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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 $J(q)>0$.

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.

Use the Fourier transform conventionThe zero mode vanishes because the compositional order parameter has zero spatial average. Parseval identity and the Fourier transform of a derivative turn the quadratic part of the dimensionless free energy intoHere ${\sum }_{\mathbf{q}}^{+}$ is the positive-wavevector sum for a real field. Each independent complex amplitude has density proportional to $e^{-G(q)|{\varphi }_{\mathbf{q}}{|}^2}$, so its elementary Gaussian integral gives the static structure factorwhenever $G(q)>0$.

The stationary points of the Brazovskii model kernel obeyBecause $\kappa <0<\gamma$, the nonzero minimum is the nonzero-wavevector soft-mode sphereAt this wavevector,The first Gaussian field theory divergence therefore occurs at