# Non-abelian gauge fields 3-vertex and 4-vertex?

1. Jun 26, 2015

### moss

I want to understand the 'vertex factor' of 3-bosons field and 4-bosons field but get confused.
(I know the lagrangian and have computed the interaction vertices already) only need to understand the vertex factor.

In other words, I want to learn how 3-boson vertex and 4-boson vertex are computed in Feynman diagrams.

It is, for example, given in Peskin&Schrioeder on page 507 but I dont get it as to what is permuted there, the color indices a,b,c or the lorentz indices and the momentum?

any help/hint is much appricaited !
Thanks.

2. Jun 28, 2015

### vanhees71

Ok, let's see how to sort this out. The gauge-covariant derivatives, acting on the quark fields read
$$\mathrm{D}_{\mu}=\partial_{\mu} - \mathrm{i} g t^a A_{\mu}^a,$$
where $t^a=\lambda^a/2$ are the generators of the su(3) Lie algebra in its fundamental representation.

The non-Abelian Faraday tensor is given by
$$F_{\mu \nu}^a=\partial_{\mu} A_{\nu}^a-\partial_{\nu} A_{\mu}^a + g f^{abc} A_{\mu}^b A_{\nu}^c.$$
Since SU(3) is a semisimple Lie group, the structure constants, defined by
$$[t^a,t^b]=\mathrm{i} f^{abc} t^c$$
can be made totally antisymmetric by a clever choice of the Lie algebra's basis, and this is the case for our choice of this basis $t^a$.

$$\mathrm{L}=-\frac{1}{4} F_{\mu \nu}^a F^{a \mu \nu} + \overline{\psi}(\mathrm{i} \mathrm{D}_{\mu} \gamma^{\mu} -M) \psi,$$
where $\psi$ consists of 6 quarks with three colors each. The $t^a$ contained in $\mathrm{D}_{\mu}$ act on the color indices. $M$ is a diagonal real matrix in flavor space.
When quantizing the theory, you have to introduce Faddeev-Popov ghosts, but these do not affect the quark-gluon and the three- and four-gluon vertices (at least not for the usually used gauges). Thus you can read off the Feynman rules immideately from the Lagrangian. The three- and four-gluon vertices come from the first term in the Lagrangian. You have to multiply out the terms. The four-gluon contribution to $\mathrm{L}$ reads
$$\mathrm{L}_{4g}=-\frac{1}{4} f^{eab} f^{ecd} A_{\mu}^a A_{\nu}^b A_{\rho}^{c}A_{\sigma}^{d} g^{\mu \rho} g^{\nu \sigma}.$$
Now to read off the Feynman rule for the corresponding vertex, you have to symmetrize/anti-symmetrize the coefficient in front of the four fields according to the symmetry of the objects. The $f^{abc}$ are totally antisymmetric, and this gives the 12 expressions with the corresponding signs in the Feynman rule given in Fig. 16.1.