= Gaussian variational kernel for a gradient quartic interaction
{c}
{title2=$J(q)=\bar a+\bar\kappa q^2+\gamma q^4$}
For $H=\int[a\phi^2/2+\kappa|\nabla\phi|^2/2+\gamma(\nabla^2\phi)^2/2+B\phi^2|\nabla\phi|^2]$ 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 $\langle H_4\rangle_0=4BS_0S_2/V$. The <Feynman-Bogoliubov inequality> therefore has stationary kernels
$$
J(q)=a+\kappa q^2+\gamma q^4+\frac{4B}{V}(S_2+q^2S_0).
$$
Thus both the mass and gradient coefficient shift. In the <thermodynamic limit>, their <self-consistency equations> are
$$
\bar a=a+2B\int_{|k|<\Lambda}\frac{k^2}{\bar a+\bar\kappa k^2+\gamma k^4}\frac{d^dk}{(2\pi)^d},
\quad
\bar\kappa=\kappa+2B\int_{|k|<\Lambda}\frac1{\bar a+\bar\kappa k^2+\gamma k^4}\frac{d^dk}{(2\pi)^d}.
$$
A 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 $J(q)>0$.
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