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I Superficial degree of divergence for scalar theories

  1. May 3, 2016 #1

    CAF123

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    I have a few questions regarding the derivation of the degree of divergence for feynman diagrams. The result is $$D = [g_E] - \sum_{n=3}^{\infty} V_n [g_n]$$ (following notation in Srednicki, ##P118##)

    I am trying to understand what ##[g_E]## is here? Since in this set up we are summing over all possible scalar theories ##n \in [3, \infty)##, we will have a tree level local interaction diagram corresponding to the case where ##E=i## where ##i## is some element of the set ##[3,\infty)##. So is it right to say that ##[g_E]## denotes a particular element of the set ##[3,\infty)##?

    If that is correct, then in the theory $$\mathcal L = \frac{1}{2} Z_{\phi} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}Z_m m^2 \phi^2 - \frac{1}{k!} Z_g g \phi^k$$ (i.e a theory where we now include only one of the elements in the above set) the formula for ##D## now becomes ##D = [g_E] - v_k [g]## and in this case ##[g_E] = [g]##, with ##v_k## the number of times a vertex shows up in some diagram? Is that correct understanding?
     
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  3. May 3, 2016 #2

    nrqed

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    Hi,

    Well, I would not say that we are "summing over all possible scalar theories", that's misleading. We are consider one diagram in a single theory. Now, this diagram has a certain number of external lines, let's call it "k", and ##[g_E]## is the dimension of the coefficient of the term with k powers of the scalar field in the potential. For example, if there are 7 external lines, ##[g_E]## is the dimension of the coefficient of ##\phi^7##. Then the sum is over all the vertices in the diagram.
     
  4. May 4, 2016 #3

    CAF123

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    Hi nqred,

    Srednicki begins by writing down $$\mathcal L = - \frac{1}{2} Z_{\phi} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}Z_m m^2 \phi^2 - \sum_{n=3}^{\infty} \frac{1}{n!} Z_n g_n \phi^n$$ which seems to me to be a summation over all possible scalar theories, i.e in such a theory we have a ##\phi^3, \phi^4...## etc interaction. Is it not? Suppose we have a diagram with E external legs. Then in ##D = [g_E] - \sum_{n=3}^{\infty} V_n [g_n]##, the first term ##[g_E]## corresponds to the tree level local interaction vertex that occurs when ##E=i## for some i in the interval [3,infinity]. So we can write $$D = [g_E] - V_i[g_i] - \sum_{n \neq i}^{\infty} V_n [g_n] = (1-V_i)[g_E] - \sum_{n \neq i}^{\infty} V_n [g_n]$$ Each of the ##[g_n]## have a different mass dimensionality in a fixed number of space time dimensions.

    Am I thinking about this wrongly? Edit: The last equality I wrote must be wrong because it doesn't reproduce the correct D for some diagrams. I was thinking of the one loop box correction to the ##\phi^3## local vertex interaction. In d=6, the coupling has null mass dimension and we only have one theory (phi^3) so Srednicki's formula reduces to ##D = [g_E] - V_3 [g_3]##. In this case, ##V_3 = 4## and ##[g_E]## = 0. But this doesn't recover the D=-2 you get by power counting.

    Thanks!
     
    Last edited: May 4, 2016
  5. May 4, 2016 #4

    nrqed

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    Hi again!

    Well, it is just a question of terminology. The way I see it is that we are considering one theory (because we are dealing with a single lagrangian) but this theory may contain an infinite number of terms. Even if there is an infinite number of terms it is still one theory.



    Yes, we can write this.

    Not necessarily! One can also have ##\phi^4,\phi^5 \ldots ## terms! The theory will be nonrenormalizable but it is still perfectly well defined as an effective field theory

    Watch out. This box diagram has four external lines. Therefore ##[g_E]## is the dimension of the coefficient of an ##\phi^4## interaction would have (note that it does not matter if there is such an interaction in the Lagrangian or not, ##[g_E]## is the dimension of the coefficient of such an interaction if it was present) and for a ##\phi^4## interaction in 6 dimensions, the coefficient would have dimension -2. So for your diagram, ##[g_E]=-2## and therefore D=-2.
     
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