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Double line notation - gluon propagator

+ 3 like - 0 dislike
66 views

Consider the following Feynman diagram for a gluon (there is momentum $k$ flowing between the two points):

enter image description here

Normally, the factor that goes along with this is $G^{ab}_{\mu\nu}(k) \ = \ \delta_{ab} \frac{- i }{ k^{2} - i \epsilon_{+} } \left( \eta_{\mu \nu} + ( \xi - 1 ) \frac{ k_{\mu} k_{\nu} }{ k^{2} } \right)$.

Using t'Hooft's double line formalism, the above diagram can be drawn as: enter image description here

My task is to give a guess as to what the corresponding factor should be. I $think$ that it's $$ G^{(\bar{i}, j)(\bar{l},k)}_{\mu\nu}(k) \ = \ \delta_{jl} \delta_{ik} \frac{- i }{ k^{2} - i \epsilon_{+} } \left( \eta_{\mu \nu} + ( \xi - 1 ) \frac{ k_{\mu} k_{\nu} }{ k^{2} } \right) $$

However, this seems a little too simple and I'm worried that I'm missing out on a detail. When I google the "t'Hooft double line formalism" I find myself a lot of material, however I find that the discussion on the vertex factors is lacking (the same goes for the three- and four- gauge boson vertices). Can somebody either help me understand if I have the correct factor or point me in the direction of some literature that can?

(I should mention that the parameter $\xi$ describes the gauge we're working in)

EDIT: I forgot to say that I'm talking about the $U(N)$ gauge group.

This post imported from StackExchange Physics at 2017-02-15 08:33 (UTC), posted by SE-user Greg.Paul
asked Dec 11, 2016 in Theoretical Physics by Greg.Paul (15 points) [ no revision ]
retagged Feb 15

1 Answer

+ 4 like - 0 dislike

First of all, there is a mistake in the indices on your diagram. $k$ must be barred, while $l$ is not. The upper unbarred index is equivalent to the lower barred index and vice versa, because one can raise and lower indices using the group (Killing) metric, so one can use them interchangeably. In what follows only the unbarred are in use.

Second, for the $U(N)$ your formula is OK, but for $SU(N)$ one has to replace $\delta_i^k \delta_j^l$ by $\delta_i^k \delta_j^l-\frac{1}{N} \delta_i^l \delta_j^k$ due to tracelessness condition. See, for example, this review.

This post imported from StackExchange Physics at 2017-02-15 08:33 (UTC), posted by SE-user Andrey Feldman
answered Dec 11, 2016 by Andrey Feldman (600 points) [ no revision ]

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