= Solution
The <Goldstone theorem> states that every spontaneously broken generator of a continuous internal global symmetry in a Lorentz-invariant quantum field theory produces a massless scalar particle.
For the classical proof, let $V(\phi)$ be the invariant scalar potential and let $v_i$ be a vacuum. Infinitesimal invariance gives
$$
\frac{\partial V}{\partial\phi_i}(T^a)_{ij}\phi_j=0.
$$
Differentiate with respect to $\phi_k$ and evaluate at the stationary point $v$, where $\partial_iV(v)=0$. The scalar mass matrix then obeys
$$
(M^2)_{ki}(T^a)_{ij}v_j=0,
\qquad
(M^2)_{ki}=\frac{\partial^2V}{\partial\phi_k\partial\phi_i}\bigg|_v.
$$
For every broken generator, $(T^a v)_i\neq0$, so $T^av$ is a zero eigenvector of the <hessian matrix>. It is a massless fluctuation tangent to the <vacuum manifold>.
For the quantum proof, spontaneous breaking means that some local field has
$$
\langle0|[\phi_i(0),Q_a]|0\rangle=i(T^a)_{ij}\langle0|\phi_j|0\rangle\neq0.
$$
Write $Q_a=\int d^3x\,j_a^0(x)$ and insert a complete set of momentum eigenstates into the current-field correlation function. <Current conservation> and Lorentz covariance imply that a scalar intermediate state couples as
$$
\langle0|j_a^\mu(0)|\pi_b(p)\rangle=if_{ab}p^\mu.
$$
The nonzero equal-time commutator requires a pole at $p^2=0$; otherwise the conserved-current spectral integral vanishes at zero momentum. Thus a massless <Goldstone boson> exists for every independent broken direction.
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