0
$\begingroup$

In Ch.12 of the textbook An Introduction to Quantum Field Theory by Peskin and Schroeder, on P.408 the renormalization conditions are given at the energy scale $M$

$$\mathrm{dressed\ 4\ point\ vertex}=-i\lambda\ \ \ \mathrm{at}\ (p_1+p_2)^2=(p_1+p_3)^2=-M^2 .\tag{12.30}$$

It was made clear later on P.410 above Eq.(12.35) that, the coupling $\lambda$ is the renormalized coupling, which should be a function of the energy scale, and thus motivated the derivation of Callan-Symanzik equation.

However, further forward into the chapter, where a simple calculation of $\beta$ and $\gamma$ functions are carried out, in Eq.(12.44), it becomes clear that $\lambda$ is used as if it is the bare coupling.

To be specific, one calculated the four-point Green function by employing previous result (10.20) and obtains

$$G^{(4)}=[-i\lambda+(-i\lambda)^2[iV(s)+iV(t)+iV(u)]-i\delta_\lambda]\cdot\Pi\frac{i}{p_i^2} ,$$

with

$$\delta_\lambda=(-i\lambda)^2\cdot 3V(-M^2).\tag{12.44}$$

Although the above procedure indeed guarantees the condition (12.30) up to the order $\lambda^2$, my question is why one can use the renormalized coupling to recalculate the coupling normalization? Does this lead to any contradiction by repeating the same procedure? I am quite confused.

$\endgroup$
  • $\begingroup$ That's precisely the distinction between a renormalizable and non-renormalizable theory. If you need "only" a finite number of counter terms, it is renormalizable, otherwise it is not. So in short: you ask a very important question but in a renormalizable this happens to work out. The proof that theories with couplings with nonnegative mass dimensions are renormalizable is not a one-liner, but you can make this plausible with dimensional analysis. $\endgroup$ – marmot May 6 at 1:36
  • $\begingroup$ Thanks for the comment, would you please be more specific about where I can further read about this? I was asking in a more "naive" way, if one uses the renormalized coupling to renormalize again, why it is consistent? I understand why the dimension of the coupling is related to renormalizability (as explained in the textbook of Peskin and Schroeder), but honestly I cannot quite get its connection to my question. Again, thx! @marmot $\endgroup$ – gamebm May 6 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.