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 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