Past exam of the short options paper of the physics Prelims of the University of Oxford / 2022 / S03 / 2

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.