# The renormalization group

1. Oct 23, 2013

### nikol

When I am reading about the Wilson approach to renormalization in Chapter12.1 of Peskin & Shroeder I am wondering why are you allowed only to contract the $\hat{\phi}$ field (this is the field that carries the high-momentums degrees of freedom)as they show in equation 12.10, I thought that we should add all the contractions, between all the 4 fields and here is the term they are making the example of:
$\int \mathcal{D}\hat{\phi}exp\left(-\int d^{d}x \frac{\lambda}{4}\phi^{2}\hat{\phi}^{2}\right)$

2. Oct 23, 2013

### The_Duck

Here we are only doing the path integral over the $\hat\phi$ field, and not integrating over $\phi$. You can only get internal lines in a Feynman diagram for fields you are integrating over. So we end up writing down Feyman diagrams where all the external lines are $\phi$ fields (because we only care about the interactions of low-energy particles) and all the internal lines are $\hat\phi$ fields (because we are only integrating over the high-momentum modes).

3. Oct 23, 2013

### nikol

Thank you I think I almost understand. Another thing I am noticing is that while integrating over the high degrees of freedom $\hat{\phi}$ the dependence of the large cutoff $\Lambda$ goes into the coefficients (see for example the expression of $\mu$ in formula 12.11 or for $\lambda^{'}$ in 12.29). Are we to assume that we will no longer have any cases that as $\Lambda->\infty$ that will not cause any of those coefficients to go to infinity? and if so is that only valid for theories that are renormalizable?

4. Oct 24, 2013

### andrien

you will have the effect of it in higher order terms,where you will have two vertices for example.just see 12.13.