# Amplitudes in renormalized perturbation theory

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.

-

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 $\lambda$.
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? – Jonathan Gleason Nov 16 '13 at 20:26
@JonathanGleason The conterterm is of order $\lambda$ and the couplings to make the amplitude one-loop will introduce addtional $\lambda$s. – Neuneck Nov 18 '13 at 11:29
@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? – Jonathan Gleason Nov 19 '13 at 1:58
@JonathanGleason I was looking at your $\delta_\lambda$ from your previous answer, which seems to be proportional to $\lambda$. – Neuneck Nov 19 '13 at 8:05
@Neuneck Yes, but I thought the point is that the total result should be proportional to $\lambda ^2$, not just $\lambda$. – Jonathan Gleason Nov 19 '13 at 21:07