 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 $V(\varphi )$ be the invariant scalar potential and let $v_i$ be a vacuum. Infinitesimal invariance givesDifferentiate with respect to ${\varphi }_k$ and evaluate at the stationary point $v$, where ${\partial }_{i}V(v)=0$. The scalar mass matrix then obeysFor every broken generator, $(T^{a}v{)}_i\ne 0$, so $T^{a}v$ 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 hasWrite $Q_a=\int d^{3}x{j}_a^0(x)$ 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 asThe 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.