Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 337 / 2 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

The first term is a gauge choice for the Spin coherent-state Berry phase. For a closed spin trajectory, changing the surface used to evaluate the solid angle changes the action by

Δ S_{B} = 4 π S .

The path-integral phase

e^{i S_{B}}

must be independent of that choice, so

e^{i 4 π S} = 1.

Hence

2 S \in Z,

and the spin is integer or half-integer.

Solved by gpt-5.6-sol high.

ii

Solution

Write

θ = π /2 - ϑ

and

ϕ = φ

. To quadratic order,

cos θ = ϑ + O (ϑ^{3}), sin^{2} θ = 1 + O (ϑ^{2}) .

After dropping the total derivative

S φ_{t}

, the quadratic Lagrangian density is

L_{2} = Sϑ φ_{t} - u (ϑ_{x}^{2} + φ_{x}^{2}) - K ϑ^{2} .

Its Euler--Lagrange equations are

S φ_{t} + 2 u ϑ_{xx} - 2 K ϑ = 0, - S ϑ_{t} + 2 u φ_{xx} = 0.

For modes proportional to

e^{i (k x - ω t)}

, nontrivial amplitudes require

ω^{2} (k) = \frac{4 u k ^{2} ( u k ^{2} + K )}{S ^{2}} .

Thus

ω \sim \frac{2 u K}{S} ∣ k ∣ (k \to 0), ω \sim \frac{2 u}{S} k^{2} (∣ k ∣ \to \infty) .

At short wavelength the anisotropy is negligible and the quadratic dispersion is that of a conventional ferromagnetic magnon. At long wavelength, easy-plane anisotropy makes the out-of-plane fluctuation the conjugate density of the in-plane phase; eliminating it produces the linear Goldstone sound mode characteristic of a superfluid.

Solved by gpt-5.6-sol high.

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