha |\ell ⟩+i\beta e^{i\theta }|u⟩)|H⟩$, so the two output amplitudes areConsequentlyRepeated counts while scanning $\theta$ determine the complex quantum coherence $\gamma$. The interference visibility is $2|\gamma |$, and the fringe position determines the relative quantum phase $\arg \beta -\arg \alpha$ whenever both components are nonzero.

A fringe alone is not enough to determine an arbitrary photon polarization quantum state: it fixes $|\alpha \beta |$ but not which of $|\alpha {|}^2$ and $|\beta {|}^2$ is larger. Indeed, $|H⟩$ and $|V⟩$ both give the flat fringe $P_1=1/2$. For a general pure quantum state, the two possible population differences areResolve this ambiguity by temporarily detecting the two paths before BS2. Their relative counts give $h=|\alpha {|}^2$ and $1-h=|\beta {|}^2$ using the same polarizing beam splitter and detectors. Taking $\alpha =\sqrt{h}$ as a global-phase choice, the reconstructed quantum state for $h>0$ isIf $h=0$, it is $|V⟩$ up to a global phase. This completes interferometric polarization tomography. The counts are obtained from many identically prepared photons; they do not identify an arbitrary quantum state from one photon.

A full fringe scan is unnecessary. Use the same components to make three fixed projective measurements, on separate batches of identically prepared photons. Direct path detection after the polarizing beam splitter measures the $H/V$ populations. Recombine after matching the arm photon polarizations, once with $\theta =0$ and once with $\theta =\pi /2$, to measure the two quadratures of the quantum coherence.

In the $H/V$ basis, write the density matrix in terms of its Bloch vector as $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$. The previous formula becomesThus three measured relative frequencies giveand determineThe two interferometric settings analyze the diagonal linear polarizations $(|H⟩\pm |V⟩)/\sqrt{2}$ and the circular polarizations $(|H⟩\pm i|V⟩)/\sqrt{2}$, with detector labels fixed by the calibrated quantum phase convention. This is informationally complete three-observable qubit tomography: direct measurements in three bases replace fitting an entire phase-dependent fringe. It also reconstructs a mixed input, for which $|r|\le 1$, although the source here is assumed pure.

Simply measuring transmission through one fixed polarizing beam splitter would determine only $r_z$. Rotating a linear analyzer through different angles can additionally determine $r_x$, but cannot distinguish the two signs of $r_y$ without a relative-phase measurement. For arbitrary, possibly elliptical photon polarization, all three settings above are needed; no extra analyzer element beyond the described path-phase control and photon polarization rotator is required.

Without the upper-arm rotation, the photon polarization still labels the path. Before BS2 the quantum state isAfter the balanced beam splitter, the polarization-valued output amplitudes areThe detectors do not resolve photon polarization. Since $⟨H|V⟩=0$, the cross terms vanish in the Born rule: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).