Spontaneous Symmetry Breaking & the Higgs
Mexican-hat potentials, Goldstone bosons, and how gauge bosons gain mass.
A symmetry of the Lagrangian need not be a symmetry of the vacuum. For a complex scalar $\phi$ with
$$V(\phi) = -\mu^2 |\phi|^2 + \lambda |\phi|^4, \quad \mu^2, \lambda > 0,$$the minimum lies on the circle $|\phi| = v = \mu/\sqrt\lambda \neq 0$. Picking any point on the circle breaks the global $U(1)$. Goldstone's theorem: each broken continuous symmetry yields a massless boson (the angular fluctuation).
If the symmetry is gauged, the would-be Goldstone is eaten by the gauge field, which acquires a mass $m_A = ev$ — the Higgs mechanism. The Standard Model uses this to give mass to $W^\pm, Z$ while preserving electromagnetic $U(1)$. The measured Higgs vacuum expectation value
$$v \approx 246 \text{ GeV}$$sets the electroweak scale, fixing $M_W \approx \tfrac{1}{2} g v$, $M_Z \approx \tfrac{1}{2}\sqrt{g^2 + g'^2}\, v$.
Interactive: ball on the Mexican hat
Roll the ball around the brim of the rotation-symmetric potential. Choosing any one direction breaks the rotational symmetry of the ground state.