= Solution
Put $q^a=(E,0,0,k)$ with $E=\sqrt{k^2+m^2}$. An orthonormal real basis of <polarization vectors> is
$$
\boxed{\epsilon_1^a=(0,1,0,0),\quad
\epsilon_2^a=(0,0,1,0),\quad
\epsilon_L^a=\left(\frac{k}{m},0,0,\frac{E}{m}\right)}.
$$
Each <polarization vector> obeys $q_a\epsilon_\lambda^a=0$. The <Minkowski metric> gives $\epsilon_\lambda\cdot\epsilon_{\lambda'}=\delta_{\lambda\lambda'}$, 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> is
$$
\sum_{\lambda=1,2,L}\epsilon_\lambda^a\epsilon_\lambda^b
=\eta^{ab}+\frac{q^aq^b}{m^2}\qquad(q^2=-m^2).
$$
In 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.
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