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The renormalization group

  1. Oct 23, 2013 #1
    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 [itex]\hat{\phi}[/itex] 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:
    [itex]\int \mathcal{D}\hat{\phi}exp\left(-\int d^{d}x \frac{\lambda}{4}\phi^{2}\hat{\phi}^{2}\right)[/itex]
  2. jcsd
  3. Oct 23, 2013 #2
    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).
  4. Oct 23, 2013 #3
    Thank you I think I almost understand. Another thing I am noticing is that while integrating over the high degrees of freedom [itex]\hat{\phi}[/itex] the dependence of the large cutoff [itex]\Lambda[/itex] goes into the coefficients (see for example the expression of [itex]\mu[/itex] in formula 12.11 or for [itex]\lambda^{'}[/itex] in 12.29). Are we to assume that we will no longer have any cases that as [itex]\Lambda->\infty[/itex] that will not cause any of those coefficients to go to infinity? and if so is that only valid for theories that are renormalizable?
  5. Oct 24, 2013 #4
    you will have the effect of it in higher order terms,where you will have two vertices for example.just see 12.13.
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