= 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 $\varphi$ with $\langle0|[Q^a,\varphi(0)]|0\rangle\ne0$ and write $Q^a=\int d^3x,J_0^a$. Translation invariance gives
$$
\langle0|[J_\mu^a(x),\varphi(0)]|0\rangle
=\int\frac{d^4p}{(2\pi)^4}e^{-ipx}\rho_\mu(p).
$$
Lorentz 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 with
$$
\langle0|J_\mu^a(0)|\pi^b(p)\rangle=if^{ab}p_\mu,
$$
which is the required Goldstone boson. Independent broken generators give independent massless modes in the ordinary relativistic case.
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