Use $\partial_x\mapsto iq$ under the <Fourier transform> and the reality conditions $\phi(-q)=\phi(q)^*$ and $p(-q)=p(q)^*$. With $\Psi=(\phi,p)$, the quadratic free energy is
$$
F=\frac12\sum_q\Psi_i(q)G_{ij}(q)\Psi_j(-q),
$$
where
$$
\boxed{
G(q)=
\begin{pmatrix}
a+\kappa q^2+\gamma q^4&i\lambda q\\
-i\lambda q&\nu
\end{pmatrix}.}
$$
The opposite imaginary off-diagonal entries make $G(q)$ a <Hermitian matrix>. Reversing the Fourier-sign convention reverses both of those signs without changing any correlator.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-344/1/e/solution