Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S3 / 3

For a fixed propagation direction and spatial mode, a photon has two independent transverse polarization amplitudes. Orthogonal horizontal and vertical states span its two-dimensional Hilbert space: $\alpha |H⟩+\beta |V⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Polarization analyzers measure bases in this space, and phase plates implement transformations of its relative amplitudes. Additional spatial or frequency modes enlarge the space and must not be silently discarded.

Choose the upper route for horizontal polarization, the lower route for vertical polarization, and absorb fixed optical phases into the path kets. The normalized output isThis is entanglement between two degrees of freedom of one photon. Interchanging the path convention interchanges the labels but not any probabilities.

Orthogonal path projectors give an upper-path probability $1/2$. A successful non-destructive lower-path detection projects onto $|V⟩|{\downarrow }_1⟩$ and its conditional polarization is therefore pure vertical. Retaining the detection record removes the upper amplitude for that run; it would be incorrect to continue evolving the original coherent two-path state as though the detection had not occurred.

A quarter-wave plate needs a specified axis angle. If its fast axis makes angle $\eta$ to horizontal, choose its Jones matrix $Q=R(\eta )\mathrm{diag}(1,i)R(-\eta )$. ThenAfter the preceding successful lower-path detection the total state is $(aH+bV)|{\downarrow }_1⟩$. The next crystal sends it toFor axes at $45^{\circ }$, the state before splitting is circular up to phase, $a=(1-i)/2$, $b=(1+i)/2$, and the probability is $1/2$. Detection in ${\downarrow }_2$ always leaves vertical polarization. If no earlier lower-path measurement was actually performed, retain the untouched upper amplitude: the final state is $(|H,{\uparrow }_1⟩+a|H,{\uparrow }_2⟩+b|V,{\downarrow }_2⟩)/\sqrt{2}$, and the last probability is $|b{|}^2/2$, or $1/4$ at $45^{\circ }$. At axis angle zero the conditional probability is instead one, a concrete demonstration that an unspecified orientation does not determine a unique number.