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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 305 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-305/1/a/solution