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:
- 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.
- 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.
- 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:
- 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?
- 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?
- How does the superficial degree of divergence say anything about the amplitude? I thought it only describes the rate of divergence of the diagram.