= Solution
The <little group> of a massive four-momentum is $SO(3)$, so a massive spin-one one-particle state has $2s+1=3$ polarizations. For $p^\mu=(E,0,0,p)$ and metric signature $(+---)$, one convenient basis is
$$
\epsilon_\pm^\mu=\frac1{\sqrt2}(0,1,\pm i,0),
\qquad
\epsilon_L^\mu=\frac1{M_X}(p,0,0,E).
$$
All three obey $p_\mu\epsilon^\mu=0$ and $\epsilon^*\mathbin\cdot\epsilon=-1$. Their completeness relation is
$$
\sum_{\lambda=\pm,L}\epsilon_\mu^{(\lambda)}epsilon_\nu^{(\lambda)*}
=-\eta_{\mu\nu}+\frac{p_\mu p_\nu}{M_X^2}.
$$
For a massless momentum, the finite-helicity representations of the $ISO(2)$ little group carry only the two <helicity> states $\lambda=\pm1$. Gauge equivalence removes the timelike and longitudinal polarizations; correspondingly, $\epsilon_L$ has no finite $M_X\to0$ limit.
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