# Non-covariant parts of the propagator?

1. Sep 16, 2008

### ismaili

I don't quite understand what's the origin of the non-covariant parts of the propagator.

The propagator can be calculated to be
$$\Delta_{\ell m}(2\pi)^{-4}\int d^4q\frac{P_{\ell m}(q)\,e^{iq\cdot(x-y)}}{q^2+m^2-i\epsilon}\quad\cdots(*)$$
where
$$P_{\ell m}(q)\equiv 2\sqrt{\mathbf{q}^2+m^2}\sum_{\sigma}u_\ell(\mathbf{p},\sigma)u^*_m(\mathbf{p},\sigma)\quad\cdots(**)$$ is the spin sum of the field coefficients (coefficients of annihilation operator).
Eq(*) is an integral over all the possible four momentum $$q$$, but the polynomial $$P_{\ell m}(q)$$ defined by eq(**) is "on-shell", i.e. $$q^2=-m^2$$; hence, we have to extend the definition of the polynomial $$P_{\ell m}$$. Note that if $$q$$ is on-shell, we can always write $$P_{\ell m}(q)$$ as a polynomial "linear" in $$q^0$$. Hence, we define the generalized polynomial as a linear function of $$q^0$$, and the polynomial transforms covariantly, i.e.
$$P^{(L)}_{\ell m}(\Lambda q) = D_{\ell\ell'}(\Lambda)D^*_{mm'}(\Lambda)P^{(L)}_{\ell'm'}(q)$$, where $$D(\Lambda)$$ is certain representation of the Lorentz group. Now consider the polynomial of a massive vector field, $$P^{(L)}_{\ell m}(p) = \eta_{\mu\nu} + m^{-2}q_\mu q_\nu$$, since there is a quadratic term in $$q^0$$, hence actually the correct polynomial should be $$P^{(L)}_{\mu\nu} = \eta_{\mu\nu} + m^{-2}\left[q_\mu q_\nu - \delta^0_\mu\delta^0_\nu(q^2+m^2)\right]$$.
We substitute this polynomial into eq(*), hence we get the propagator for massive vector field is
$$(2\pi)^{-4}\int d^4q\frac{(\eta_{\mu\nu} + m^{-2}q_\mu q_\nu)e^{iq\cdot(x-y)}}{q^2+m^2-i\epsilon} + m^{-2}\delta^{(4)}(x-y)\delta^0_\mu\delta^0_\nu$$, the second term is the non-covariant part of the propagator.

What I don't understand is, why should we define the generalized polynomial in a way like what he did? Why the linear in $$q^0$$ so important? I'm confused and I don't get the logic and the reasoning, is there someone who can instruct me? These stuff is contained in p.277 or so basically. Any ideas would be appreciated.

2. Feb 6, 2012

### makehc

You've probably already resolved this, but:

the extension of the polynomial doesn't need to be defined to be linear in q^0. Weinberg uses that form purely to simplify the manipulation of moving the derivative operators to the left of the theta functions in (6.2.12) (Note the second term's dependence on P^(1)_{lm}.) Then, he rewrites the result in terms of a covariant polynomial and a non-covariant term in (6.2.20,.21).