Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 69 / 1

Spontaneous symmetry breaking occurs when the statistical Hamiltonian is invariant under a symmetry but a chosen equilibrium pure state is not. A small selecting field or a boundary condition chooses one of several states related by the symmetry. For a finite unconstrained system the symmetric ensemble generally has zero order parameter; spontaneous order is defined by taking the thermodynamic limit before removing the selecting field. A ferromagnet with a selected magnetization direction is the standard example.

If the broken symmetry is continuous, nearby symmetry-related states cost no energy for a uniform displacement along the manifold of states. Locality and a nonzero stiffness then make a slowly varying displacement cost gradient energy rather than a mass term. These gapless long-wavelength fluctuations are Goldstone modes. In the conventional relativistic internal-symmetry setting their count is the number of broken generators. Nonrelativistic dynamics can pair broken generators, as in a ferromagnet; gauge symmetries and broken space-time symmetries require additional qualifications. A discrete broken symmetry alone does not produce a Goldstone mode.

For an interface with a selected mean height, the continuous displacement is a rigid translation normal to the mean plane. The tension gives a stiffness for spatially varying translations. The following calculation concerns this static Goldstone sector. It determines the excitation energy and equilibrium fluctuations, but a frequency dispersion requires a separate inertial or dissipative dynamics. In low dimensions thermal roughness also limits the existence of a strictly localized flat interface in an infinite system.

Put $D=d-1$ for the membrane's internal dimension and assume $\sigma >0$. The small-slope expansion isThe zeroth-order term is the projected area and is independent of the height field. Discarding it changes only an overall normalization of the partition function. Thus the leading fluctuation energy and its statistical weight areOne obtains this Gaussian field theory by discretizing the height field on a microscopic mesh, integrating every height with its Boltzmann weight, and then writing the continuum measure. A microscopic cutoff is retained for quantities dominated by short wavelengths. Quadratic order here really means small slopes: a large rigid height shift is still free because the energy has no dependence on $h$ itself.

The uniform height mode has zero energy. Fix its mean, impose a boundary condition, or divide out the integral over global translations to normalize the partition function. Height differences are independent of that choice. The approximation is the tension-only part of small-slope elastic membrane energy; there is no bending rigidity or confining potential in this part.

Use the Fourier transform conventionParseval's identity and $\nabla \mapsto i\mathbf{q}$ give the diagonal quadratic formIn a finite periodic box one can equivalently use orthonormal discrete Fourier modes, for which each independent real amplitude has a positive stiffness $\sigma q^2$. Fourier orthogonality eliminates all cross terms between different modes. The zero mode is removed as explained in part (a).

The stiffness vanishes as $q\to 0$, so capillary waves at long wavelength have arbitrarily low energy. The statistical Hamiltonian is invariant under $h(\mathbf{x})\mapsto h(\mathbf{x})+a$ for any constant $a$. Selecting a particular mean membrane height breaks this continuous normal-translation symmetry. A varying local translation therefore has a gradient cost but no restoring mass: these are Goldstone modes. The diagonal static kernel is $q^2$; it should not be identified with a dynamical frequency without specifying the surrounding fluid or membrane inertia.

Each nonzero mode is Gaussian, giving the covarianceConsequently the capillary height-difference correlation isThe difference removes the zero mode and makes the integral infrared-finite at each fixed separation. Where required, cut off $q$ at $\Lambda \sim 1/a$ with microscopic length $a$. The question's $d$ is the embedding dimension, so the momentum integral has dimension $D=d-1$, not $d$.

For $d=2$, $D=1$, use ${\int }_{-\infty }^{\infty }dq(1-\cos qr)/(2\pi q^2)=|r|/2$. HenceThe root-mean-square relative height grows as $r^{1/2}$: the interface wanders increasingly far from a fixed height.

For $d=3$, $D=2$, angular integration givesThe Bessel function expression separates the universal logarithmic growth from a cutoff-dependent constant. The membrane is logarithmically rough, so its absolute height fluctuations are unbounded in the thermodynamic limit even though long-wavelength slopes need not be large.

For $d=4$, $D=3$, the massless Green function at nonzero separation is $1/(4\pi r)$. Thuswhere $B_{\infty }$ is finite once the microscopic cutoff is supplied. For a spherical sharp cutoff, $B_{\infty }=k_{B}T\Lambda /({\pi }^2\sigma )$. Relative height fluctuations saturate at large distance, allowing a localized interface height selected by boundaries. The roughness distinction is linear growth in $d=2$, logarithmic growth in $d=3$, and bounded long-distance fluctuations in $d=4$. The short-distance cutoff and the small-slope assumption remain necessary; bounded height fluctuations do not imply a perfectly flat microscopic membrane.

Assume the pinning coefficient $t>0$. Here $t$ belongs to an actual energy, so its Boltzmann coefficient is $\beta t$. The quadratic kernel and covariance becomeTherefore the pinned capillary-wave correlation isThe normal-translation symmetry is now explicitly broken, and the formerly massless Goldstone mode has a restoring term. For $r\ll \xi$, but beyond the microscopic cutoff, the unpinned behavior is recovered. For $r\gg \xi$, the connected height covariance decays and $B_t$ approaches $2⟨h^2⟩$, so roughness no longer grows without bound even for $d=2,3$.

In internal dimension one the exact continuum answer isIn internal dimension two the connected covariance at $r>0$ is $k_{B}T{K}_0(r/\xi )/(2\pi \sigma )$, where $K_0$ is a modified Bessel function; its zero-separation value needs the cutoff and the saturation value grows as $(k_{B}T/\pi \sigma )\log (\xi /a)$ for $\xi \gg a$. In internal dimension three the connected covariance is approximately $k_{B}T{e}^{-r/\xi }/(4\pi \sigma r)$ beyond the cutoff. Thus pinning replaces the long-distance massless tails by exponential decay. A negative $t$ would destabilize the quadratic statistical Hamiltonian rather than bind the membrane.