# In a $\phi^4$ theory, does the single vertex Feynman diagram (with no internal propagators) contribute to the scattering amplitudes? [closed]

If I have an interaction term in the Lagrangian of the form:

$$L_{int}=g\frac{1}{4}\phi^4 + g'\phi^3\chi + \dots$$

How does the trivial diagram (i.e. just a cross with a vertex at the center) contribute to the scattering amplitude? $$(2\pi)^4\delta^4(p_3+p_4-p_1-p_2)A=\langle \phi(p_3)\phi(p_4)|T|\phi(p_1)\phi(p_2)\rangle$$

where $A$ is the scattering amplitude and $S=1-iT$. It seems like it should contribute a factor proportional to $g$ since there are no internal propagators, but I'm confused as how to calculate an numerical factors.

## closed as unclear what you're asking by ACuriousMind♦, Martin, Danu, Sebastian Riese, Daniel GriscomFeb 20 '16 at 18:24

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• It's the tree-level Feynman diagram, why would you think it does not contribute? – ACuriousMind Feb 13 '16 at 15:46
• In your normalization the amplitude will simply be $6i g$. There is a symmetry factor of $4!$ in the diagram. – Prahar Feb 13 '16 at 15:48
• Just realized that one normally has a factor of $4!$ dividing the $\phi^4$ term, not just $4$, which is where my mistake came in. Thanks! – user138901 Feb 14 '16 at 15:16