Spontaneous symmetry breaking

Remarks

Since $U$ is invariant, every point of the orbit $G\phi_0$ is again a minimum point, so the ground states include a copy of the homogeneous space $G/H$. The same definition applies to a field theory with a constant field $\phi \colon \R^{1,3} \to V$ and Lagrangian density $\frac12 \abs{\partial\phi}^2 - U(\phi)$, where $G$ acts by orthogonal maps.

Example: $V = \C = \R^2$ with $G = U(1)$ acting by multiplication, and $U(\phi) = \lambda (\abs{\phi}^2 - v^2)^2$ with $\lambda, v > 0$. The minimum points form the circle $\abs{\phi} = v$, and the stabilizer of $\phi_0 = v$ is trivial, so $U(1)$ is broken to $\set{1}$.

Non-example: the same $V$ and $G$ with $U(\phi) = m^2 \abs{\phi}^2 + \lambda \abs{\phi}^4$ and $m^2 > 0$. The only minimum point is $\phi_0 = 0$, which is fixed by all of $U(1)$, so the symmetry is not broken.

In quantum theory the definition is stated for the vacuum state: the symmetry is spontaneously broken if the vacuum is not invariant under it. This needs infinitely many degrees of freedom; in quantum mechanics with finitely many degrees of freedom, tunnelling usually makes the ground state unique and symmetric.