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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-69/2/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 69 / 2 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Choose

m (x_{∥}, x_{⊥}) = \int \frac{d q _{∥} d ^{d - 1} q _{⊥}}{( 2 π ) ^{d}} m (q_{∥}, q_{⊥}) e^{i (q_{∥} x_{∥} + q_{⊥} \cdot x_{⊥})} .

For a real scalar field

m (- q) = m (q)^{*}

. The parallel derivative becomes

i q_{∥}

, and the transverse Laplacian becomes

- q_{⊥}^{2}

. The latter is squared in the energy, producing

q_{⊥}^{4}

, not

q_{⊥}^{2}

. Thus

β H = \frac{1}{2} \int_{q} [K q_{∥}^{2} + L q_{⊥}^{4} + t] ∣ m (q) ∣^{2} - h m (0) .

The zero-momentum source is understood in a finite box or as a uniform-field distribution before taking the thermodynamic limit. With an orthonormal box transform the source is

- h V m_{0}

. The positive coefficients and

t > 0

give a normalizable Gaussian field theory, with mean uniform field

\overline{m} = h / t

and connected Fourier covariance

[K q_{∥}^{2} + L q_{⊥}^{4} + t]^{- 1}

.

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