Massless spin 1 polarization

Question

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?

Answer 1

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 $p^\mu M_\mu = 0$.

Your suggestion to restrict the ISO(2) to group members which mix only the transverse polarizations is not viable because it would jeopardize the unitary and irreducible representation of a massless spin 1 particle.