Goldstone's theorem

Remarks

Proof idea: for $X$ in the Lie algebra $\mathfrak{g}$, write $X\phi$ for the derivative of $t \mapsto e^{tX}\phi$ at $t = 0$. Differentiating $U(e^{tX}\phi) = U(\phi)$ at $t = 0$ gives $dU_\phi(X\phi) = 0$ for all $\phi$. Differentiating this identity in $\phi$ at $\phi_0$, and using $dU_{\phi_0} = 0$, gives $\mathrm{Hess}\, U(\phi_0)(X\phi_0) = 0$. The linear map $X \mapsto X\phi_0$ has kernel the Lie algebra $\mathfrak{h}$ of $H$, so its image, which lies in the kernel of the Hessian, has dimension $\dim G - \dim H$.

Example: for $U(\phi) = \lambda(\abs{\phi}^2 - v^2)^2$ on $\C^n = \R^{2n}$ with $G = SO(2n)$ and $\phi_0 = (v, 0, \dots, 0)$, the stabilizer is $SO(2n - 1)$. The theorem gives $\dim SO(2n) - \dim SO(2n-1) = 2n - 1$ massless directions, the tangent directions of the sphere of minima; the radial direction has mass squared $8\lambda v^2$.

Goldstone, Salam and Weinberg proved a quantum version in 1962. In a Lorentz-invariant quantum field theory, let a continuous symmetry be generated by a conserved local current. If the vacuum is not invariant, then the theory contains massless particles. The hypotheses matter. In gauge theories the Goldstone modes can be removed by a gauge transformation and the gauge fields become massive, which is the Higgs mechanism. In two spacetime dimensions a continuous global symmetry cannot be spontaneously broken in a quantum field theory, by a theorem of Coleman, so no Goldstone bosons arise there.

Going deeper

For a student

The classical statement is pure calculus: the orbit of the ground state is a manifold of minima, and its tangent space lies in the kernel of the Hessian. The number of Goldstone bosons is the dimension of the orbit, $\dim G/H$. Pions are approximate Goldstone bosons of the chiral symmetry of quantum chromodynamics, which is broken spontaneously and also slightly broken by quark masses, so pions are light but not massless. The theorem would rule out symmetry breaking in the electroweak theory because no massless scalars are seen; the Higgs mechanism shows how gauge invariance evades it.