For the $O(2)$ model write
$$
\boldsymbol\phi=(v+\rho)(\cos\theta,\sin\theta),
\qquad
v^2=-\frac{\mu^2}{4g}.
$$
At long distances the massive radial field $\rho$ can be neglected, leaving the <Goldstone-mode effective free energy>
$$
F_\theta=\frac{\gamma v^2}{2}\int d^dx\,(\nabla\theta)^2.
$$
Therefore
$$
\langle\theta(\mathbf x)\theta(\mathbf y)\rangle
=\frac1{\gamma v^2}\int\frac{d^dk}{(2\pi)^d}
\frac{e^{-i\mathbf k\cdot(\mathbf x-\mathbf y)}}{k^2}.
$$
The mode is massless, so its <correlation length> is infinite. For $d>2$ its large-distance Green function is
$$
\langle\theta(\mathbf x)\theta(\mathbf y)\rangle
\sim\frac{\Gamma(d/2-1)}{4\pi^{d/2}\gamma v^2}\,r^{2-d}.
$$
For $d=2$ it is $-(2\pi\gamma v^2)^{-1}\log(r/a)$ up to an infrared-dependent constant, and in $d=1$ it is $-|r|/(2\gamma v^2)$ up to such a constant.

The invariant diagnostic is the <phase-difference variance>
$$
\langle[\theta(\mathbf r)-\theta(0)]^2\rangle
=\frac2{\gamma v^2}\int\frac{d^dk}{(2\pi)^d}
\frac{1-\cos(\mathbf k\cdot\mathbf r)}{k^2}.
$$
It diverges linearly in $d=1$ and logarithmically in $d=2$, destroying true long-range order, but approaches a finite infrared limit for $d=3,4$. Thus the continuous-symmetry ordered phase exists for $d=3,4$ and not for $d=1,2$, in agreement with the <Mermin-Wagner theorem>. The lower critical dimension is $d_{\mathrm{lc}}=2$; the two-dimensional $O(2)$ model can instead show <Berezinskii–Kosterlitz–Thouless transition>[quasi-long-range order below a BKT transition].

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 303 / 3 / b / Solution

{c}

For a broken $O(2)$ symmetry with order-parameter magnitude $v$, the long-wavelength phase field has free energy
$$
F_\theta=\frac{\rho_s}{2}\int d^dx\,(\nabla\theta)^2,
$$
where $\rho_s$ is the stiffness, equal to $\gamma v^2$ in the simplest Landau-Ginzburg model.


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