Past exam of the short options paper of the physics Prelims of the University of Oxford / 2022 / S03 / 1

In the double-slit experiment, each photon is detected as a localized event carrying energy $h\nu$, while repeated events build the interference pattern of coherent probability amplitudes from the two slits. Acquiring information that distinguishes the paths suppresses the interference; splitting the classical intensity into two incoherent alternatives would not reproduce the fringes.

Put one plate before each slit at normal incidence, with the same known refractive index $n$ and illumination of vacuum wavelength $\lambda$. Relative to the plate-free reference, their thickness difference $\Delta d=d_1-d_2$ produces an optical path difference $(n-1)\Delta d$ and a phase shiftObserve the displacement of the full interference fringe pattern, or scan a controllable reference phase and estimate $\Delta \varphi$. This compares the plates directly; a detectable phase displacement establishes unequal thickness under the common-material assumption. For fringe spacing $\lambda D/b$, the displacement is $-D(n-1)\Delta d/b$, where $b$ is slit separation and $D$ the screen distance, with sign set by the slit labeling.

A differential phase measurement can resolve a small difference sensitively and avoid determining two large absolute thicknesses. It does not guarantee a universally faster or more photon-efficient test: noise, coherence, alignment, and required accuracy matter. Moreover a null phase shift at one wavelength only impliesso equality requires a prior bound, a broadband or multiple-wavelength measurement, or continuously counting interference fringes as the plates are introduced. If the refractive indices differ, this apparatus measures optical thickness difference rather than geometrical thickness difference.

Take equal arm lengths, no plate, and coherent illumination. At a coated beam splitter, choose the real-amplitude convention where transmission has amplitude $1/\sqrt{2}$ and reflection has amplitude $+1/\sqrt{2}$ from one side and $-1/\sqrt{2}$ from the other; which reflection carries the minus sign is a basis convention. The two paths interfere constructively at detector 1 and destructively at detector 2 in the diagram's labeling. ThusFor example, writing the beam splitter as $B=2^{-1/2}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ and the two equal-reflection mirrors as $M=-\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, one obtains $BMB(1,0{)}^T=(0,-1{)}^T$. The first output component propagates downwards to detector 2; the nonzero second component propagates upwards to detector 1. The overall minus sign is an unobservable global phase. The redistribution is interference of amplitudes, rather than two successive independent coin tosses.

In the stated ordered propagation basis, a downward input is transmitted downward and reflected upward, giving the first column $(1,1{)}^T/\sqrt{2}$. An upward input is reflected downward with the opposite reflection phase and transmitted upward, giving the second column $(-1,1{)}^T/\sqrt{2}$. ThereforeEqual mirror reflection phases exchange the two directions. Choosing a phase of $\pi$ per reflection givesEach path encounters one mirror, so $M$ describes both arms together, not two successive reflections of each photon. Changing its common phase leaves all detection probabilities unchanged. Direct multiplication gives $B^{†}B=M^{†}M=I$, so both are unitary matrices. They preserve the norm of a quantum state and hence total probability for these lossless optical elements.

Ideal detectors are represented by measurement matrices, but detection itself is not unitary evolution within the two traveling modes. With $e_1=(1,0{)}^T$ and $e_2=(0,1{)}^T$, the projective measurement operators areThey give Born rule probabilities $⟨\psi |P_j|\psi ⟩$ and sum to $I$. An absorbing detector maps the incident photon to an excited detector and vacuum field; a complete physical description includes those additional degrees of freedom.

For a monochromatic photon of vacuum wavelength $\lambda$, a plate of thickness $d$ replaces air by glass and adds phase $\varphi =2\pi (n-1)d/\lambda$. Neglect losses and reflections, or assume antireflection coatings. Between the first beam splitter and mirror 1, the affected arm is the second basis component, so $D_{\varphi }=\mathrm{diag}(1,e^{i\varphi })$. The successive quantum states, omitting common propagation phases, areThe first output component feeds detector 2 and the second feeds detector 1, givingThe imbalance gives $p_1-p_2=\cos \varphi$. Thus the measured probabilities determine only $\varphi$ up to $\varphi \mapsto \pm \varphi +2\pi m$, and hence only a corresponding family of thicknesses. There is no unique absolute thickness from this one measurement. The zero-plate limit recovers $p_1=1$, $p_2=0$.

Adding a known phase $\pi /2$ in the same arm changes the observed phase to $\varphi +\pi /2$. Combining measurements with and without that extra plate determines both quadratures. Explicitly, $p_1-p_2=\cos \varphi$ and $p_1^{(\pi /2)}-p_2^{(\pi /2)}=\cos (\varphi +\pi /2)=-\sin \varphi$. Hence the ordered pair determines $\varphi$ modulo $2\pi$, removing the sign ambiguity of a cosine alone.

It still leaves the integer fringe order undetermined:Thus the thickness remains periodic with period $\lambda /(n-1)$. A known thickness interval shorter than this period, a suitable multiple-wavelength measurement, or continuous phase tracking supplies the missing information. If $n$ is unknown, a single wavelength only determines the excess optical thickness $(n-1)d$ modulo $\lambda$.