Past exam of the physics A course of the University of Cambridge / 2011 / Paper 2 / Section C / C11

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.