For a real field, $\phi_{-\mathbf k}=\phi_{\mathbf k}^*$. The stated <Fourier transform> gives
$$
\nabla\phi(\mathbf x)=\int\frac{d^dk}{(2\pi)^d}\,i\mathbf k e^{i\mathbf k\cdot\mathbf x}\phi_{\mathbf k}.
$$
Using
$$
\int d^dx\,e^{i(\mathbf k+\mathbf q)\cdot\mathbf x}
=(2\pi)^d\delta^{(d)}(\mathbf k+\mathbf q)
$$
in both quadratic terms yields
$$
F[\phi]
=\frac12\int\frac{d^dk}{(2\pi)^d}
(\gamma k^2+\mu^2)|\phi_{\mathbf k}|^2.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 303 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-303/1/b/i/solution