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QFT Counter Terms example calculation?

  1. Jul 1, 2010 #1
    I'm reading a QFT text right now and to fully understand the physical perturbation theory method I would like anyone to suggest a refrence or supply an example of a calculation using the physical perturbation theory:

    As an example to start a discussion consider.
    L = \frac{1}{2}((\partial \phi)^2 - m_p^2 \phi^2) - \frac{\lambda_p}{4!}\phi^4 + A(\partial \phi)^2+B \phi^2 + C\phi^4

    how might one go about calculating the propagator out to say second order in lambda obviously in bare qft i would arrive at

    [tex]D(k) = \frac{1}{k^2 - m_0^2 + i \epsilon}[/tex] + divergent terms

    but i am unsure how to use the physical theory to this end
  2. jcsd
  3. Jul 1, 2010 #2


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    This is how I understand it:

    You want your theory to become a free field theory for [itex]t \rightarrow\pm\infty[/itex]. For an interacting theory this means you need counterterms.

    For example, you demand that the vacuum expectation value of your field vanishes:

    <0| \phi(x) | 0 > = 0

    So you take your bare theory and calculate to first order this object (by taking a functional derivative of Z with respect to the source J and put the source to zero; this means that you get all the diagrams with one source, in which the source is "differentiated away"). You'll find out that it's NOT zero. So you have to add a counterterm to make it zero, because otherwise the earlier mentioned boundary conditions are not satisfied.

    Which text do you use? Srednicki has a nice discussion about this, in chapter 9 :)
  4. Jul 1, 2010 #3
    I'm using many books, Mainly "Quantum Field Theory in a NutShell" by Anthony Zee. I enjoy the focus on physics rather than math presented in this book, however I am also using Peskin & Schroeder for more mathematical rigor. I have access to Srednicki as well and will look at the reference you suggested, thank you for that. I'm going to have to verify your claim and will post once i am in agreement with your statement thank you for your help
  5. Jul 1, 2010 #4
    <0|\phi(x)|0> = \frac{1}{Z(J=0)}\int D\phi \phi(x) e^{\frac{i}{2}\int (\partial \phi^2 - m^2 \phi^2) - \frac{\lambda}{4!}\phi^4}

    <0|\phi(x)|0> \approx \frac{1}{Z(J=0)}\int D\phi \phi(x)(1-\frac{ i \lambda}{4!}\int\frac{\delta^4}{\delta J(w)^4}d^4w) e^{\frac{i}{2}\int (\partial \phi^2 - m^2 \phi^2) + J\phi}|_{J=0}

    <0|\phi(x)|0> \approx \frac{1}{Z(J=0)} \frac{1}{i} \frac{\delta}{\delta J(x)} \int D\phi (1-\frac{ i \lambda}{4!}\int\frac{\delta^4}{\delta J(w)^4}d^4w) e^{\frac{i}{2}\int (\partial \phi^2 - m^2 \phi^2)+J\phi}|_{J=0}

    \approx \frac{1}{Z(J=0)} \frac{1}{i} \frac{\delta}{\delta J(x)} (1-\frac{ i \lambda}{4!}\int\frac{\delta^4}{\delta J(w)^4}d^4w) Z(J=0)e^{-\frac{i}{2} J_x D_{xy}J_y}

    \approx \frac{1}{i}\frac{\delta}{\delta J(x)} (1-\frac{ i \lambda}{4!}\int\frac{\delta^4}{\delta J(w)^4}d^4w)e^{-\frac{i}{2} J_x D_{xy}J_y}|_{J=0}

    first term is

    \frac{1}{i}\frac{\delta}{\delta J(x)}e^{\frac{-i}{2} J_x D_{xy}J_y}|_{J=0}

    since all odd order functional derivs evaluated at [tex]J=0[/tex] will give zero this is zero as well the second term is too in fact all terms are in this calculation example...

    the second term has [tex]\frac{\delta^5}{\delta J(w)^4 \delta J(x)}[/tex] hence will give you zero and all orders do this pattern repeats

    [tex]\frac{\delta^{4n+1}}{\delta J(w)^{4n} \delta J(x)}[/tex]

    4n+1 is always odd hence this is zero for all orders as it should be with only one source this has no where to propagate to, sorry your understanding is incorrect though I do appreciate your effort
    Last edited: Jul 1, 2010
  6. Jul 2, 2010 #5


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    Yes, I'm sorry; I considered phi to the 3 instead of phi to the 4; in phi to the 4 you have an explicit Z2 symmetry, so indeed your vev should be zero :)
  7. Jul 3, 2010 #6
    so...still how might i calculate the propagator out to first order using the physical theory?
  8. Jul 3, 2010 #7
    I, too, am just learning this stuff at the moment using Zee. I haven't got as far as you phymath, but I had a quick skim of the renormalisation chapter. I agree with you that it would be nice to see at least the first order calculation spelled out explicitly.

    As far as I can guess, it must be something like this:

    A first order correction to the propagator would be a diagram like III.3.1a
    Write the integral corresponding to this diagram, but with the counter terms in place, then in the momentum space version, do the trick of evaluating the counter terms (cutoff dependent) by expanding the integral in powers of k and successively differentiating (twice at a time) with respect to k and setting k to zero.

    However, that's just a guess - it would be good if someone with access to Zee's book who understood this stuff could spell out the steps involved...
  9. Jul 3, 2010 #8
    good to see someone else finds this section of Zee's book lacking. While I haven't finished the entire calculation the first reference to Srednicki does offer an example calculation of the mass renormalization using the physical pert theory. Mark Srednicki does generously offer his pre-textbook version of his book on qft you can find at the website below


    if you look at e-page 109 "Loop Corrections to the Propagator" you can find the example we are looking for, just know his "m" is Zee's [tex]m_p[/tex] I'm going to go through it and if i find it helpful I'll try and retype the more friendly Zee-like calculation. Good to see someone else is having the same interests!

    PS- he is not using a phi-4 theory as his example scalar theory
    Last edited: Jul 3, 2010
  10. Jul 3, 2010 #9
    Thanks for the tip - I'll have a look tomorrow. I've looked stuff up in the Srednicki preprint before - I think I'll buy a proper copy - it looks like a very nice treatment of the subject.
  11. Jul 4, 2010 #10


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    Hi, I'll have a look at it and if I did the calculation I'll post it here, but the weather here is really good for Dutch standards so I've got some priorities now ;)

    Zee's informal treatment about renormalization (Confusio et. al.) is very nice, but he could be a little more explicit; it's one of these big things in QFT people are struggling with.

    The nice thing about Srednicki is that he introduces renormalization right away, instead of presenting bare field theory and noticing later on that it needs to be adjusted. In Srednicki's treatment renormalization is a very natural thing to do, which can be a little bit obscured if you first learn bare field theory and after some chapters learn that it's not the whole story.
  12. Jul 5, 2010 #11
    I did the Srednicki calculation its very straight forward and works I suggest doing it, however it brought me a new question, when do I use the "crossed" Feynman diagrams? i mean by crossed the physical ones such as

    -----k-->--X----k-->-- = [tex]-i(A k^2 + B m^2)[/tex]

    I dont know when i should use this, i think i should use it any time i have two propagators and something in between then, I should include this diagram but im unsure if thats true
  13. Jul 5, 2010 #12
    I noticed over the weekend that Peskin and Schroeder treat renormalised perturbation theory in section 10.2. Towards the end of that section they give the rules for when to use the "crossed" diagrams. They also give the explicit calculation of the one loop correction to the propagator in the [itex]\phi^4[/itex] case.
  14. Jul 5, 2010 #13


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    You should look at which diagrams contain infinities. The A and B are chosen in such a way to cancel the infinities.

    So, draw the diagrams up to a certain order, calculate them explicitly, and every time you need to work away an infinity in your calculation this corresponds to a certain counterterm, which you draw as a Feynman diagram with a cross as above.

    I don't have the Zee book right now with me, but I would guess that in phi-4 theory your first infinity comes with the simplest loop diagram (an external line, then one loop, and then an external line). In four dimensions this diagram diverges. So doing the calculation (and choosing a certain renormalization scheme) fixes the A and B by demanding that the answer should be finite (the infinities are in the coupling parameter used in the Lagrangian, right?). After that you draw this as two external lines connected by a cross.
  15. Jul 6, 2010 #14
    I'm looking at Peskin and Schroeder's calculation of the one loop correction to the propagator (section 10.2), trying to understand how they chose the set of diagrams to include in the computation (the five diagrams just above eq 10.20). I understand that they're working to order [itex]{\lambda^{2}}[/itex] so I can see why they have the five they've chosen, but why not have additional terms -for example the first graph but with a cross in one of the legs ?

    I assume things like this are excluded because they're too high order in [itex]\lambda[/itex] ? I don't understand what order in [itex]\lambda[/itex] the counterterms are. Obviously the [itex]-i\delta_{\lambda}[/itex] is of order [itex]\lambda[/itex], but what order is the [itex]-i(p^2\delta_{z}-\delta_{m})[/itex] diagram and why?
  16. Jul 6, 2010 #15
    Note that in the google version the diagrams of eq 10.20 are the corrections to the coupling const [tex]\lambda[/tex] not corrections to the propagator, the propagator only has diagrams of 1 line and 1 line with loops in them (making 2 external lines) such as

    ------ + ----()---- , the () is a loop lol

    So I believe the argument is this:

    say we want to calculate the 1-loop correction to the propagator, ok we draw all the diagrams at this order. for example in my version of P&S p. 328 equation 10.29 (not the e-page if you have the electronic version) we have a one loop on the line well this will diverge by the power counting theorem (2 external lines -> D = 4- 2 = 2) thus for the divergent terms we add in a counter term diagram. hence the crossed diagram.

    so to address what you were specifically asking there's no "order" to the counter terms they will be the same order as the diverging diagram because they will absorb the infinities at that order, in my P&S eq 10.20 on p. 326 the diagrams above (1 4 pt vertex + 3 1-loop diagrams) you would add 1 crossed 4 pt vertex to absorb the infinites from the divergence of the 3 1-loop diagrams hence you see equation 10.21 at the end there is a [tex]-i\delta_{\lambda}[/tex] corresponding to the diagram

    take everything i said cautiously obviously since i started the thread im in the same boat as you are but thats my understanding so far

    PS lets work with the Google version of P&S so we have the same thing to work with

    Last edited: Jul 6, 2010
  17. Jul 6, 2010 #16
    You are absolutely right - that section is dealing with the vertex correction !!

    The reason I was attempting to associate a coupling constant "order" to the counterterms was something I read in Srednicki (online version) - in the Feynman rules (starting just after equation 10.14), rule 9 says that each of the coefficients in the counter term is of order [tex]g^2[/tex]. However, I think your statement that they're effectively of the same order as the divergence they're trying to deal with sounds right.

    Thanks for the discussion incidentally - I'm running slightly ahead of myself here - I've only got as far as finishing chapter 7 in Zee (trying to teach myself, doing the exercises as I go along) !
  18. Jul 6, 2010 #17
    ok so i think this is resolved for now, Now to start a new thread: what the hell is this Faddeev-Popov determinant?! ahhh lol
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