Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 344 / 1

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.

Average the trial field over an integer number of periods. The squared-field, squared-gradient and squared-Laplacian averages are $A^2/2$, $q_0^2{A}^2/2$ and $q_0^4{A}^2/2$, while $⟨{\varphi }^2|\nabla \varphi {|}^2⟩=A^4{q}_0^2/8$. Hence the single-mode smectic mean-field free energy isFor the physically stabilizing case $B>0$, minimization givesThe amplitude tends continuously to zero at $a_c$, so this mean-field approximation predicts a continuous phase transition. The ordered minimum has free-energy density $-(a_c-a{)}^2/(8B{q}_0^2)$.

A nonzero $B$ alone is not sufficient for this conclusion: $B<0$ makes the displayed quartic amplitude energy unbounded below and requires higher-order stabilization. The continuous-transition statement uses $B>0$ and the stipulated fixed wavevector $q_0$.

Apply the Jensen inequality to the convex exponential function under the trial equilibrium measure:The given identity therefore implies $e^{-F}\ge e^{-F_0-⟨H-H_0{⟩}_0}$. Taking minus the logarithm reverses the inequality and proves the Feynman-Bogoliubov inequalityThe units $\beta =1$ make $H$ and $F$ dimensionless here. Equality holds when $H-H_0$ is constant on the support of the trial measure. Optimizing the trial parameters tightens an upper bound, which is the basis of the Gaussian variational approximation.

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.

Differentiate the bound with respect to one $J(q)$, using $\partial S_0/\partial J(q)=-J(q{)}^{-2}$ and $\partial S_2/\partial J(q)=-q^{2}J(q{)}^{-2}$:Thus an interior minimizing Gaussian variational kernel for a gradient quartic interaction hasUse the half-space mode density given in the question. In the thermodynamic limit, the coupled self-consistency equations areThe admissible solution must have $J(q)>0$. The coefficient $\gamma$ is unchanged because the interaction's derivative contributes only a constant and a $q^2$ term.

The formal continuum integrals require the coarse-graining ultraviolet cutoff $\Lambda$ in three dimensions: the first radial integrand tends to a constant at large $k$, so taking $\Lambda =\infty$ would leave $\bar{a}$ divergent. The low-wavevector fluctuation argument is independent of this short-distance regularization.

In the usual three-dimensional isotropic setting, the transition is expected to be discontinuous, by the Brazovskii fluctuation-induced first-order transition mechanism.

If $\bar{\kappa }<0$, write $q_{\ast }^2=-\bar{\kappa }/(2\gamma )$ and $r=\bar{a}-{\bar{\kappa }}^2/(4\gamma )$. Then $J(q)=r+\gamma (q^2-q_{\ast }^2{)}^2$. A continuous finite-wavevector transition would require $r\to 0$ with $q_{\ast }>0$. But the soft-mode shell integral behaves asThe positive fluctuation corrections in the self-consistency equations diverge in that limit. In particular $\bar{\kappa }=\kappa +2B{I}_0$ cannot stay at a finite negative value while $I_0$ diverges. Thus the isotropic disordered state cannot lose stability through a zero-mass mode on a fixed nonzero shell in this approximation.

Ordering is instead expected through a free-energy crossing with a finite-amplitude smectic phase, giving a first-order phase transition. This is a reasoned fluctuation prediction for the finite-wavevector transition; the disordered Gaussian variational approximation alone does not locate the coexistence point or calculate the amplitude jump. A zero-wavevector instability or a lower-dimensional setting would require a separate analysis.