Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$$ T_{\mu \nu} = \frac{q_\mu q_\nu}{M^2} $$


$$ T_{1\lambda} = T_{\lambda1} = T_{2\lambda} = T_{\lambda2} = 0 $$

$$ T_{00} = \frac{|q|^2}{M^2} \left( \frac{q_0-E_q}{M} \right)^2 $$

$$ T_{ss} = 1 + \frac{q_0^2-E_q^2}{M^2} + \frac{|q|^2}{M^2} \left( \frac{q_0-E_q}{M} \right)^2 $$

$$ T_{0s} = T_{s0} = - \frac{|q|}{M} \left( \frac{q_0-E_q}{M} \right) - \frac{E_q|q|}{M^2}\left(\frac{q_0-E_q}{M}\right)^2 $$

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

$$ T_{ss} \ni \frac{q_0^2-E_q^2}{M^2} = \frac{q_0+E_q}{M} \frac{q_0-E_q}{M} $$


$$ T_{0s} = T_{s0} \ni - \frac{|q|}{M} \left( \frac{q_0-E_q}{M} \right) $$

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.

share|cite|improve this question
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. – Trimok Jan 2 '14 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. – jwimberley Jan 2 '14 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. – Trimok Jan 2 '14 at 20:05
Moreover, I don't understand you relations $q_\mu \epsilon^\mu(q)$.Here $q_\mu$ is on-shell, so $q_0= E_q$ – Trimok Jan 2 '14 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. – jwimberley Jan 2 '14 at 20:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.