The scalar 3-point function (1-loop correction) for $\phi^3$ theory could be found using the following diagram:

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$$ iV_3 = (-ig)^3\int\frac{d^4k}{(2\pi)^4}\frac{i^3}{[k^2-m^2][(k+p_1)^2-m^2][(k+p_1+p_2)^2-m^2]}. $$

Does this look right? On my solution, it was written as

$$ iV_3 = (-ig)^3\int\frac{d^4k}{(2\pi)^4}\frac{i^3}{[k^2-m^2]^3}. $$

Therefore I'm confused about whether we should include the external momenta when finding the scalar 3-point function for $\phi^3$ theory. These momenta are included in the 2-point function, so maybe the reason for dropping external momenta here is for simplicity and approximation purpose?

  • 2
    $\begingroup$ Your loop integral looks fine. No idea why the external momenta are missing in your "solution". $\endgroup$
    – Hyperon
    Dec 27, 2022 at 14:58
  • $\begingroup$ @Hyperon Thank you!! $\endgroup$
    – IGY
    Dec 27, 2022 at 15:20

2 Answers 2


It really depends on what you are interested in. If you really want a three-point function $\langle \phi(x_1)\phi(x_2)\phi(x_3)\rangle$, be it in position or momentum space, then the external lines do matter.

If you are interested in the 1PI graph which will give you corrections to the vertex then the external lines should be amputated. The rationale is that you are really correcting the vertex so that the full three-point function will end up being the result of attaching the external lines to the corrected vertex.


OP's 1-loop diagram makes sense for generic external momenta (that satisfy total momentum conservation). This diagram is e.g. done in Srednicki [1]. Returning to OP's title question, it seems relevant to mention that Srednicki considers a renormalization condition (16.14) where the external momenta are zero.


  1. M. Srednicki, QFT, 2007; chapter 16. A prepublication draft PDF file is available here.

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