= Solution
When all masses are equal, the free energy depends on the real vector $\boldsymbol\phi$ only through <inner products>, so its symmetry is the <orthogonal group> $O(N)$. The mean-field potential is
$$
V(\boldsymbol\phi)=\frac12\mu_0^2\boldsymbol\phi^2+g_0(\boldsymbol\phi^2)^2.
$$
For $\mu_0^2>0$, its unique minimum is $\boldsymbol\phi=0$, which preserves $O(N)$. For $\mu_0^2<0$, the minima form the <sphere>
$$
\boldsymbol\phi^2=-\frac{\mu_0^2}{4g_0}.
$$
Choosing one minimum leaves the subgroup $O(N-1)$ that fixes its direction, so <spontaneous symmetry breaking> gives $\boxed{O(N)\to O(N-1)}$.
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