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Propagator counterterm in phi^4

  1. Aug 9, 2006 #1
    Hello--I'm looking at Peskin p.324-323 where he describes the renormalization of [tex]\phi^4[/tex] theory. I'm a little confused about the Feynman rules that one gets out of the lagrangian with counter terms.

    My question in a nutshell: The propagator is given by [tex]\frac{i}{p^2 - m^2}[/tex], why is it that the counter term looks like the inverse of this, namely [tex]i(p^2\delta_Z - \delta_m)[/tex], when they come from terms in the lagrangian that have identical form?

    That is to say, the lagrangian contains the terms:

    [tex]\frac{1}{2}(\partial_\mu \phi_r)^2 - \frac{1}{2}m^2\phi_r^2 + \frac{1}{2}\delta_Z(\partial_\mu \phi_r)^2 -\frac{1}{2}\delta_m\phi_r^2[/tex]

    The first two yield the propagator with junk in the denominator, while the last two yield a counterterm with junk in the numerator.

    Why is this?

    I can "read off" the rules for the counter terms, and it makes sense that it puts stuff in the numerator. Similarly, I know that the propagator is given by the Green's function of the free theory, which is why it yields something in the denominator. But am I thinking about it too much if I think it's really strange?
  2. jcsd
  3. Aug 9, 2006 #2


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    You are asking a valid question.
    You could say that the full progator is
    [tex]\frac{i}{(1 + \delta Z) p^2 -(1 +\delta m) m^2}[/tex], but in order to use this you would need to now [itex] \delta Z [/itex] and [itex] \delta m [/itex] to all orders.

    The point is that those counterterms coefficients are expansions in the coupling constant, designed to render all the calculations finite, order by order in lambda, right? So, schematically,
    [tex] \delta m = c_1 \lambda + c_2 \lambda^2 + \dots [/tex]
    (and the same for [itex] \delta Z [/itex]) and one calculates the coefficients c_1, c_2 ... perturbatively.

    So, if you are working to one-loop, you could say that the valif propagator to one-loop is
    [tex]\frac{i}{(1 + d_1 \lambda) p^2 -(1 + c_1 \lambda) m^2}[/tex],
    but in some sense, this is not consistent since it really contains contributions to all orders in lambda. So if you expand around lambda=0, you get
    [tex]\frac{i}{ p^2 - m^2} +\frac{i}{ p^2 - m^2} (i p^2 d_1 \lambda - c_1 \lambda) \frac{i}{ p^2 - m^2} [/tex],
    which corresponds to using the rule for the corrections in the numerator (the previous expression essentially amounts to summing to all order the one loop correction to the propagator which can be done because it's a simple geometric series).

    So, to summarize, you *could* formally put the corrections due to the counterterms in th edenominator, but since they are calculated perturbatively, it is consistent to treat the corrections as "interactions", which puts them in the numerator when one writes down the expressions for the Feyman integrals.

    Hope this makes sense.

  4. Aug 9, 2006 #3
    Excellent, that clears up everything! Thanks Patrick.
  5. Aug 10, 2011 #4
    Hello, first of all thanks to Patrick for the nice explanation. I was struggling with this today also, but couldn't quite get it to work out.

    I tried to go back to square one and show how if one does perturbation theory about the rescaled Lagrangian those counter-terms would show up in the series expansion. Trying to go about it canonically, I very quickly hit a wall because the interaction Hamiltonian no longer satisfied

    [tex]H_{int} = -L_{int}[/tex]

    due to the presence of field derivatives. I tried pressing on, but things got very ugly quickly, and even if I succeeded in writing out the interaction Hamiltonian, I don't think Wick's theorem would work anymore either. Does anyone know how to go about doing perturbation theory with interaction terms that involve derivatives of the field? For the record, I got this for the interaction Hamiltonian (corresponding to eq 10.18 on page 324 of Peskin ... minus the lambda term as that is just a product of fields so easy to deal with):

    [tex]H_{int} = \frac{1}{2} \int \mathrm{d}^3 x \:\:\:\: \delta_Z (\dot{\phi})^2 + \delta_Z (\nabla \phi)^2 + \delta_m \phi^2[/tex]
  6. Aug 30, 2011 #5
    Well, sorry for the double post but I wanted to try one more time to see what you think about field derivatives in the interaction terms. After pressing on without knowing in detail (only vaguely) how the momentum terms crop up in the Feynman rules, I eventually got to the section on non-Abelian gauge theory, where I thought it would be best if I go back and figure this out once and for all so I can truly understand how the non-Abelian gauge boson Feynman rules come about. Here's what I did.

    First of all, for an example problem, I considered the non-linear sigma model, but with only one field for simplicity (don't have to deal with too many indicies and delta terms). So the Lagrangian was:

    [tex]\mathcal{L} = \frac{1}{2g^2} (\partial_\mu \pi)^2 + \frac{1}{2g^2} (\pi \partial_\mu \pi)^2[/tex].

    I then considered the four point function using the path integral approach of section 9.2 in Peskin. If we label the momentum [itex]k_1[/itex] through [itex]k_4[/itex], and expand out the interaction exponential to first order, then we have for example 4 terms where we contract momentum 3 and 4 with the field derivative terms to first order. This gives a term:

    [tex]\frac{-2i}{g^2} (2 \pi)^4 \delta^{(4)} (\Sigma k_i) \;\;\; \prod_{i = 1}^{4} \int \frac{\mathrm{d}^4 k_i}{(2 \pi)^4} \frac{i g^2 e^{-i k_i \cdot x_i} }{k_i^2 + i \epsilon} k_3 \cdot k_4 [/tex]

    Now this looks like exactly what I want (even down to the combinatoric factor of 2 that comes from the Feynman rule in figure 13.1 with [itex]i = j = k = l[/itex]), with the momentum conserving delta function and 4 propagators. But, the [itex]k_3 \cdot k_4[/itex] should be combined into the momentum integrals, meaning the 3rd and 4th propagators are no longer the desired Feynman propagator. How would I go about justifying the Feynman rule from this point exactly?
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