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

For a uniform field with $r^2=\mathbf{\varphi }\cdot \mathbf{\varphi }$, the potential isFor $T>T_c$, one has ${\mu }^2>0$, so ${\mathbf{\varphi }}_0=0$. The Hessian has $N$ positive equal eigenvalues, and all $N$ modes are gapped. The full $O(N)$ symmetry is unbroken.

For $T<T_c$, one has ${\mu }^2<0$, andThe vacuum manifold is $S^{N-1}$. Choosing one point breaks $O(N)$ spontaneously to $O(N-1)$. The radial fluctuation has squared mass $-2{\mu }^2$, while the $N-1$ tangent directions are gapless Nambu-Goldstone bosons, one for each broken continuous generator modulo the unbroken subgroup.

For the $O(2)$ model writeAt long distances the massive radial field $\rho$ can be neglected, leaving the Goldstone-mode effective free energyThereforeThe mode is massless, so its correlation length is infinite. For $d>2$ its large-distance Green function isFor $d=2$ 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 varianceIt 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 quasi-long-range order below a BKT transition.

For generic ${\mu }^2\ne {\lambda }^2$, the symmetry is $O(1)\times O(2)$: an independent sign flip of ${\varphi }_1$ and orthogonal transformations of $({\varphi }_2,{\varphi }_3)$. When ${\mu }^2={\lambda }^2$, it is enhanced to $O(3)$.

If ${\mu }^2,{\lambda }^2>0$, the unique ground state is ${\mathbf{\varphi }}_0=0$. The full $O(1)\times O(2)$ symmetry remains unbroken and every mode is gapped.

If ${\mu }^2<0$ and ${\lambda }^2>0$, the minima areThe discrete $O(1)\cong {\mathbb{Z}}_2$ symmetry is spontaneously broken, while $O(2)$ remains intact. There is no Goldstone mode because only a discrete symmetry is broken.

If ${\mu }^2>0$ and ${\lambda }^2<0$, the minima areThe $O(1)$ factor remains unbroken, while $O(2)$ is spontaneously broken to the $O(1)$ subgroup that fixes a chosen point on the circle. The one-dimensional vacuum circle gives one Goldstone mode.

If ${\mu }^2={\lambda }^2<0$, the enhanced $O(3)$-symmetric potential has the sphere of minimaChoosing a ground state breaks $O(3)$ to $O(2)$. The vacuum manifold $O(3)/O(2)\simeq S^2$ has dimension two, so there are two Goldstone modes.