 Solution

Goldstone's theorem states that every spontaneously broken generator of a continuous global internal symmetry gives a massless scalar excitation in a Lorentz-invariant quantum field theory, subject to the standard locality and positivity assumptions.

Choose a local field $\phi$ with $⟨0|[Q^a,\phi (0)]|0⟩\ne 0$ and write $Q^a=\int d^{3}x,J_0^a$. Translation invariance givesLorentz covariance forces the contribution of a spin-zero intermediate state to have ${\rho }_{\mu }(p)=p_{\mu }\rho (p^2)$. Current conservation then implies $p^2\rho (p^2)=0$. The nonzero equal-time integral demanded by the order parameter rules out $\rho =0$, so the spectral density must contain support at $p^2=0$. Hence there is a massless one-particle pole withwhich is the required Goldstone boson. Independent broken generators give independent massless modes in the ordinary relativistic case.