# Massless spin 1 polarization

Reading Schwartz's book on Quantum Field Theory, one finds on page 120 that the forward polarization for a massless spin 1 field is $$\epsilon_{f}\propto p_{\mu}$$. This is equivalent to $$A_{\mu}=\partial_{\mu}\phi$$ and hence it corresponds to pure gauge and is therefore not normalizable.

When he talks about massless spin fields later on page 126, he follows a similar train of thought to the massive case and assumes a mixing between polarization vectors given by some linear combination of them, since they are closed under the Lorentz group. However the longitudinal mode now became the forward polarization, i.e. $$\propto p_{\mu}$$ in the limit of $$m\rightarrow 0$$.

Using the same argument as was used for the massive field case, that the time-like polarization ($$p_{\mu}$$) doesn't mix with the other three $$\epsilon_{i}$$ as it corresponds to pure gauge, why can't we use the same argument to the massless case and say that the two physical $$\epsilon_{i}$$ won't mix with $$p_{\mu}$$?

$${\rm ISO}(2)$$ is the little group for the massless case. Are there no group members that mix only the physical transverse polarizations, as in the massive case where we have $$SO(3)$$, which mixes only 3 non time-like polarizations?

• Please try to pose questions with the best grammar possible and exposing ideas in the clearest way possible. Keep the quality of the questions high. Aug 14, 2020 at 8:57

The theory of induced representations requires, for a massless spin 1 particle, the little group preserving the momentum $$p^\mu$$ to be ISO(2). However, as there are members of the ISO(2) which mix the transverse polarizations $$\epsilon^\mu_1$$ and $$\epsilon^\mu_2$$ with the forward polarization $$\epsilon^\mu_f \propto p^\mu$$, in order to have Lorentz invariant scattering matrix elements $$\mathcal M = \epsilon^\mu M_\mu$$, it is requested the Ward identity $$p^\mu M_\mu = 0$$.