Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 72 / 1 / Solution

Spontaneous symmetry breaking occurs when the Hamiltonian has a symmetry that an equilibrium pure thermodynamic phase does not share. For a ferromagnetic XY model, a nonzero magnetization selects an orientation even though the zero-field Hamiltonian is rotation invariant. The order of limits matters: with $V$ the volume and $h$ a field selecting the orientation, the spontaneous magnetization isA finite system at zero field instead averages over symmetry-related orientations and has zero vector magnetization. Below an ordered transition, that symmetric mixture must be distinguished from a selected pure thermodynamic phase.

A Goldstone mode is a soft fluctuation tangent to the continuously degenerate family of equilibrium orientations. For the XY model, the uniform change $\theta \mapsto \theta +ϵ$ costs no energy. Local changes cost only gradients, so the Goldstone-mode effective free energy starts with $\frac{1}{2}{\rho }_s\int (\nabla \theta {)}^2$, with no restoring term proportional to ${\theta }^2$. A Fourier transform mode therefore has a stiffness proportional to $q^2$, becoming arbitrarily soft at long wavelength. The dimensionless phase stiffness here is $\bar{K}=\beta {\rho }_s$. These are the spin waves of the angle field. Their large thermal fluctuations can invalidate the proposed ordered phase in low dimension; the existence of a symmetry of the Hamiltonian alone does not guarantee its spontaneous breaking.