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?


1 Answer 1

  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.


$^1$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 $\lambda$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.