Spin-1 Propagator and polarization vectors

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 4K views
TriTertButoxy
Messages
190
Reaction score
0
Hi. I am stuck.
By inverting the spin-1 differential operator I was able to derive (quite easily) the propagator for the spin-1 field (in a spontaneously broken gauge theory) in the [itex]R_\xi[/itex] gauges for the arbitrary gauge parameter [itex]\xi[/itex]. The result is
[tex] \tilde{D}^{\mu\nu}(p)=\frac{-i}{p^2-m^2+i\epsilon}\left(g^{\mu\nu}-(1-\xi)\frac{p^\mu p^\nu}{p^2-\xi m^2+i\epsilon}\right).[/tex]
But now, when I try to calculate the two-point correlator using the mode expansion for the spin-1 field, I can't quite get the same answer.
[tex] \langle 0|T(\hat{A}_\mu(x)\hat{A}_\nu(y))|0\rangle=\int \frac{d^4p}{(2\pi)^4}\frac{ie^{-i p.(x-y)}}{p^2-m^2+i\epsilon}\left(\epsilon_\mu^{[0]}(\mathbf{p})\epsilon_\nu^{*[0]}(\mathbf{p})-\sum_{\lambda=1,2,3}\epsilon_\mu^{[\lambda]}(\mathbf{p})\epsilon_\nu^{*[\lambda]}(\mathbf{p})\right)[/tex]
I need to do a polarization sum, but can't quite figure out how to get the gauge-dependence in there. What are the polarization vectors? and how do I derive them?
 
Last edited:
Physics news on Phys.org
Ok. I found the answer in Greiner, Field Quantization 1996, chapter 7. The answer is very long, and I do not expect anyone to answer. Thanks to those who have thought about this problem.