= Solution
For a null momentum $q^\mu$, impose transversality
$$
q_\mu\epsilon^\mu(q,\lambda)=0.
$$
This removes one of four vector components. The remaining equivalence
$$
\epsilon^\mu\sim\epsilon^\mu+\alpha q^\mu
$$
removes a second component, leaving two transverse polarizations. For momentum along the third axis they may be chosen as
$$
\epsilon_\pm^\mu=\frac1{\sqrt2}(0,1,\pm i,0),
$$
which carry helicity $\pm1$.
The equivalence is a <gauge redundancy>: vectors that differ by a multiple of $q^\mu$ describe the same physical state rather than distinct measurable configurations. A Lorentz transformation of a chosen transverse representative can require a compensating gauge transformation. Therefore the amplitude
$$
\mathcal M=\epsilon_\mu\mathcal M^\mu
$$
must be invariant under $\epsilon_\mu\mapsto\epsilon_\mu+\alpha q_\mu$. For arbitrary $\alpha$, this is exactly the <Ward identity>
$$
q_\mu\mathcal M^\mu=0.
$$
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