As an exercise for myself, I have been working on rewriting the massive vector boson propagator (unitary gauge). I have run into a problem interpreting some of the terms that stick around when the propagator is rewritten this way. Here's what I have: I've taken the unitary gauge vector propagator
$$ D_{\mu\nu}(q) = \frac{i}{q^2 - M_W^2 + i \varepsilon} \left( -g_{\mu \nu} + \frac{q_\mu q_\nu}{M_W^2} \right) $$
and projected it into its helicity components. The convention I am using is that
$$ \epsilon^1_\mu(\vec q) = (0,1,0,0) \\ \epsilon^2_\mu(\vec q) = (0,0,1,0) \\ \epsilon^0_\mu(\vec q) = \frac{1}{M}(|q|,0,0,E_q) \\ \epsilon^s_\mu(\vec q) = \frac{1}{M}(E_q,0,0,|q|) $$ where $E_q = \sqrt{M^2+|q|^2}$. These are an orthonormal set, where
$$ \epsilon^{\lambda}_\mu \epsilon^{\lambda' \mu} = - \eta_{\lambda} \delta_{\lambda \lambda'} $$ where $\eta_\lambda = 1$ for $\lambda = \pm,0$ and $-1$ for $\lambda = s$, the scalar polarization. So, it's easy to show that
$$ X_{\mu \nu} = \sum_{\lambda,\lambda'} X_{\lambda,\lambda'} \epsilon^{\lambda}_\mu \epsilon^{\lambda' }_{\nu} \Rightarrow X_{\mu \nu} \epsilon^{\lambda \mu} \epsilon^{\lambda' \nu} = \eta_\lambda \eta_{\lambda'} X_{\lambda \lambda'} $$
In particular, $$ - g_{\mu \nu} \Rightarrow g_{\lambda \lambda'} = \frac{\eta_\lambda' \delta_{\lambda \lambda'}}{\eta_{\lambda} \eta_{\lambda'}} = \frac{\delta_{\lambda \lambda'}}{\eta_{\lambda}} = \eta_{\lambda}\delta_{\lambda \lambda'} $$ Where I've run into difficulty is breaking up the transverse term.
$$ q_\mu \epsilon^{1\mu}(\vec q) = 0 \\ q_\mu \epsilon^{2\mu}(\vec q) = 0 \\ q_\mu \epsilon^{0\mu}(\vec q) = \frac{|q|}{M}(q_0-E_q) \\ q_\mu \epsilon^{s\mu}(\vec q) = \frac{1}{M}(q_0 E_q - |q|^2) $$
The last of these can be rewritten
$$ q_\mu \epsilon^{s\mu}(\vec q) = M + \frac{E_q}{M}(q_0-E_q) $$
So, the helicity components of
$$ T_{\mu \nu} = \frac{q_\mu q_\nu}{M^2} $$
are
$$ T_{1\lambda} = T_{\lambda1} = T_{2\lambda} = T_{\lambda2} = 0 $$
$$ T_{00} = \frac{|q|^2}{M^2} \left( \frac{q_0-E_q}{M} \right)^2 $$
$$ T_{ss} = 1 + \frac{q_0^2-E_q^2}{M^2} + \frac{|q|^2}{M^2} \left( \frac{q_0-E_q}{M} \right)^2 $$
$$ T_{0s} = T_{s0} = - \frac{|q|}{M} \left( \frac{q_0-E_q}{M} \right) - \frac{E_q|q|}{M^2}\left(\frac{q_0-E_q}{M}\right)^2 $$
The term equal to $1$ in the scalar polarization term cancels out the corresponding scalar polarization term $g_{ss}$. Furthermore, all of the terms proportional to $(q_0-E_q)^2$ I understand. They cancel out the pole in the propagator, since
$$ \frac{1}{q^2 - M^2 + i \epsilon} = \frac{1}{q_0^2 - E_q^2 + i \epsilon} \sim \frac{1}{q_0 - E_q + i \epsilon} \frac{1}{q_0 + E_q} $$
and after canceling out the pole they retain a factor $q_0 - E_q$ which forces them to be 0 while on-shell. These are explicitly off-shell corrections. However, I'm not sure how to interpret the terms
$$ T_{ss} \ni \frac{q_0^2-E_q^2}{M^2} = \frac{q_0+E_q}{M} \frac{q_0-E_q}{M} $$
and
$$ T_{0s} = T_{s0} \ni - \frac{|q|}{M} \left( \frac{q_0-E_q}{M} \right) $$
Naively, it appears to me that these cancel out the pole at $q_0 = E_q$, but the remaining portion does not vanish at at $q_0 = E_q$ (or at least as at $q_0$ approaches $E_q$) I think that a careful analysis of the behavior of these terms the pole might shed light on this, or that maybe it is a gauge artifact, but I am stuck.
