Past exam of the physics A course of the University of Cambridge / 2010 / Paper 2 / Section C / C10

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.