4
$\begingroup$

I want to obtain the Feynman rule for the 3-gluon vertex, but looking at the result I don't really know how to tackle it.

enter image description here

The relevant term in the Lagrangian is

\begin{equation} \mathcal{L}_\text{QCD} \supset \frac{1}{2}g_sf^{abc}(\partial_\mu G_\nu^a - \partial_\nu G_\mu^a)G_b^\mu G_c^\nu \end{equation}

and the Feynman rule should be something similar to

\begin{equation} -ig_sf_{abc}[g_{\mu\nu}(p-q)_\sigma + g_{\nu\sigma}(q-r)_\mu + g_{\sigma\mu}(r-p)_\nu]. \end{equation}

I assume that it has something to do with the most general way to write it with the tensors available, but I don't know how to reason it. Could you give me some orientation?

Improve this question
$\endgroup$
3
  • $\begingroup$ Hint: particle 4-momenta $p$, $q$ and $r$ arise after the Fourier transform of the coordinate-space vertex (which should be easy to read off from the Lagrangian) as $p_{\mu} = -i \partial_{\mu}$. $\endgroup$
    – Prof. Legolasov
    Commented Dec 9, 2018 at 15:36
  • $\begingroup$ I would find an answer to this question useful. Altough it is clear that one should Fourier transform the fields. The result is not inmediately clear after that. Both the relative sign of the momenta and the particular index contractions don't arise that easily. $\endgroup$
    – P. C. Spaniel
    Commented Mar 1, 2019 at 1:15
  • $\begingroup$ One way to see this is to consider taking functional derivatives of the (free) generating function. Essentially you take your term in the interaction lagrangian, Fourier transform and swap out your fields for functional derivatives with the momenta and indices given in your diagram. The Feynman rule is just all possible terms you get from applying this derivative. $\endgroup$
    – acernine
    Commented Apr 15, 2023 at 21:08

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.