Past exam of the physics A course of the University of Cambridge / 2010 / Paper 2 / Section C

For a time-independent Hamiltonian operator, a symmetry operator $U$ satisfies $UH{U}^{†}=H$, or $[U,H]=0$. Thus applying $U$ to an energy eigenstate preserves its energy. Commuting Hermitian observables can be diagonalized within each degenerate energy eigenspace, allowing states to be labelled by symmetry quantum numbers.

A continuous unitary symmetry $U(\epsilon )=e^{-i\epsilon Q/\hslash }$ has Hermitian generator $Q$. Differentiating the invariance condition gives $[H,Q]=0$. The Heisenberg equation of motion then yieldswhen $Q$ has no explicit time dependence. spatial translations connect to momentum conservation, rotations to angular-momentum conservation, and time-translation invariance to energy conservation. A central potential energy conserves all components of angular momentum, but they cannot all be assigned sharp simultaneous values because they do not commute; $H,L^2,L_z$ are a compatible choice. Discrete reflection symmetry gives parity labels without a continuously variable generator. A quantum degeneracy can be protected by a symmetry, but a quantum degeneracy alone does not identify its physical origin.

In Compton scattering, a photon transfers energy and momentum to an initially stationary, approximately free electron. With incident and scattered wavelengths $\lambda ,{\lambda }^{\prime }$ and photon deflection angle $\theta$, momentum conservation givesCombine this with $hc/\lambda +m{c}^2=hc/{\lambda }^{\prime }+\sqrt{m^2{c}^4+p_e^2{c}^2}$ to obtainThe photon loses energy to electron recoil; the shift is independent of incident wavelength in this free-electron model and is largest at backscattering. At energies comparable with atomic binding, initial electron motion, binding and coherent scattering from the atom require additional treatment. Classical elastic scattering from a fixed charge alone does not produce this recoil wavelength shift.

In Arthur Holly Compton's spectral test, molybdenum-target X-rays scattered from graphite at $90^{\circ }$ were compared with the primary spectrum using the same slits, crystal and tube potential energy. A calcite crystal resolved the shifted line by Bragg's law; lines on both sides established angular zero. The crystal glancing angle on the plotted axis differs from the photon scattering angle. The compared curves had different intensity scales, and the paper did not directly observe the recoil electrons in this X-ray test. These controls support the wavelength comparison without converting it into a direct recoil-track measurement.^[ref]

For independent angular momenta ${\mathbf{J}}_1,{\mathbf{J}}_2$, set $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Since different subsystems commute, $\mathbf{J}$ satisfies the same angular-momentum algebra. The product basis $|j_1{m}_1⟩|j_2{m}_2⟩$ diagonalizes $J_1^2,J_2^2,J_{1z},J_{2z}$; the coupled basis diagonalizes $J_1^2,J_2^2,J^2,J_z$. Its labels obeyThe Clebsch-Gordan coefficients are the unitary change of basis between them. Start from the maximum-weight state $j=j_1+j_2,m=j$, apply $J_{-}=J_{1-}+J_{2-}$ with wavefunction normalization, and construct the remaining multiplets by orthogonality.

For two spin-one-half systems, the symmetric triplet state is $|\uparrow \uparrow ⟩$, $(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, $|\downarrow \downarrow ⟩$, with $j=1$; the antisymmetric singlet state $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ has $j=0$. The dimension identity ${\sum }_j(2j+1)=(2j_1+1)(2j_2+1)$ checks completeness. Coupling orbital and intrinsic spin similarly gives total atomic angular momentum and organizes spin-orbit energy splittings. Exchange symmetry is an additional restriction when the constituents are identical.

quantum tunnelling is transmission through a finite region in which the classical kinetic energy $E-V$ is negative. The wavefunction need not vanish there: the Schrödinger equation permits evanescent solutions. The transmitted particle retains its conserved energy; it does not borrow energy to surmount the barrier.

For a rectangular barrier of height $V_b$ on $0<x<d$, with $0<E<V_b$ and zero exterior potential energy, define $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_b-E)}/\hslash$. A left-incident scattering solution has the three piecesContinuity of $\psi ,{\psi }^{\prime }$ at both boundaries imposesEliminating $b,c,r$ givesEqual incident and exit velocities justify $T=|t{|}^2$; different exterior potential energies require the velocity ratio. Probability-current conservation gives $R+T=1$. Although either real exponential alone has zero current, their complex-coefficient combination can carry the current crossing the barrier.

For $\kappa d\gg 1$, $T\simeq 16E(V_b-E)V_b^{-2}{e}^{-2\kappa d}$. A slowly varying barrier generalizes this to the WKB approximation $T\sim \exp [-2{\int }_{x_1}^{x_2}\sqrt{2m(V-E)}dx/\hslash ]$, with matching prefactors and turning-point limitations. This exponential sensitivity explains why very small geometric or energy changes can produce large changes in a tunnelling rate.

In alpha decay, a quasibound alpha particle penetrates the nuclear Coulomb barrier; penetration probability, attempt frequency and nuclear preformation determine the rate. In a scanning tunnelling microscope, a bias between a conducting tip and sample gives current approximately proportional to $e^{-2\kappa d}$. Piezoelectric positioning and constant-current feedback probe surfaces, while vibration isolation and electronic-state dependence limit a purely geometric interpretation. Atomic-scale steps and surface reconstruction were resolved in the original surface-imaging study.^[ref] In a Josephson junction, coherent tunnelling of Cooper pairs permits $I=I_c\sin \phi$, a phase-dependent supercurrent through a weak link. Thus the same barrier physics explains natural decay and underpins sensitive surface and superconducting devices.