= 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,
$$
\mathbf m_0=\lim_{\mathbf h\to0^+}\lim_{V\to\infty}\langle\mathbf m\rangle_{V,\mathbf h}\ne0.
$$
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 $q\to0$. \b[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.
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