When you have a symmetry, we can find the conserved current $J$ and the cut-off charge $Q_f = \int_\Sigma f J$, where $\Sigma$ is a spatial hypersurface and $f:\Sigma\rightarrow\mathbb{R}$ is a Schwarz function. Note that we cannot use $f=1$ in infinite volume, which is the case we must consider since there is no spontaneous symmetry breaking in finite volume.
If the symmetry is spontaneously broken, then there is a realisation of the theory above some vacuum state $|0\rangle$ and an n-point operator $\phi_n(x_1,...,x_n)$ which is not preserved by the symmetry. Using the OPE, we can argue that we can also find a 1-point operator $\phi(x)$ which is not preserved, ie.
$$\langle 0|[Q_f,\phi(0)]|0\rangle\neq0.$$
for a suitable $f$. This implies
$$\langle 0| [J(x),\phi(0)] |0\rangle\neq 0$$
for some $x$. Now look at $M_+(x) = \langle 0| J(x)\phi(0) |0\rangle$, $M_-(x) = \langle 0| \phi(0)J(x) |0\rangle$. By what I just said, $M_+ \neq M_-$.
Using the spectral decomposition of the Hilbert space of the theory we can write
$$M_{\pm}=\int_{V_+}K_\pm(x,p)dp ,$$
the integral being over the forward light cone in momentum space. Lorentz invariance requires these guys are of the form
$$K_\pm = p e^{\pm ipx}\rho_\pm (-p^2).$$
We can write
$$M_\pm (x) = i \frac{d}{dx} \int_0^\infty \rho_\pm (m^2) W_m (\pm x)dm^2$$
where $W_m$ is the Klein-Gordon propagator
$$\int_{O^+_m}e^{ipx}dp$$
the integral being over the positive mass shell with mass $m$.
Now Let $M=M_+ -M_-$. For spacelike $x$, $W_m(x)=W_m(-x)$, so we have
$$M(x) = i \frac{d}{dx} \int_0^\infty (\rho_+ (m^2) - \rho_- (m^2))W_m(x)dm^2$$
By Lorentz invariance, for spacelike $x$, $\phi(0)$ and $J(x)$ commute, so $M(x)=0$. The formula above then implies $\rho_+ = \rho_- := \rho$. In general then,
$$M(x) = i \frac{d}{dx} \int_0^\infty \rho(m^2)(W_m(x)-W_m(-x))dm^2.$$
Now we take the derivative of both sides and use the Klein-Gordon equation $\frac{d}{dx}^2 W_m = -m^2W_m $ :
$$\frac{d}{dx}M(x) = \int_0^\infty m^2 \rho(m^2)(W_m(x)-W_m(-x))dm^2.$$
This should be zero since the current is conserved. This then implies (take the Fourier transform)
$$p^2 \rho(-p^2)=0.$$
The only possibility is $\rho(m^2)=c \delta(m^2)$, where c is some (necessarily nonzero, since the commutator above must be nonzero) constant. We have shown that all the contributions to the 2-point function $\langle 0| J(x)\phi(0)|0\rangle$ come from intermediate states of zero mass. Thus, $L^2(O_0^+)$ is contained in the discrete spectrum of the Hilbert space as a subrepresentation of the Poincare group. in other words, the spectrum includes a massless scalar field : it's the Goldstone boson!
In conclusion, the Noether charge of a spontaneously broken symmetry (not a gauge symmetry, for which the argument does not apply) generates the Goldstone boson when acting on the noninvariant ground state.
Reference : IAS course on fields and strings vol 2, Witten's lecture 1
