Gaussian variational approximation

A Gaussian variational approximation chooses a quadratic trial kernel and minimizes a variational upper bound on the interacting free energy. The optimized kernel often equals the bare kernel plus a self-consistent, wavevector-independent mass shift.

For $H=\int [a{\varphi }^2/2+\kappa |\nabla \varphi {|}^2/2+\gamma ({\nabla }^2\varphi {)}^2/2+B{\varphi }^2|\nabla \varphi {|}^2]$ and an even positive trial kernel $J(q)$, define $S_0={\sum }_q^{+}1/J(q)$ and $S_2={\sum }_q^{+}{q}^2/J(q)$ using a positive-wavevector sum for a real field. Isserlis theorem gives $⟨H_4{⟩}_0=4B{S}_0{S}_2/V$. The Feynman-Bogoliubov inequality therefore has stationary kernelsThus both the mass and gradient coefficient shift. In the thermodynamic limit, their self-consistency equations areA coarse-graining ultraviolet cutoff $\Lambda$ is necessary for the first integral in three dimensions: its large-$k$ radial integrand tends to a constant. Physical solutions require $J(q)>0$.

For an exact dimensionless Hamiltonian $H$ and trial Hamiltonian $H_0$, the free energies obeywhere $F_0=-\log Z_0$ and the expectation uses the trial Gibbs distribution $Z_0^{-1}{e}^{-H_0}$.

The identity $e^{-F}=e^{-F_0}⟨e^{-(H-H_0)}{⟩}_0$ and the Jensen inequality for the convex exponential function give $\log ⟨e^{-(H-H_0)}{⟩}_0\ge -⟨H-H_0{⟩}_0$. Taking minus the logarithm proves the bound.