When $b=\kappa_1=0$, stationarity in $\phi$ gives
$$
\frac{\delta F}{\delta\phi}=a\phi+\frac c2p^2=0,
\qquad
\phi=-\frac{c}{2a}p^2.
$$
Substituting this value, or completing the square in the Gaussian integral over $\phi$, gives
$$
F[p]=\int\left[\frac A2p^2+\frac{\widetilde B}{4}p^4+\frac\kappa2|\nabla p|^2\right]d\mathbf r,
\qquad
\boxed{\widetilde B=B-\frac{c^2}{2a}}.
$$
When $A>0$ and the quartic term is negligible, each Fourier mode is Gaussian with the <Ornstein--Zernike correlation function>
$$
\boxed{S_p(q)=\frac{k_BT}{A+\kappa q^2}}
$$
up to the chosen Fourier normalization. Its correlation length is $\xi_p=\sqrt{\kappa/A}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 1 / e / i / Solution

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A Gaussian field with free-energy kernel $A+\kappa q^2$ has structure factor proportional to $(A+\kappa q^2)^{-1}$ and correlation length $\sqrt{\kappa/A}$.


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 Incoming links: Ornstein--Zernike correlation function