Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 303 / 3 / c

The $O(2)$ rotation symmetry requires the quadratic dependence on $({\varphi }_2,{\varphi }_3)$ to be proportional to ${\varphi }_2^2+{\varphi }_3^2$, soThe quartic terms involving only that pair must be proportional to $({\varphi }_2^2+{\varphi }_3^2{)}^2$, and the terms coupling it to ${\varphi }_1$ must be proportional to ${\varphi }_1^2({\varphi }_2^2+{\varphi }_3^2)$. With the convention that the symmetric double sum counts off-diagonal terms twice, this giveswhile ${\mu }_1^2$, $g_{11}$, and the common mixed coupling are unrestricted. The stated ${\mathbb{Z}}_2$ symmetry imposes no further relation because every term is already even in ${\varphi }_1$.

Let $r_1={\mu }_1^2$, $r_2={\mu }_2^2={\mu }_3^2$, and ${\rho }^2={\varphi }_2^2+{\varphi }_3^2$. With every $g_{ij}=g>0$, the uniform potential isFor $r_1>0,r_2>0$, the minimum is the origin and ${\mathbb{Z}}_2\times O(2)$ is unbroken. If $r_1<\min (0,r_2)$, then ${\varphi }_1^2=-r_1/(4g)$ and $\rho =0$; the ${\mathbb{Z}}_2$ factor is broken, $O(2)$ remains, and there is no Goldstone mode. If $r_2<\min (0,r_1)$, then ${\varphi }_1=0$ and ${\rho }^2=-r_2/(4g)$; the first ${\mathbb{Z}}_2$ remains while $O(2)$ is broken to the reflection fixing the chosen direction, producing one Goldstone mode.

The positive coordinate half-axes are continuous transition lines. For $r_1=r_2<0$, the potential has an enhanced $O(3)$ symmetry and a sphere of minima; $O(3)\to O(2)$ gives two Goldstone modes on this line. Crossing the negative diagonal exchanges the two ordered phases and gives a first-order line at mean-field level.