Past exam of the phase transitions and collective phenomena course of the University of Cambridge / 2012 / Paper 1 / 2 / Solution

Spontaneous symmetry breaking means that a pure thermodynamic phase is less symmetric than the Hamiltonian even when the selecting external field is removed. For example,The direction of the infinitesimal field is held fixed in this limit. Reversing the limits in a finite symmetric sample normally gives zero. For the XY model, an ordered state chooses one direction from a circle of energetically equivalent directions.

A Goldstone mode is a long-wavelength fluctuation along this continuously degenerate family. A uniform angle rotation costs no energy; a slowly varying one costs only gradient energy. In the stated static phase-only model the quadratic kernel is $\bar{K}{q}^2$, tending to zero as $q\to 0$. There is one independent soft order-parameter phase coordinate for broken $O(2)$ rotation. The static Hamiltonian does not specify a dynamical dispersion $\omega (q)$; that requires a dynamical law. Magnitude fluctuations instead have a restoring local curvature and have been excluded from this low-energy model.