2
$\begingroup$

Before I put forward my actual question, I think it will be useful to set the context in a clear way and that involves my understanding of a few very basic things of Chiral Perturbation Theory.

Chiral Orders

Chiral Lagrangian is of the following structure:

$$ { \cal L } = { \cal L }_{p^2} + { \cal L }_{p^4} + { \cal L }_{p^6} + ... $$

At lowest tree level ( O($p^2$ ) matrix elements won't produce any infinities and we are ok, but when we want to calculate an amplitude at O($ p^4$) what we do is as follows:

  1. We calculate the 1-loop diagrams which are 4th order in momentum and these loops contain vertices which are derived from $ { \cal L }_{p^2} $, and of course the loops produces divergent terms.

  2. We then consider the tree level contributions from $ { \cal L }_{p^4} $ and the coefficients of these $ p^4 $th order operators, known as the low energy constants absorb the infinities encountered in the above step (1), and everything becomes finite.


Now I come to my actual question. Of course the parameters of the theory are getting renormalized automatically by the standard procedure discussed above, then, do we have to separately think about the renormalization of "mass", "coupling constant" etc ?

And what is the role of wave function renormalization here ?

Any help will be highly appreciated.

Many thanks.

$\endgroup$
2
  • $\begingroup$ I believe the low energy constants are essentially coupling constants. $\endgroup$
    – ragnar
    Jul 3, 2015 at 21:26
  • $\begingroup$ @ragnar I believe even mass is also a coupling constants that gives the strength of self interaction. So yes, low energy constants are also but they play the role of counter terms too. $\endgroup$
    – quanta
    Jul 3, 2015 at 21:46

0

Your Answer

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