# In a scalar QFT, how are Feynman diagrams drawn if the interaction Lagrangian has multiple terms?

In my introductory course for QFT we have covered many different interaction Lagrangians using scalar fields, for example $$\phi^3$$ theory. However, so far we've only covered Lagrangians with a single interaction term. Consider now an expression of the sort $$\mathcal{L}_I=-\frac{a}{3!}\phi^3 - \frac{b}{4!}\phi^4.$$

As you can see it has two terms. I don't fully understand how I would then draw the Feynman diagram for a simple process such as $$\phi(p_1)+\phi(p_2) \to \phi(p_3) + \phi(p_4) .$$

In $$\phi^4$$ theory this can be computed at first order while it requires second order in $$\phi^3$$. If both interactions are considered, how would one draw the diagrams and therefore compute the result? I assume I would need one vertex for the $$\phi^4$$ term and two for the $$\phi^3$$ term. Does this mean that every diagram has three vertices? Or does it mean that you can either have diagrams with one vertex or diagrams with two vertices? Or does it mean something different?

• draw the Feynman diagram for a simple process An infinite number of Feynman diagrams contribute to that “simple process”. The two interaction terms mean that there are two kinds of vertices that can appear in them. Calculate to whatever order in $a$ and $b$ that you want, or calculate to a given loop order. Commented May 1 at 0:53

Briefly speaking, since the interaction terms enter additively in the action $$S[\phi] ~:=~\underbrace{S_2[\phi]}_{\text{quadratic part}} + \underbrace{S_{\neq 2}[\phi]}_{\text{interactions}}, \tag{1}$$ their effects enter multiplicatively/distributively in the path integral/the generating functional for all diagrams \begin{align} Z[J]~=~&\int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S[\phi] +J_k \phi^k \right)\right\} \cr ~\sim~& \underbrace{\exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}}_{\text{bag of vertices}} \underbrace{\exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}}_{\text{bag of propagators}} , \end{align}\tag{2} cf. my Phys.SE answer here. Correlation functions behave very similarly. The upshot is that all possible diagrams should be included in calculations -- usually organized perturbatively according to the various coupling constants.