This is the <Gibbs--Bogoliubov--Feynman inequality>. In units with $k_BT=1$, $H$ is the exact dimensionless Hamiltonian, here the full interacting free-energy functional, and $H_0$ is a freely chosen trial Hamiltonian. Its <partition function> and free energy are
$$
Z_0=\int\mathcal D\phi\,e^{-H_0[\phi]},
\qquad
F_0=-\log Z_0.
$$
The notation
$$
\langle X\rangle_0
=Z_0^{-1}\int\mathcal D\phi\,X[\phi]e^{-H_0[\phi]}
$$
means an expectation in the trial Gibbs ensemble. The bound is therefore
$$
F=-\log\int\mathcal D\phi\,e^{-H}
\leq F_0+\langle H-H_0\rangle_0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 1 / c / Solution

Choose the translationally invariant <Gaussian variational approximation>
$$
H_0=\sum_{\mathbf q}^{+}J(q)|\phi_{\mathbf q}|^2,
\qquad J(q)>0.
$$
The superscript $+$ means that one member of each pair $\{\mathbf q,-\mathbf q\}$ is included; this is necessary because a real field obeys $\phi_{-\mathbf q}=\phi_{\mathbf q}^*$. The constrained zero mode is omitted.

For each independent complex mode,
$$
\int_{\mathbb R^2}d(\operatorname{Re}\phi_{\mathbf q})\,d(\operatorname{Im}\phi_{\mathbf q})
e^{-J(q)|\phi_{\mathbf q}|^2}
=\frac{\pi}{J(q)}.
$$
Consequently,
$$
F_0=\sum_{\mathbf q}^{+}\log\frac{J(q)}{\pi},
\qquad
\langle H_0\rangle_0=\sum_{\mathbf q}^{+}1,
\qquad
\left\langle\sum_{\mathbf q}^{+}G(q)|\phi_{\mathbf q}|^2\right\rangle_0
=\sum_{\mathbf q}^{+}\frac{G(q)}{J(q)}.
$$
Substitution into the <Gibbs--Bogoliubov--Feynman inequality> gives
$$
\widetilde F
=\sum_{\mathbf q}^{+}
\left[
\log\frac{J(q)}{\pi}-1+\frac{G(q)}{J(q)}
\right]
+g\int d^d\mathbf r\,\langle|\phi(\mathbf r)|^3\rangle_0,
$$
which is the required upper bound.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 1 / d / Solution

{c}

For an exact dimensionless Hamiltonian $H$ and trial Hamiltonian $H_0$, the free energies obey
$$
F\leq F_0+\langle H-H_0\rangle_0,
$$
where $F_0=-\log Z_0$ and the expectation uses the trial Gibbs distribution $Z_0^{-1}e^{-H_0}$.


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