In non-spontaneously broken QFT like QED the gauge bosons cannot have a mass due to gauge symmetry (follows from Ward identity). Also they have only 2 polarizations.
However in a spontaneously broken gauge theory the gauge boson becomes massive after the symmetry breaking. For example consider the standard U(1) example: $$L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + D_\mu\phi^\dagger D^\mu\phi - V(\phi)$$ where $V(\phi)$ is a Mexican hat potential. After the symmetry breaking the photon is massive and has 3 polarizations.
What is the difference between both cases? Why is it relevant whether the theory is spontaneously broken or not (I guess it has something to do with the Ward identity)? And why are we so sure that this cannot happen in our 'normal' theories like QED (i.e. that in the end the photon becomes massive).
Edit: Maybe I should reformulate the question a bit. The definition of a spontaneous broken theory is that $Q|\Omega\rangle \neq 0$, where $Q$ is the charge operator as given by the symmetry and $|\Omega\rangle$ the groundstate. Why does the charge of the groundstate of my theory matter at all? I can formulate gauge invariance only as a Heisenberg operator symmetry: $\hat{A^\mu} \sim \hat{A^\mu} + \partial^\mu \hat{\alpha}$. Therefore the photon should still be massless. Obviously this cannot be true because we have plenty of examples where gauge bosons become massive (e.g. W boson or superconductors).
I guess that something goes wrong with the argument that the photon cannot be massive by gauge invariance. Since I don't know a nice proof of this (Only argumentations like this by resumming feynman diagrams) I guess something goes wrong there. The question is about what exactly goes wrong? And more important: Why is the mass of the photon protected in the normal case, but not in spontaneous symmetry breaking case. What's the crucial difference?
Edit 2: It took me a while to figure it out, but I think I found a suitable nonperturbative argumentation why the photon cannot become massive. Instead of unitarity gauge work in the usual Lorentz gauge (and Feynman gauge). Then the equal time commutation relations read $[A^\mu(x), A^\nu(0)] = 0$ and $[\dot{A}^\mu(x), A^\nu(0)] = i \eta^{\mu\nu}$. Also the equation of motion is given by $\partial_\alpha\partial^\alpha A^\mu = J^\mu$. Therefore apply the equation of motion on the propagator $\langle 0|TA^\mu(x)A^\nu(0)|0\rangle$:
\begin{align} \partial_\alpha \partial^\alpha \langle 0|TA^\mu(x)A^\nu(0)|0\rangle &= \partial_\alpha \langle 0|T(\partial^\alpha A^\mu(x))A^\nu(0)|0\rangle + \partial_\alpha \left(\delta(t) \delta_{0\alpha} \langle 0|[A^\mu(x),A^\nu(0)]|0\rangle\right)\\ &= \langle 0|T(\partial_\alpha\partial^\alpha A^\mu(x))A^\nu(0)|0\rangle + \delta(t) \langle 0|[\partial_0 A^\mu(x),A^\nu(0)]|0\rangle\\ &= \langle 0|TJ^\mu(x)A^\nu(0)|0\rangle + i\eta^{\mu\nu}\delta^4(x) \end{align}
The commutator come from time ordered product and can be replaced by there equal time value because of the $\delta(t)$ in front. After a Fourier transformation this reads (call the Fourier transformed propagator $D^{\mu\nu}(p)$):
$$-p^2 D^{\mu\nu}(p) = \int d^4x e^{-ipx} \langle 0|TJ^\mu(x)A^\nu(0)|0\rangle + i\eta^{\mu\nu}$$
For $J^\mu = 0$ (free case) this simply gives the free photon propagator $D^{\mu\nu}(p) = \frac{-i\eta^{\mu\nu}}{p^2}$. Now if $J^\mu \neq 0$ multiply both sides with $p_\mu$ (which translates to $\partial_\mu$ for $J^\mu$)
\begin{align}-p^2 p_\mu D^{\mu\nu}(p) &= -\int d^4x e^{-ipx} \langle 0|T(\partial_\mu J^\mu(x))A^\nu(0)|0\rangle - \int d^3x e^{-ipx} \langle 0|[J^0(x),A^\nu(0)]|0\rangle + ip^\nu\\ &= - \int d^3x e^{-ipx} \langle 0|[J^0(0,\vec{x}),A^\nu(0)]|0\rangle + ip^\nu \end{align}
here I used $\partial_\mu J^\mu = 0$. Now as long as $[J^0(0,\vec{x}),A^\nu(0)] = 0$ (this is for example the case in the theory above since $J^0$ does not contain any $\partial_0 A^0$) we have $$-p^2 p_\mu D^{\mu\nu}(p) = ip^\nu$$ or $$p_\mu D^{\mu\nu}(p) = -i\frac{p^\nu}{p^2}$$ which means that $D^{\mu\nu}(p)$ must have a pole at $p^2 = 0$ (i.e. the photon is massless). Clearly this it not satisfied by the propagator for a massive particle $\sim \frac{1}{p^2 - m^2}$
Note that this argumentation never used any property of the ground state of our theory. In the Higgs mechanism one expands the field around its vev and ignores higher orders. After that the photon is massive (in lowest order of perturbation theory) which is a clear contradiction to my derivation above. So my first guess would be that the mass of the photon will again cancel when taking into account all feynman diagrams.
