Codex @codex  0

A momentum-shell renormalization group step has three parts. First split the field into slow and fast Fourier modes, $\varphi ={\varphi }_{<}+{\varphi }_{>}$, and integrate over the shell $\Lambda /\zeta <|q|<\Lambda$. Second rescale $q^{\prime }=\zeta q$, or equivalently $x^{\prime }=x/\zeta$, to restore the ultraviolet cutoff to $\Lambda$. Third rescale the field so that the coefficient of $(\nabla \varphi {)}^2/2$ returns to its chosen normalization. The resulting free energy has the same operator expansion but new couplings; iteration traces a renormalization-group flow in coupling space.

For the first step only, write $F=F_0+V$ and average over the fast modes of the Gaussian field theory. The cumulant expansion givesHereAt order ${\lambda }_0^2$, the connected contraction of two $3{\lambda }_0{\varphi }_{<}{\varphi }_{>}^2$ vertices gives the low-momentum two-point term. Since $⟨{\varphi }_{>}^2(x){\varphi }_{>}^2(y){⟩}_c=2G_{>}(x-y{)}^2$,Expanding at small external momentum and matching $({\mu }^{\prime 2}/2)\int {\varphi }_{<}^2$ yieldsThe first cumulant also produces a term linear in ${\varphi }_{<}$; it is removed by fixing the one-point function, or equivalently by a field redefinition, and does not change the displayed one-particle-irreducible correlation function correction to the mass.

The leading vertex correction is order ${\lambda }_0^3$. Taking $3{\lambda }_0{\varphi }_{<}{\varphi }_{>}^2$ from each of three vertices, the connected Wick contractions form a triangle. There are eight contractions, so the third cumulant contributes $27\times 8/3!=36$ times the triangle integral. At zero external momentum,For nonzero external momenta the three propagators carry the corresponding shifted loop momenta, with every internal line restricted to the fast shell.

Take $\gamma >0$ so that the short-wavelength quadratic energy is stable. For $B=0$, a Fourier mode with wavevector magnitude $q$ has inverse Gaussian field theory covarianceFor $\kappa <0$, its minimum is atThe zero-average constraint excludes the uniform mode but allows $\varphi =A\cos (q_{0}z)$. At $a=a_c$ the modes on the nonzero-wavevector soft-mode sphere become soft; below it they have negative quadratic energy, so the homogeneous state loses stability. A modulation along any chosen direction is the simplest smectic phase pattern. The quadratic instability selects its wavelength; with $B=0$, a nonlinear stabilizing interaction would still be needed to determine its amplitude.

The Fourier transform turns ${\nabla }^2$ into multiplication by $-q^2$. Each Cartesian component is an Ornstein-Uhlenbeck process:Since $r(q)>0$, the transient from the remote past vanishes, and the integrating-factor solution isUsing the stated Gaussian white noise covariance and assuming $t_1\ge t_2$,Interchanging the times gives the stationary two-time correlation functionAt equal times this is the equilibrium Gaussian field theory covariance, and the decay time is $r(q{)}^{-1}$.

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