[Physics] Deriving Feynman rules from a Lagrangian for vertex factors for “more complicated” interactions

feynman-diagramsparticle-physicsquantum-chromodynamicsquantum-field-theorysupersymmetry

I am trying to derive Feynman rules from a given Lagrangian and I got stuck on some vertex factors. What for example is the vertex factor that corresponds to the four-scalar interaction that is decribed by the following Lagrangian?

\begin{equation}
L = -\frac{1}{4} g_3^2 \phi^\dagger \lambda^a \phi \chi^\dagger \lambda^a \chi + \frac{2}{9} g_1^2 \phi^\dagger \phi \chi^\dagger \chi \,,
\end{equation}

where $\phi,\chi$ are complex scalar (color triplet) fields, $\lambda^a$ are the Gell-Mann matrices, and $g_1,g_3$ are the coupling constants corresponding to $\text{U}(1)$ and $\text{SU}(3)$ respectively.

If we would have only had the second term here, say, then the vertex factor would simply be found by "dropping" the fields and multiplying by $i$. But now there are two terms contributing, and in the first term the Gell-Mann matrices even mix the color components of the scalar triplets. So how do I proceed in this case?

And could anyone give me some general strategies on how to derive vertex factors for "complicated" interactions? For example, I also find it tricky to get the sign right if there is a derivative in an interaction.

(If you are interested in the context of this Lagrangian, for $\phi = \tilde{u}_R$ and $\chi = \tilde{d}_R$ this Lagrangian describes the interaction between two up squarks and two down squarks in a supersymmetric theory.)

Best Answer

You can compute the feynman rule for the $\phi$-$\phi$-$\chi$ vertex by taking $$e^{-i \int \mathrm d^4x L_\mathrm{full} }\frac{\delta}{\delta \phi^a} \frac{\delta}{\delta \phi^b} \frac{\delta}{\delta \chi^c} e^{i\int \mathrm d^4x L_\mathrm{full}}$$ where $L_\mathrm{full}$ is the sum of the free and interaction Lagrangeans and afterwards remove any propagator connecting to external degrees of freedom.