Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 337 / 1

A nonrelativistic Landau-Ginzburg theory for a complex order parameter with global $U(1)$ symmetry isIn the broken phase $r<0$, write $\psi =\sqrt{n}{e}^{i\theta }$. The first term isup to the sign convention for the phase. Thus the momentum conjugate to $\theta$ is $-n$: this is number-phase conjugacy. Canonical quantization gives the corresponding local commutator, while the spatial integral relates the global phase to total particle number.

Consequently a state with a sharply selected phase, as used to describe spontaneous symmetry breaking, cannot be an eigenstate of particle number. It is a coherent superposition of charge sectors with number fluctuations. Conversely, an exact finite-volume number eigenstate has no definite phase; the phase-selected broken-symmetry states emerge in the thermodynamic limit.

Letwhere $dg$ is Haar measure. Left invariance gives $hK{h}^{-1}=K$ for every $h\in SU(2)$. Since the spin-$S$ representation is irreducible, Schur lemma impliesInserting this coherent-state resolution between $N$ short time steps and taking the continuum limit gives, for a noninteracting spin with zero Hamiltonian,up to the normalization generated by $c$. A Hamiltonian would add $-i\int ⟨g|H|g⟩dt$ in the exponent.

Fordirect differentiation and $S_3|\uparrow ⟩=S|\uparrow ⟩$ giveThe $\dot{\psi }$ term is a total derivative and records only the arbitrary phase used to represent a ray. It cancels against the endpoint-state phases, so the physical Spin coherent-state path integral depends only on the path on the two-sphere. For example, choosing the section $\psi =-\varphi$ gives the Spin coherent-state Berry phase proportional toChanging the surface used to fill a closed path changes its solid angle by $4\pi$. Single-valuedness of the path-integral phase requiresThus the Wess-Zumino coefficient is quantized in integer or half-integer units.

Let $H|n⟩=E_n|n⟩$, $\beta =1/T$, $Z={\sum }_n{e}^{-\beta E_n}$, and $A_{nm}=⟨n|A|m⟩$. Inserting two energy resolutions into the commutator and carrying out the one-sided Fourier integral gives the Lehmann representationIf the trace in the question is intentionally unnormalized, the same formula holds without $1/Z$.

For $H=\epsilon a^{†}a$,Thereforeand the Green function requested literally for $A=a^{†}$ isPhysically, a number-conserving oscillator has no anomalous response connecting two creation operators. The nonzero normal retarded propagator pairs annihilation with creation: $G_{a{a}^{†}}^R(\omega )=1/(\omega -\epsilon +i{0}^{+})$, whose pole is the one-quantum excitation at energy $\epsilon$.