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In $\phi^n$ theory in Peskin & Schroeder the superficial degree of divergence is: $$D = d - V[\lambda] - \big(\frac{d-2}{2}\big)N \tag{10.13}$$ where $d$ is the dimension, $V$ is the number of vertices and $N$ the number of external lines. Take $N$ to be fixed. We have three possibilities:

  1. If $[\lambda] < 0$ then for all but finitely many $V$, $D > 0$ so an infinite number of diagrams diverge. Such theories are said to be non-renormalizable.
  2. If $[\lambda] > 0$ then for all but finitely many $V$, $D < 0$ so only a finite number of diagrams diverge. Such theories are said to be super-renormalizable.
  3. If $[\lambda] = 0$ then $D$ is always constant at all orders in pertubation theory. Such theories are said to be renormalizable.

Peskin & Schroeder say in a super-renormalizable theory

Only a finite number of Feynman diagrams superficially diverge

and in a renormalizable theory

Only a finite number of amplitudes superficially diverge; however, divergences occur at all orders in perturbation theory

My questions are the following:

  1. Why do we only consider $[\lambda]$ when categorizing a theory as non-renormalizable, super-renormalizable, or renormalizable? For example how do we know the contribution of $[\lambda]$ won't be counteracted by increasing $N$ so that the superficial degree of divergence remains nonpositive? In other words, why do we neglect the effect of $N$ or take it to be constant?
  2. I don't understand Peskin & Schroeder's explanation for the difference between renormalizable and super-renormalizable theories. How does a constant superficial degree of divergence (i.e. when $[\lambda] = 0$) imply only a finite number of amplitudes superficially diverge but that divergences still occur at all orders in perturbation theory?
  3. How does the superficial degree of divergence say anything about the amplitude? I thought it only describes the rate of divergence of the diagram.
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2 Answers 2

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  • It should stressed be that Peskin & Schroeder are here using the old Dyson definitions of renormalizability. For a more general derivation of eq. (10.13), see e.g. this Phys.SE post, which also answers several of OP's questions.

  • OP's main question seems to be the following. Assume that the spacetime dimension $d>2$ and that the theory only has one (scalar) field $\phi$ with a single coupling constant $\lambda$.

    An $N$-point (connected) amplitude/correlation function is a sum of infinitely many (connected) Feynman diagrams with a fixed number $N$ of external legs.

    If $[\lambda]\geq 0$ [corresponding to the (super)renormalizable case], then only for finitely many natural numbers $N \leq \frac{d}{[\phi]}$, where $[\phi]=\frac{d-2}{2}$, the $N$-point amplitude can contain Feynman diagrams$^1$ with $D\geq 0$.

    (The above argument can be generalized to a theory with finitely many types of fields and coupling constants.)

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$^1$ NB: Be aware that due to possible (UV) divergent subdiagrams, the actual (UV) divergencies may be worse than indicated by $D$.

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Qmechanic's answer is very good and tackles your main question well. I want to however add an important detail to your first question.

Namely, we generally do not just consider $[\lambda]$ when determining a theory to be renormalizable or not, this is only sufficient in special cases (which $\phi^n$ belongs to). Two examples where just looking at the dimension of the coupling is not sufficient are non-abelian gauge theories in 4 dimensions, or nonlinear sigma models in two dimensions. There, the mass dimensions of the coupling suggest that the theory is renormalizable, but in the process of renormalizing we might have to add additional terms to the action that are not present initially.

More specifically, when renormalizing a theory we want to absorb all divergences appearing into a renormalization of the constants in the bare lagrangian, such as mass or coupling constants. We do this by adding counterterms to the lagrangian that render correlation functions finite and then absorb these counterterms into redefinitions of the constants. But for us to do that, all possible counterterms must be of the same structure as the terms already present in the lagrangian, as we do not want to change the theory by adding new interactions.

In the example of non-abelian gauge theories, the reason why power counting is not enough is that a priori, one is not protected against counterterms that give rise to gluon masses, ghost masses, or 4-gluon couplings with wrong gauge-index structures. All of which are not gauge invariant.

To really prove renormalizability in these theories, one has to look at the structure of the renormalized action. In the two examples I mentioned, the Zinn-Justin equation, obtained from ward identities, tells us that the theory is renormalizable. In the case of non-abelian gauge theories, it is precisely the BRST symmetry that gives us the relevant Zinn-Justin equation and tells us that the above mentioned counterterms indeed do not show up.

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  • $\begingroup$ Thank you, this is very interesting. So does power counting not work for non-abelian gauge theories? $\endgroup$
    – CBBAM
    Mar 28 at 22:13
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    $\begingroup$ I updated the answer with some more info about the case of non-abelian gauge theories $\endgroup$
    – mika
    Mar 29 at 14:01
  • $\begingroup$ Thank you! So the problem in non-abelian gauge theories is that the counterterms alter the theory (by altering/introducing masses, coupling constants...etc)? I'm curious how these masses are introduced as I thought rewriting the Lagrangian in terms of renormalized terms and counterterms is just bookkeeping, so how does it introduce, for example, new masses? $\endgroup$
    – CBBAM
    Mar 29 at 19:33
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    $\begingroup$ Well, in the end the masses/etc. are not introduced. However, we are searching for a proof of renormalization, thus we have to prove that they are not introduced at any order. Just calculating, say, the necessary 2 loop diagrams and saying "look, no masses need to be introduced here" does not tell us they are not needed at 3 loops. $\endgroup$
    – mika
    Apr 1 at 8:28

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