Past exam of the short options paper of the physics Prelims of the University of Oxford / 2022 / S03 (Quantum ideas)

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 $\lambda$.

A Stern-Gerlach experiment uses a nonuniform magnetic field to exert force $F_z\simeq {\mu }_z{\partial }_{z}B$ on a magnetic dipole. The two resolved beams reveal discrete projections of spin, rather than a continuum of classical dipole orientations. For the ground-state silver chemical atom, filled shells contribute no net angular momentum and the outer $5s$ electron has orbital angular momentum zero, leaving an effective spin-$1/2$ magnetic moment. This provides evidence for the electron's two-valued intrinsic spin.

Neutral silver chemical atoms can pass through the apparatus without the large charge-dependent Lorentz force that would deflect and broaden a free-electron beam. Neutrality isolates the small magnetic-gradient force far more cleanly. The later parts use an ideal spin filter model; a literal free-electron apparatus requires additional control of charge forces.

A vertical Stern-Gerlach spin measurement is a $z$-spin measurement. The Born rule givesThus this measurement cannot distinguish the two preparations, even by repeating the same measurement on many identically prepared copies: it discards their relative-phase difference in this basis.

A horizontal Stern-Gerlach spin measurement measures $S_x$. The inputs are its orthogonal eigenstates, soThe output beam identifies the input with certainty in a single ideal measurement.

For an axis halfway from $+z$ toward $+x$, let $|n+⟩=\cos (\pi /8)|\uparrow ⟩+\sin (\pi /8)|\downarrow ⟩$. The Born rule givesThe negative-outcome probabilities are complementary. Both outcomes can occur for both inputs, so one result cannot distinguish them perfectly. Repeated measurements on independently prepared copies distinguish their different statistics with arbitrarily high confidence. Reversing the diagonal direction exchanges the two numbers. This separates single-copy certainty from statistical distinguishability.

For a normalized quantum state, $|\alpha {|}^2+|\beta {|}^2=1$, and the Born rule gives $\boxed{p(z+)=|\alpha {|}^2,p(z-)=|\beta {|}^2}$. If the displayed coefficients are not normalized, divide both by $|\alpha {|}^2+|\beta {|}^2$. A quantum superposition is a coherent linear combination: relative phase affects other spin measurements, unlike an incoherent statistical mixture with the same $z$-measurement probabilities.

Every pure qubit quantum state is an eigenstate of spin along some axis. Its Bloch vector isFor real normalized $\alpha ,\beta$, choose $\theta$ modulo $2\pi$ with $\alpha =\cos (\theta /2)$ and $\beta =\sin (\theta /2)$, allowing an irrelevant global sign. ThenEquivalently, $S_{\mathbf{n}}=(\hslash /2)\left(\begin{array}{cc}{\alpha }^2-{\beta }^2& 2\alpha \beta \\ 2\alpha \beta & {\beta }^2-{\alpha }^2\end{array}\right)$. A Stern-Gerlach spin measurement on this axis yields its positive result with certainty, so the state is not a nontrivial quantum superposition in that measurement basis.

Send a beam through a vertical Stern-Gerlach spin measurement apparatus, spatially resolve its two spin components, and block the negative-$z$ beam. The transmitted particles are prepared in $\boxed{|\uparrow ⟩}$. For an initially unpolarized beam, half the intensity survives ideally. This is conditional quantum state preparation by a projective measurement and filtering.

The first Stern-Gerlach spin measurement transmits $x+$ with probability $|⟨x+|\uparrow ⟩{|}^2=1/2$. The selected quantum state $|x+⟩$ then yields $z-$ with conditional probability $|⟨\downarrow |x+⟩{|}^2=1/2$. Relative to the original beam,Among particles reaching the second apparatus, the negative-$z$ fraction is $1/2$; the requested fraction uses the original intensity and is $1/4$.

Let ${\theta }_j=j\pi /N$ and retain the positive spin beam at every stage. The overlap of successive positive eigenstates isThus the unconditional survival probability, relative to the original intensity, isThe last retained quantum state is $|{\psi }_{\pi }^{+}⟩=|\downarrow ⟩$, so its final negative-$z$ result is certain conditional on survival and adds no further factor. In particular $p_2=1/4$, $p_{10}\simeq 0.7805$, and $p_{100}\simeq 0.9756$. Since $\log \cos x=-x^2/2+O(x^4)$,This is a changing-measurement-axis form of the quantum Zeno effect: frequent selections steer the surviving spin through a continuous succession of nearby eigenstates.

A pure bipartite quantum state is entangled if it cannot be written as a tensor product $|a⟩\otimes |b⟩$. More generally a mixed quantum state is entangled if its density matrix cannot be expressed as a convex mixture of product density matrices. Two applications are quantum teleportation, which transfers an unknown qubit using a shared entangled pair plus classical communication, and superdense coding, which uses a shared entangled pair and transmission of one qubit to communicate two classical bits. Neither permits faster-than-light communication.

Measuring the first qubit applies $P_0=|0⟩⟨0|\otimes I$ or $P_1=|1⟩⟨1|\otimes I$. The Born rule givesThe normalized conditional quantum states areEach expression applies only when its denominator is nonzero; an impossible outcome has no defined conditional state.

When $p_0,p_1>0$, the observed relative frequencies converge, for large sample size, to the conditional probabilitiesEquality says that the second binary outcome has the same distribution whichever first outcome occurred; this is independence of these two particular measurement results. Cross-multiplication givesFor finite samples the two empirical fractions fluctuate; they are estimates, not exact deterministic equalities. If $p_0=0$ or $p_1=0$, the missing conditional fraction is undefined, but one outcome is constant, so the joint measurement results are independent and the displayed modulus identity still holds.

This criterion only addresses the computational-basis statistics. For examplehas all four joint probabilities equal to $1/4$, yet is maximally entangled: its first-qubit density matrix is $I/2$ and $\alpha \delta =-1/4\ne 1/4=\beta \gamma$. Moreover classical correlations of mixed separable states also exist, for example $\frac{1}{2}|00⟩⟨00|+\frac{1}{2}|11⟩⟨11|$. Correlations in a single basis alone neither certify nor exclude quantum entanglement.

Expand a normalized product state asThus $\alpha =ac$, $\beta =ad$, $\gamma =bc$, $\delta =bd$, givingConversely the nonzero coefficient matrix $C=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$ has determinant zero precisely when it has rank one, so the condition also suffices for a pure two-qubit state to factor.

For any fixed pair of local measurement settings with operators $E_i,F_j$, the Born rule factorizes:Hence a product state has independent local results in every pair of local bases. This assumes independent settings and local measurements; shared classical randomization of preparations or a subsequent joint interaction can generate correlations. A separable mixture is not necessarily a single product state.

Alice's outcome is random. Conditional on knowing it, Bob may assign a different quantum state, but without that classical record he sees the same density matrix. For the displayed pure two-qubit state, his reduced density matrix before Alice's measurement isAfter Alice's computational-basis measurement with the outcome unannounced, the weighted mixture $p_0{\rho }_{B|0}+p_1{\rho }_{B|1}$ is exactly this same matrix. More generally, for any local trace-preserving operation with measurement operators $K_a$,The equality follows from the cyclic property of the partial trace for operators acting on Alice's subsystem. Bob's every local outcome probability therefore remains unchanged. This is the no-communication theorem: $\boxed{\text{entanglement alone cannot send information faster than light}}$. Conditional state changes become usable correlations only when the classical outcomes are compared.