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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.

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

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    $\begingroup$ It's the tree-level Feynman diagram, why would you think it does not contribute? $\endgroup$ – ACuriousMind Feb 13 '16 at 15:46
  • $\begingroup$ In your normalization the amplitude will simply be $6i g$. There is a symmetry factor of $4!$ in the diagram. $\endgroup$ – Prahar Feb 13 '16 at 15:48
  • $\begingroup$ 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! $\endgroup$ – user138901 Feb 14 '16 at 15:16