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https://i.stack.imgur.com/PdDNf.jpg

I am currently studying the one-loop renormalization structure for $\phi^4$ theory and I can’t seem to understand what’s stopping me from making one loop diagrams (I have drawn them and they’re the last pictures in the imgur) using the counterterm vertex rules in renormalized perturbation theory for a two particle scattering process at the one-loop level. As a reference, I'm referring to pages 324-326 of Peskin & Schroeder. I’ve expanded out the $S$ matrix elements but still can’t see why these counterterm loop diagrams aren’t taken into account in the amplitude.

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Once you renormalize your theory, you should use the counterterms in Feynman diagrams when computing the amplitudes at loop level. In fact, that's pretty much what they are meant for. If you compute the amplitudes considering the loop diagrams alone, without the counterterms, your expressions will lead to divergences due to the infinite integrals over the momenta happening in the loops. The counterterms are meant to remove these divergences and leave your final computation with a finite (i.e., renormalized) result.

As Peskin & Schroeder mention at the top of page 326, you adjust the counterterms precisely so to balance the divergences coming from loops while satisfying the renormalization conditions. The final goal then is exactly to use these counterterms in calculations as a way to get a finite result.

As for the diagrams you drew, they end up being of higher order. Since the counterterm vertex must balance out the loop contributions, it is itself of one-loop order. Hence, at one-loop at least, those diagrams won't enter the computation because they would be of order $\mathcal{O}(\lambda^3)$ and higher.

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  • $\begingroup$ Just to clarify my understanding, (while referring to the amplitude Feynman diagrams on page 326) we first draw all the Feynman diagrams up to the one loop level which are the first four Feynman diagrams drawn. We then add appropriate counter terms(last diagram) while satisfying renormalization conditions to balance loop divergences. Which is why I wouldn’t want to add one loop diagram with a counter term vertex -i\lambda__{z} and a -i\lambda vertex as as it wouldn’t satisfy the renormalization condition and I would get a divergence again within the loop right ? $\endgroup$
    – Spoonszzz
    Aug 8, 2021 at 2:29
  • $\begingroup$ @Spoonszzz Ohhhh I apologize. I understood your question wrongly (I think I looked at the wrong image). Those diagrams won't enter the computation because they'll be of higher order (since the counterterm vertex is balancing the loops, it should itself be of one-loop order, even though it doesn't have a loop in the drawing). I will update my answer. $\endgroup$ Aug 8, 2021 at 4:13
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Counter-term diagrams enter in one order higher than regular diagrams, because they are designed to take care of divergences that first enter at 1-loop level. If you are expanding in $\lambda$ for example, the lowest order contribution to the counter-term is $O(\lambda^2)$. Hence the CT loop diagrams you've drawn will contribute, but they only need to be taken into account at 4th order in perturbation theory.

For example, see Chapter 5 of these notes (https://www.damtp.cam.ac.uk/user/dbs26/AQFT.html) for one particular choice of $\delta\lambda$ (eqn 5.15).

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  • $\begingroup$ Shouldn’t counter-term diagrams enter one order lower than regular diagrams instead of one order higher so that the renormalization condition is a satisfied if I’m understanding correct?. Also I’ve looked at the notes you’ve posted and it helped my understanding thanks. My fault was misunderstanding the purpose of counter terms and putting a lot of thought into the CT diagram itself and not it’s purpose but now I realize why the CT diagrams I’ve drawn aren’t taken into account at 1-loop level and instead are taken into account at 3-loop level $\endgroup$
    – Spoonszzz
    Aug 8, 2021 at 3:32
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    $\begingroup$ What I mean is that the CT diagrams contribute at higher levels (numbers of loops) than you would expect if the CT vertex was a regular vertex, due to $\delta\lambda$ having quadratic dependence on the coupling. Hopefully my wording isn't too vague! Also I'm glad the notes helped! $\endgroup$
    – Campbell
    Aug 8, 2021 at 3:49

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