Derivation of the Yang-Mills 3 gauge boson vertex

  • #1

Summary:

I can't seem to derive the vertex rules from the Yang-Mills lagrangian. I struggle to properly identify the origin of each momentum and the indices associated.
Hello everyone,

I am stuck in the derivation of the three gauge-boson-vertex in Yang-Mills theories. The relevant interaction term in the Lagrangian is


$$\mathcal{L}_{YM} \supset g \,f^{ijk}A_{\mu}{}^{(j)} A_{\nu}{}^{(k)} \partial^{\mu} A^{\nu}{}^{(i)} $$


I have rewritten this term using the total asymmetry of the structure constants:

$$\mathcal{L}_{YM} \supset \dfrac{g}{6} f^{ijk} \left[ A_{\mu}{}^{(j)} A_{\nu}{}^{(k)} (\partial^{\mu} A^{\nu}{}^{(i)} - \partial^{\nu} A^{\mu}{}^{(i)}) + A_{\mu}{}^{(i)} A_{\nu}{}^{(k)} (\partial^{\mu} A^{\nu}{}^{(j)} - \partial^{\nu} A^{\mu}{}^{(j)}) + A_{\mu}{}^{(k)} A_{\nu}{}^{(i)} (\partial^{\mu} A^{\nu}{}^{(j)} - \partial^{\nu} A^{\mu}{}^{(j)})\right] $$


Now consider the following diagram (from Peskin & Schroeder, section 16.1): View attachment 250501

This is where I'm stuck: I know that the derivative of the field will make the momenta appear in the expression. The problem is that I do not understand which gauge field momentum appears from which derivative, and how to go from this expression to the answer, which is

$$ g f^{abc} \left[ g^{\mu \nu} (k-p)^\rho + g^{\nu \rho} (p-q)^\mu + g^{\rho \mu} (q-k)^\nu \right] $$

where the momenta and indices are taken according to the attached diagram.

Huge thanks for any help you might bring!
 

Answers and Replies

  • #2
Orodruin
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In all honesty, writing out the anti-symmetrization explicitly will not help you much, it will rather serve to complicate things.

The main thing to keep in mind here is that there are three different possibilities for which field to connect to the derivative term. Each of these possibilities contribute with its own term to the Feynman rule - where the momentum is the momentum corresponding to that possibility - and in each of these possibilities you have two different possibilities for connecting the other fields so in total you have six terms. Just collecting these six terms in an appropriate manner should give you the correct Feynman rule.
 

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