Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 337 (Applications of Quantum Field Theory)

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 $\epsilon$.

Under Fermi-surface renormalization-group scaling, tangential momentum $k$ is fixed whileIn frequency space the quadratic action is invariant whenequivalently, in the time representation $t\mapsto s^{-1}t$ and $\psi (t,k,\ell )\mapsto s^{-1/2}\psi (t,k,\ell )$. The free Fermi liquid action is therefore marginal.

The deformation $\mu (k){\psi }^{†}\psi$ has no factor of frequency or normal momentum, so it is relevant. It shifts the zero of the quasiparticle energy from $\ell =0$ to $\ell =\mu (k)/v_F(k)$ and hence deforms the Fermi surface.

A generic quartic interaction is irrelevant because momentum conservation fixes a normal component and leaves an additional positive power of $s$. It can be marginal only when all four momenta remain on the Fermi surface while satisfyingFor a smooth generic Fermi surface, the robust possibilities are forward or exchange scattering and the opposite-momentum BCS channel ${\mathbf{k}}_2=-{\mathbf{k}}_1$, ${\mathbf{k}}_4=-{\mathbf{k}}_3$.

The tree-level Cooper-pair scattering amplitude is simply $V$. Define the density of states at the Fermi surface, in the normalization of the question, byIn the one-loop Cooper diagram the two internal fermions have opposite momenta. After integrating over internal frequency, the remaining normal-energy integral containsConsequentlyand summing the leading geometric series gives the running BCS theory couplingFor repulsive $V>0$ it flows logarithmically toward zero. For attractive $V<0$, its denominator vanishes at the Cooper instability scaleAt this scale the normal Fermi liquid becomes unstable: opposite-momentum fermions form Cooper pairs, a superconducting or superfluid condensate develops, and a gap opens.

Put $\xi =v_F(k)\ell$ and use the same ${\nu }_F$. With a symmetric cutoff $|\xi |<\Lambda$, the BCS gap equation becomesThe frequency integral is $1/[2\sqrt{{\xi }^2+{\Delta }^2}]$, soThusThe gap and the one-loop strong-coupling scale have the same nonperturbative exponential dependence; their order-one prefactors depend on the cutoff convention and microscopic completion. The renormalization-group divergence is therefore the normal-state signal of the paired, gapped phase found from the self-consistent gap equation.