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

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.