Past exam of the physics A course of the University of Cambridge / 2011 / Paper 2 / Section C

identical particles have the same intrinsic properties. Exchanging their bookkeeping labels cannot distinguish a physical preparation or change a measurable prediction. This is stronger than merely failing to track classical trajectories. For ordinary bosons and fermions, the many-particle wavefunction has respectively bosonic exchange symmetry or fermionic exchange symmetry under every particle exchange operator, acting on spatial and spin coordinates together. The Spin-statistics theorem connects integer spin to bosons and half-integer spin to fermions; this connection uses relativistic quantum-field assumptions, rather than following from the nonrelativistic Schrödinger equation alone.

For one-particle states $u,v$ satisfying orthonormality, the two alternatives form a quantum superposition:Its probability density contains an exchange interference term, absent from an incoherent mixture of labelled alternatives. The antisymmetric expression vanishes when both one-particle states are identical: one complete one-particle state can contain at most one fermion. This Pauli exclusion principle explains electron shell filling, the electronic Fermi sea and degeneracy pressure. Opposite spins are different one-particle states, so two electrons may share an atomic orbital. Spatially separated wave packets with negligible overlap can often be treated as distinguishable for local measurements; the underlying exchange symmetry remains.

In thermal equilibrium, occupation of a level of energy $ϵ$ obeysbosons can accumulate macroscopically in one state, giving Bose-Einstein condensation; fermions instead fill successive states up to the Fermi energy at zero temperature. When occupation is dilute, both expressions reduce to the Maxwell-Boltzmann distribution. The identical-particle classical partition function contains $1/N!$, avoiding spurious entropy from permutations and the Gibbs paradox. Exchange effects become conspicuous when the thermal de Broglie wavelength is comparable to the interparticle spacing.

An operational test is the Hong-Ou-Mandel effect. Down-converted photon pairs enter separate ports of a balanced beam splitter; indistinguishable alternatives for one photon per output cancel. Scanning the relative arrival delay restores coincidences away from overlap. Matching polarization, spectral filtering and spatial alignment controls distinguishability; measuring delayed accidental coincidences controls background. The observed dip tests two-photon interference, and its finite depth must not automatically be interpreted as a failure of bosonic symmetry. The original apparatus varied beam-splitter position and subtracted a separately measured accidental background.^[ref]

For observables $A,B$, nonzero $[A,B]$ obstructs a complete common eigenbasis. In a finite-dimensional system, commuting Hermitian operators can be simultaneously diagonalized; quantum degeneracy may require choosing an appropriate basis inside the degenerate eigenspace. Noncommuting observables can still share particular eigenvectors, so incompatibility is not a claim that every individual state must have nonzero spreads in both quantities.

The Robertson uncertainty relation follows by applying Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$:In particular $[x,p_x]=i\hslash$ gives $\Delta x\Delta p_x\ge \hslash /2$. These are preparation spreads, not necessarily measurement errors. A zero commutator expectation can make this bound zero even when the linear operators do not commute.

Measurement order can matter: preparing $|+z⟩$, measuring $S_x$, then measuring $S_z$ gives both z outcomes with probability $1/2$ when the intermediate result is ignored. A direct z measurement would give $+\hslash /2$ certainly. Finally the Heisenberg equation of motion $dA/dt=(i/\hslash )[H,A]+\partial A/\partial t$ relates commutation to conservation: an observable with no explicit time dependence and $[H,A]=0$ is conserved. For unbounded self-adjoint operators these statements require the appropriate common domains; a formal commutator calculation alone need not justify simultaneous spectral measurements.

In a classically forbidden region, the Schrödinger equation has evanescent solutions rather than forcing the wavefunction to vanish. Matching at both sides of a finite barrier permits nonzero transmission. For a broad barrier the WKB approximation giveswhere $C$ is a matching prefactor. Transmission is exponentially sensitive to barrier width and height. A single decaying exponential function carries no probability current; the matched scattering solution includes the combination needed for transmitted current. energy is conserved throughout.

In alpha decay, a quasibound alpha particle penetrates the nuclear Coulomb barrier. Multiplying the penetration probability by an assault frequency estimates the decay rate; the strong dependence on alpha energy explains the wide range of lifetimes. In a scanning tunnelling microscope, a voltage drives electron quantum tunnelling between a conducting tip and sample. Approximately $I\propto V{e}^{-2\kappa d}$, so constant-current feedback resolves changes in gap width, with electronic density of states also affecting the image. Piezoelectric motion, vibration isolation and stable voltage are necessary; a bright feature is not automatically a taller atom. Atomic-scale imaging was demonstrated with a vacuum-tunnelling microscope in the original surface study.^[ref] In a Josephson junction, Cooper pair quantum tunnelling through a weak link gives $I=I_c\sin \phi$, allowing supercurrent without an applied voltage and phase-sensitive interference. These are distinct applications of barrier penetration and coupling with quantum coherence, not particles borrowing energy to climb over a barrier.

Addition of two spin-one-half angular momenta gives total spin $S=0$ or $S=1$. The normalized singlet state and triplet states areThey have $S^2$ eigenvalues $0$ and $2{\hslash }^2$, and $S_z=M\hslash$. The singlet state is antisymmetric under a particle exchange operator, whereas all three triplets have symmetric spin-exchange parity. For two identical electrons the full state must be antisymmetric: a singlet state pairs with a spatial state with symmetric exchange parity, a triplet state with an antisymmetric spatial state. Thus two electrons in the same atomic orbital must form a singlet state.

The identity ${\mathbf{S}}_1\cdot {\mathbf{S}}_2=(S^2-S_1^2-S_2^2)/2$ gives $-3{\hslash }^2/4$ in the singlet state and $+{\hslash }^2/4$ in the triplet state. Hence an exchange interaction distinguishes their energies, with ordering determined by its coupling sign. The singlet state has perfect opposite-spin correlations along any common measurement axis and is entangled. The $M=0$ triplet state is also entangled, while the $M=\pm 1$ triplet states are product states along the chosen axis.