 Solution

Put $q^a=(E,0,0,k)$ with $E=\sqrt{k^2+m^2}$. An orthonormal real basis of polarization vectors isEach polarization vector obeys $q_a{ϵ}_{\lambda }^a=0$. The Minkowski metric gives ${ϵ}_{\lambda }\cdot {ϵ}_{{\lambda }^{\prime }}={\delta }_{\lambda {\lambda }^{\prime }}$, since $E^2-k^2=m^2$. The first two are transverse to the spatial momentum; the longitudinal polarization of a massive vector boson is spatially longitudinal but still orthogonal to the full four-momentum. The polarization sum for a massive vector boson isIn the rest frame all three polarization vectors are spatial unit vectors. Their three independent positive-norm states are the physical spin states of a massive spin-one particle.