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In Peskin & Schroeder's An Introduction to Quantum Field Theory, section 10.5, page 338, the book gives a two-loop renormalization example (in scalar $\phi^4$ theory).

Before we start the two-loops, let's us recall the renormalization condition in (10.19)

enter image description here

For the second renormalization condition, my understanding for two-loop case is $$\text{One-loops + two-loops + counterterms}=-i\lambda. $$

However, the book in two-loop examples only considers the two loop case. The relevant two-loop feynman diagrams are given in (10.51)

enter image description here

The sentence below (10.51) reads

The value of last diagram in (10.51) is just a constant, which we can freely adjust to absorb any divergent terms that are independent of the external momenta.

  1. Here, does the momentum independent divergence including double poles divergence? i.e. $(\frac{1}{\epsilon})^2$.

  2. However, on page 339, the book refers

    enter image description here

    which is the result from one-loop renormalization. In this case, the $\delta_\lambda$ is already specified. And we cannot freely adjust it to absorb momentum independent divergence. This is contradict with the sentence below (10.51).

So I am wondering what's wrong with my logic?

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1 Answer 1

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  1. Yes, see e.g. the first equation on p. 341.

  2. The 1-loop counterterms are of order ${\cal O}(\lambda^2)$ while the 2-loop counterterms are of order ${\cal O}(\lambda^3)$ in the perturbative formal power series for the coupling constant $\lambda$. [The order$^1$ is determined by the number of $\times$-vertices in the Feynman diagrams that a counterterm is designed to (partially) cancel.] The 1-loop and 2-loop counterterms are hence independent terms.

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$^1$The counting is more complicated if the Feynman diagram contains $\otimes$-vertices. Since the counterterm $\delta_{\lambda}$ is a power series expansion in $\lambda$ (the lowest power is quadratic), the corresponding $\otimes$-vertex doesn't contain a definite power of $\lambda$.

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  • $\begingroup$ Thank you, that’s make sense! Could you please elaborate more on how we obtain this $\mathcal{O}(\lambda^2)$ and $\mathcal{O}(\lambda^3)$ counterterms? And they are independent? $\endgroup$
    – Daren
    Commented Mar 10, 2023 at 13:07
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Mar 10, 2023 at 13:34

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