Amplitudes in renormalized perturbation theory

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This question arose while reading Peskin and Schroeder, specifically, it arose in regards to the sum of diagrams above their Eq. (10.20) on pg. 326.

The context is $\phi ^4$ theory and they are using a vertex renormalization condition to compute the counterterm $\delta _\lambda$ corresponding to the coupling constant $\lambda$. To do this, they calculate the $4$-point amplitude up to one-loop order in perturbation theory. In the process of doing this, however, they do not seem to include any one-loop diagrams involving the counterterms. Indeed, they only seem to include counterterm diagrams up to tree level.

Why is this? It seems, at least naively, that if one is doing a one-loop computation, one should compute all one-loop diagrams, and not discriminate between those Feynman rules in the 'original' theory and those that only arise during renormalization.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Jonathan Gleason
retagged Aug 23, 2014

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The bare four-point vertex is of order $\sim \lambda$, and the counterterm four-point vertex is of order $\sim \lambda^2$ (although multiplied by infinity). The perturbation is done by considering the terms with the same order in $\lambda$.

So, you need to consider a one-loop diagram with two bare vertices and a tree diagram with one counterterm at the same time.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Yuji
answered Nov 16, 2013 by (1,395 points)
This was more or less what I was expecting, but when I went to check this myself, I didn't see how $\delta _\lambda$ was of order $\lambda ^2$. Indeed, according to Peskin and Schroeder's Eq. (10.17) on pg. 324, $\delta _\lambda =\lambda _0Z^2-\lambda$, which is just of order $\lambda$ . . . or am I missing something?

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Jonathan Gleason
@JonathanGleason The conterterm is of order $\lambda$ and the couplings to make the amplitude one-loop will introduce addtional $\lambda$s.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Neuneck
@Neuneck I don't understand. The Feynman rule corresponding to the $4$-point counterterm vertex is just $-\mathrm{i}\, \delta _\lambda$, not $-\mathrm{i}\, \delta _\lambda \lambda$ or anything like this, so this entire counterterm diagram should just be of order $\lambda$. What am I missing?

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Jonathan Gleason
@JonathanGleason I was looking at your $\delta_\lambda$ from your previous answer, which seems to be proportional to $\lambda$.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Neuneck
@Neuneck Yes, but I thought the point is that the total result should be proportional to $\lambda ^2$, not just $\lambda$.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Jonathan Gleason
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