= Solution
For $\mu^2<0$ and $g>0$, minimizing the $O(N)$-invariant potential gives
$$
\phi^2=v^2=-\frac{\mu^2}{4g}.
$$
The free energy has the full $O(N)$ symmetry, but choosing one point on this sphere leaves only the rotations $O(N-1)$ fixing that point. This is <spontaneous symmetry breaking>, $O(N)\to O(N-1)$. The $N-1$ tangent directions along the sphere cost no potential energy and are the massless <Goldstone boson>[Goldstone modes] predicted by the <Goldstone theorem>.
Back to article page
