 Solution

Choose an even positive $J(q)$. Each independent complex Fourier mode of the real field has density $(J(q)/\pi )e^{-J(q)|{\varphi }_q{|}^2}$, soThe quadratic part therefore has expectation ${\sum }_q^{+}G(q)/J(q)$. For the quartic part, apply Isserlis theorem to its four jointly centered Gaussian random variables, or to their real and imaginary components. Pairing the two differentiated fields together imposes $q_2=-q_1$, and its contribution isEach of the two other pairings instead gives $-(B/V)|{\sum }_{q}q/J(q){|}^2=0$, because the summand is odd under $q\mapsto -q$. Thus only the displayed contribution remains. Each full sum is twice its positive-wavevector sum for a real field; putting $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ gives $⟨H_4{⟩}_0=4B{S}_2{S}_0/V$.

Substitute these expectations, $F_0$ and $⟨H_0{⟩}_0$ into the Feynman-Bogoliubov inequality:The zero mode is absent by the composition constraint. The counting uses one representative of every nonzero $\{q,-q\}$ pair, so no extra factor of two is assigned to an independent complex amplitude.