= 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> is
$$
\mathbf m_0=\lim_{h\to0^+}\lim_{V\to\infty}\langle\mathbf m\rangle_{V,h}.
$$
A 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 $\theta\mapsto\theta+\epsilon$ costs no energy. Local changes cost only gradients, so the <Goldstone-mode effective free energy> starts with $\frac12\rho_s\int(\nabla\theta)^2$, 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.
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