QCD gluon propagator in axial gauge, polarization sumby tobias_ Tags: axial gauge, polarization sum, propagator, qcd 

#1
Apr111, 12:17 PM

P: 2

Hi!
I have a process with multiple feynman diagrams where gluon propagators occur. When I use an axial gauge for the gluon propagator, do I have to use the same nvector for every propagator? Following this I wonder whether I can use the same nvector for every polarization sum in axial gauge or have to take different ones. Thanks, Tobias gauge field propagator in general axial gauge: [tex] G_{\mu\nu}^{ab}(q,\alpha)=\frac{i\delta^{ab}}{q^{2}}\left(g^{\mu\nu}\frac{q_{\mu}n_{\nu}+q_{\nu}n_{\mu}}{qn}+q_{\mu}q_{\nu}\frac{n^{2}+\alp ha q^{2}}{(qn)^{2}}\right) [/tex] 



#2
Apr111, 02:58 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

It seems pretty clear that in deriving the propagator above, one assumes that the vector [tex]n_\mu[/tex] doesn't vary with position. If it did, you would have to compute it's Fourier transform and the momentum space Feynman rules would be more complicated. Similarly, [tex]n_\mu[/tex] is independent of gauge indices, so there's no freedom to choose different values within a given computation.




#3
Apr111, 04:23 PM

P: 669

You can use a different gauge for the internal gluons (different from the external legs), but each internal gluon must be in the same gauge. Changing the vector n is a form of gauge transformation. 



#4
Sep2012, 04:42 AM

P: 3

QCD gluon propagator in axial gauge, polarization sum
I'm fully aware that this post is over one year old, but could someone provide a source for the above statement or at least scetch if and why this is true?



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