Peskin and Schroeder say that the Ward Identity of QED proves that non-transverse photon polarizations can be consistently ignored, but I'm confused about the details.
Setup
One starts by considering some process with an external photon whose momentum is chosen to be $k^\mu=(k,0,0,k)$ and let the two transverse polarization vectors be $\epsilon_1^\mu=(0,1,0,0)$ and $\epsilon_2^\mu=(0,0,1,0)$. The Ward identity tell us that if the amplitude for the process is $\mathcal{M}=\mathcal{M}^\mu\epsilon_\mu(k)$, where we've factored out the polarization vector for the external photon in consideration, then the amplitude obeys $\mathcal{M}^\mu k_\mu=0$, on shell. With our setup, this simply tells us $\mathcal{M}^0=\mathcal{M}^3$. If we then calculate the square of the amplitude and sum over external polarizations we'd find $|\mathcal{M}|^2=\sum_{i\in\{1,2\}}\epsilon_{i\mu}\epsilon_{i\nu}^*\mathcal{M}^\mu\mathcal{M}^{*\nu}=|\mathcal{M}^1|^2+|\mathcal{M}^2|^2$. Due to the Ward identity, this is equal to $-\eta_{\mu\nu}\mathcal{M}^\mu\mathcal{M}^{*\nu}$ and so we can make the replacement $\sum_{i\in\{1,2\}}\epsilon_{i\mu}\epsilon_{i\nu}\to -\eta_{\mu\nu}$. Peskin and Schroeder claim (pg 160-161) that this is proof that non-transverse photons can be consistently ignored.
Multiple Questions
1) P&S appear to be claiming that had we also summed over non-transverse polarizations we would find that the polarization sum just turns into $-\eta_{\mu\nu}$. However, if I use the two vectors $\alpha_1^\mu=(1,0,0,0)$ and $\alpha_2^\mu=(0,0,0,1)$ as the basis for the non-transverse polarizations then it would appear that the polarization sum would turn into $\delta_{\mu\nu}$ rather than $-\eta_{\mu\nu}$. How does one know that there should be an extra minus sign so that $|\mathcal{M}^0|^2$ comes in with a minus sign relative to all the other squared amplitudes? That is, I'd somehow know that the proper calculation is given by $\sum_{\rm all\ polarizations}|\mathcal M|^2=\mathcal{M}^\mu\mathcal{M}^{*\nu}(\epsilon_{1\mu}\epsilon^*_{1\nu}+\epsilon_{2\mu}\epsilon_{2\nu}^*-\alpha_{1\mu}\alpha_{1\nu}^*+\alpha_{2\mu}\alpha_{2\nu}^*)$, but I don't see where that minus sign would arise from. One could guess that you'd have to end up with something proportional to $\eta_{\mu\nu}$ due to Lorentz-Invariance, but that's not very satisfactory. Finally, the polarization sum is naively a sum of manifestly positive numbers, yet the P&S argument depends on some kind of cancellation between these number, so how is this possible?
2) What about scenarios where one doesn't perform the polarization sum? I thought that the polarization sum was only performed when the detector is insensitive to polarization, which is not always the case at hand. I would have thought that one could show that non-transverse polarization are unphysical without having to do a polarization sum. For example, given $\mathcal{M}^\mu$ I would have thought that we could have contracted this with one of the non-traverse polarization vectors, say $\alpha_2^\mu$, and we should find that the amplitude for this process vanishes by itself.
3) Shouldn't we be showing that we can also ignore non-transverse states in all parts of the diagram, not just on external legs? If non-transverse states can run in loops then they need to be included in physical initial and final states, due to the optical theorem, which needs to be avoided. Or are P&S claiming that they've proved that they've shown that non-transverse polarizations can be ignored in the initial and final states, so by the optical theorem they can be ignored in loops, too?