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Renormalization of QFT - the counterterms

  1. Oct 2, 2013 #1
    I would like to ask if anyone can give me a hand with the understanding of the counterterms. I am reading by myself Chapter 10 of Peskin & Shroeder and got stuck in the middle of their example of how to renormalize [itex]\phi^{4}[/itex] theory. What is puzzling me is how to obtain from the new Lagrangian (equation 10.18):
    [itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}_{r}-\frac{1}{2}m^{2}\phi^{2}_{r}-\frac{\lambda}{4!}\phi^{4}_{r}+\frac{1}{2}\delta_{Z}(\partial_{\mu}\phi)^{2}_{r}-\frac{1}{2}\delta_{m}\phi^{2}_{r}-\frac{\delta_{\lambda}}{4!}\phi^{4}_{r}[/itex]
    the first new vertex (shown on fig.10.3, the single line with the[itex] \otimes[/itex]), equal to:
    [itex]i(p^{2}\delta_{Z}-\delta_{m})[/itex]
    If you ask me, looking at terms (#4 + #5) in the above Lagrangian, this is a kinetic energy term and I imagine should bring a propagator into the Feynman rules (not a vertex!!), that should look something like this:
    [itex]\frac{i}{\delta_{Z}p^{2}-\delta_{m}}[/itex])
    Obviously I am not right and I would like to clarify what I am missing. Thanks in advance to anyone that tries to help.
     
  2. jcsd
  3. Oct 2, 2013 #2

    Bill_K

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    It's because they're treating #4 and #5 as a perturbation. Note that if you start with the full propagator, [itex]\frac{i}{p^2 + \delta_{Z}p^{2}-m^2 - \delta_{m}}[/itex], the leading term in the perturbation expansion of this is [itex]\frac{i}{p^2-m^2} (\delta_{Z} p^2 - \delta_m)\frac{i}{p^2-m^2} [/itex], which is what the diagram represents.

    In general, (A - B)-1 = A-1 + A-1 B A-1 + A-1 B A-1 B A-1 + ...
     
    Last edited: Oct 2, 2013
  4. Oct 2, 2013 #3
    Another way of looking at what Bill_K write: just as the perturbation ##(\lambda/4!)\phi^4## produces a vertex with 4 incoming lines that has the value ##-i \lambda##, the perturbation ##(\delta_m/2!)\phi^2## produces a vertex with 2 incoming lines that has the value ##-i \delta_m##. The ##\delta_Z## term produces a similar vertex except that the derivatives bring down powers of momentum.
     
  5. Oct 3, 2013 #4
    No,it is not a kinetic energy term from which you should extract the propagator.These terms represents self interaction type term.You will get propagator only by using the first two terms in the lagrangian which are free field term.You do get the vertex factor from it.You can use the general rule of obtaining the vertex factor by multiplying L by i,substitute the plane wave form for the field operators,then remove all the factors which are already taken care by normalization,external lines etc.The rest is vertex factor.You have to be a bit careful to apply this however.
    The second method is to sandwich the interaction term between <0| and |k1k2>,write the[itex] \phi[/itex] as plane wave.Now since only two [itex] \phi[/itex] appears in any term(4th or 5th),you have k1+k2=0,when you evaluate the interaction term sandwiched between <0| and |k1k2>,you also have to include the possible permutation which after differentiation yields 1/2(2(ik1)(ik2z-2δm).
    So you have 1/2[(-2k1k2δz)-2δm],use k1=-k2 and multiply the whole term by i(as per the first method).you have i(k2δzm),as required.
     
  6. Oct 3, 2013 #5
    It is just the other way around.You are supposed to build the full propagator by adding infinity of those diagram using the identity,you just mentioned.
     
  7. Oct 3, 2013 #6

    Bill_K

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    Depends on what you are trying to accomplish, doesn't it. In this case we are trying to derive the Feynman rules, not employ them. If you take a look at the page in Peskin and Schroeder that we are discussing, you'll see that they derive the rules by expanding the full propagator.
     
  8. Oct 3, 2013 #7
    I did give a derivation of the rule,but what I see on the page is that they write the lagrangian and write the feynman rules next.It is upto the person to derive them who is reading it.
     
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