Cut-off Regularization of Chiral Perturbation Theory

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1. Mar 21, 2015

quantatanu0

I was trying to learn renormalization in the context of ChPT using momentum-space cut-off regularization procedure at one-loop order using order of p^2 Lagrangian. So,

1. There are counter terms in ChPT of order of $$p^4$$ when calculating in one-loop order using Lagrangian of order $$p^2$$.

2. Divergences are of polynomial kind and logarithmic kind.

3. The counter terms always take care of polynomial divergences (and $$1/\epsilon$$ kind of div. in dimensional method)

4. The logarythmic divergence gets absorbed during coupling constant renormalization.

During my calculation using cut-off method I obtained a result where I have only logarithmic divergence and no other divergent term, then I need to understand what is the use of counter-terms in this case.

In any case, we have to consider the counter-terms in ChPT but here we are not doing dimensional regularization so no $$1/\epsilon$$ to get killed by the counter terms, and in my calculation involving cut-off regularization, I have no polynomial divergence either ! Only logarithmic divergence, then what is the use of the counter-terms here ?

2. Mar 26, 2015

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Mar 26, 2015

quantatanu0

OK, I am happy to tell you guys that I have figured it out. Here's what I do:

Let's say the counter terms introduce coupling constants (low energy constants) $$L_i , i=1,2,... n$$ and the divergent term coming from the loop calculations is $$D$$, and this can be any kind of divergence, log, polynomial, and/or any other kind (separately or together). Then:

$$L_i = L^r_i + c_i \frac{D}{n}$$ where $$c_i$$ are constants that one chooses in a way that the divergence gets cancelled by the counter terms. And this is all.