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

For the one-mode smectic phase ansatz $\varphi (\mathbf{r})=A\cos (q_{0}z)$, take $A\ge 0$ without loss of generality. Spatial averaging givesandUsing $G(q_0)=a-a_c$, the free-energy density isFor $a\ge a_c$, its minimum is $A=0$. For $a<a_c$, the nonzero stationary point isandThus the modulation amplitude grows linearly and continuously from zero below $a_c$. In the language of an order-parameter critical exponent, this nonanalytic $|\varphi {|}^3$ theory has the mean-field value $\beta =1$.

This is the Gibbs--Bogoliubov--Feynman inequality. In units with $k_{B}T=1$, $H$ 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 areThe notationmeans an expectation in the trial Gibbs ensemble. The bound is therefore

Choose the translationally invariant Gaussian variational approximationThe superscript $+$ means that one member of each pair $\{\mathbf{q},-\mathbf{q}\}$ is included; this is necessary because a real field obeys ${\varphi }_{-\mathbf{q}}={\varphi }_{\mathbf{q}}^{\ast }$. The constrained zero mode is omitted.

For each independent complex mode,Consequently,Substitution into the Gibbs--Bogoliubov--Feynman inequality giveswhich is the required upper bound.

Translational invariance makes the pointwise variance independent of $\mathbf{r}$. Applying Parseval identity to the stated Fourier convention givesIn the thermodynamic limit, the reciprocal-lattice sum becomesAt each point $\varphi (\mathbf{r})$ is a centered Gaussian random variable, so the supplied absolute third moment gives

SetThe factor two in the last expression restores the omitted negative wavevectors. For one independent mode $J_k=J(k)$,Differentiating the bound from parts (d) and (e) givesIts stationary point therefore obeys the self-consistency equationThe interaction has generated a positive, wavevector-independent mass shift.

Write the optimized kernel asNear the soft-mode sphere, the supplied asymptotic result saysThus the self-consistency equationcannot reach $\bar{a}=a_c$ for any $g>0$: the fluctuation correction diverges first. The isotropic static structure factor therefore remains finite, and the isotropic state never undergoes the continuous Gaussian instability predicted in part (a).

The modulated minimum from part (b), however, has negative free energy for sufficiently low $a$ and amplitude $A_{\ast }=\pi (a_c-a)/(8g)$. For small positive $g$, its free energy must cross that of the still locally stable isotropic variational state. At the crossing the isotropic inverse susceptibility is positive and the smectic amplitude is nonzero. The order parameter consequently jumps, which is the Brazovskii fluctuation-induced first-order transition.