# Off-shell corrections to massive vector boson propagator in polarization form

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.

• Note that there is a link between the choice of polarizations convention (gauge), and the residue at the pole of the propagator, see this previous answer, so it has to be coherent. Commented Jan 2, 2014 at 19:42
• Thanks for the link to this answer; I'm looking over it but initially I am still confused. As I understand, equation #2 that you give is true for on-shell q. This is why I expected all the above terms to die as $q_0 \to E_q$, but instead there are these extra terms which don't appear to. In the unitary gauge, I expected there to be transverse and longitudinal polarizations that survive on-shell, plus additional off-shell corrections. Commented Jan 2, 2014 at 19:49
• Yes, it is on shell, but the propagator being covariant, the off-shell expression of the residue should be the same, and I am not sure that your definition of the polarizations are coherent with your definition of the propagator. Commented Jan 2, 2014 at 20:05
• Moreover, I don't understand you relations $q_\mu \epsilon^\mu(q)$.Here $q_\mu$ is on-shell, so $q_0= E_q$ Commented Jan 2, 2014 at 20:10
• I'm still digesting your second comment. Regarding, $q_\mu \epsilon^\mu$, no, I'm considering a general $q_\mu$, so it could be off-shell, as in a semi-leptonic decay through a virtual W. Commented Jan 2, 2014 at 20:21