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

[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 don't 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.

 

[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 $t^a$.

The classical QCD Lagrangian reads
$$\mathrm{L}=-\frac{1}{4} F_{\mu \nu}^a F^{a \mu \nu} + \overline{\psi}(\mathrm{i} \mathrm{D}_{\mu} \gamma^{\mu} -M) \psi,$$
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.

In a more graphical way you can think of the vertex as the contraction of the four gluon fields in the interaction Lagrangian written above with the external-field points with the external lines amputated. Then you can collect all the combinations by just building up this Feynman diagram. Note that Peskin/Schroeder uses the usual convention to give the sum of all these contractions. This you have to take into account when using the Feynman rules and determining the symmetry factors of the diagram.