# Why is QED renormalizable?

My understanding of renormalizability is that a theory is renormalizable if it the divergences in its amplitudes can be cancelled out by finitely many terms. I see that by adding counterterm (in the MS-bar scheme)

$$L_{ct}=-\frac{g^2}{12\pi^2}\left(\frac{2}{\epsilon}-\gamma+\ln4\pi\right),$$

the one-loop divergence of QED can be made finite. However, I do not see how this makes QED renormalizable? Surely as we work with diagrams with more loops, we will get more counterterms - given that we can have diagrams with arbitrarily many loops, do we not need an infinite number of counterterms to cancel these out?

QED has only a finite number of irreducible divergent diagrams. The main notion of divergence of a diagram is power-counting: The term every diagram represents has the form of a fraction like $$\frac{\int\mathrm{d}^n p_1\dots\int\mathrm{d}^n p_m}{p_1^{i_1}\dots p_k^{i_k}}$$ and you can compute the difference between the momentum power in the numerator and denominator and call it $$D$$. Heuristically the diagram diverges like $$\Lambda^D$$ in a momentum scale $$\Lambda$$ if $$D > 0$$, like $$\ln(\Lambda)$$ if $$D=0$$, and is finite if $$D < 0$$. This can fail - the diagram can be divergent for $$D < 0$$ - if it contains a smaller divergent subdiagram.
If you work out the general structure of $$D$$ for the diagrams of QED, you should be able to convince yourself that QED has only a finite number of divergent one-particle irreducible diagrams. That cancelling the irreducible diagrams is enough to cancel iteratively the divergences in all higher-order diagrams containing them in arbitrary combinations to all orders is a non-trivial statement sometimes called the BPHZ theorem, whose technical meaning - though not by this name - is explained by the Scholarpedia article on BPHZ renormalization.
• I am more familiar with $\phi_4^4$ rather than $QED_4$ but I don't think the statement about finitely many counterterms is correct, i.e., counterterms coming from finitely many graphs. This would correspond to a superrenormalizable theory which is not the case here. In a just renormalizable theory, there are finitely many operators in the the lagrangian which recieve counterterm contributions but they each do so from infinitely many graphs (with same external leg structure). Aug 20 '20 at 20:39