How does one prove that the current of a spontaneously broken symmetry generates a particle? I am having a hard time arguing that, after spontaneous breaking of a continuous symmetry of a field Lagrangian, local fluctuations around the vacuum can be interpreted as particles (without referring to analogies from condensed matter physics). I encountered a treatment which stated that, when the symmetry is unbroken, the corresponding current annihilates the vacuum
$$J^{\mu} | 0 \rangle = 0$$
while after spontaneous symmetry breaking, the current creates a state out of the vacuum with some momentum $k^{\mu}$
$$J^{\mu} | 0 \rangle = k^{\mu} | k \rangle $$
My question is why this second equality holds only after spontaneous symmetry breaking. 
 A: It is proved by the so called Goldstone theorem which I believed in discussed in every QFT book. In any case, the intuitive picture is provided by the very definition of what a spontaneously broken (continuous) symmetry is about. 
I am going basically to tell you what the book of Tom Banks say about (but, again, any QFT book contains it; the second volume by Weinberg has e.g. a very nice discussion about it). 
A continuous (exact) symmetry always leaves the action (that is the dynamical laws) invariant. This gives rise to a conserved current $J_\mu$ with $\partial_\mu J^\mu=0$ on the equations of motion. On the other hand, should the symmetry be broken spontaneously, the correlation functions such as 
$$\langle0|T \phi_{i_1}(x_1)\phi_{i_2}(x_2)\ldots|0\rangle $$
are not invariant under the action of the symmetry because not all the symmetry generators  annihilate the vacuum. This is possible only iff
$$
\int d^4x \partial_\mu \langle0|T J^\mu(x)\Phi(y)|0\rangle\neq 0
$$
for some local (possibly composite) operator $\Phi(x)$ charged under the symmetry for every  broken currents.
Taking the Fourier transform of this expression you get that 
$$
k_\mu \Gamma^\mu(k)=k^2 \Gamma(k)\neq 0 \qquad \mbox{for } k\rightarrow 0
$$
which means that the $J^\mu-\Phi$ two point function has a simple pole $1/k^2$ at $k=0$. In other words, the currents generate massless particles when acting on the vacuum, $J^\mu(x)|0\rangle\sim e^{ikx}k^\mu|k\rangle$
