quantum field theory - Why is renormalization necessary in finite theories?

Michael Brown made the following comment here:

The modern understanding of renormalization (due to Kadanoff, Wilson and others) is hardly controversial and has nothing really to do with infinities. It is necessary even in completely finite theories, but the fact that it fixes infinities in QFT is a bonus.

Can anyone explain what this means?

Answer

Here's one way of interpreting this statement (this is essentially an elaboration of the comment by use Learning is a mess). The basic idea behind Wilsonian renormalization is that renormalization can be regarded as the process by which we understand how the theory effectively behaves at different scales, like momentum scales. Disclaimer: the math in this answer is schematic.

Consider a quantum theory of fields $\phi$ with a hard momentum space cutoff $\Lambda$. Such a QFT would be described, in the Euclideanized functional integral picture, by its partition function \begin{align} Z(\mathbf u, \Lambda) = \int [d\tilde\phi]_0^\Lambda e^{-S[\mathbf u, \Lambda,\tilde\phi]} \end{align} Here, $\mathbf u = (u_1, \dots, u_n)$ represents the parameters on which the action of the theory depends (like coupling constants), $S[\mathbf u, \Lambda,\phi]$ is the Euclidean action, and \begin{align} [d\tilde\phi]_{k_a}^{k_b} = \prod_{k_a<|k|

Now, let any real number $0

Let's call $S_s$ the action "at scale $s$". Then in some situations, the action at scale $s$ can simply be related to the original action at scale $s=1$ by taking the couplings to depend on the scale $s$; \begin{align} S_s[\mathbf u, \Lambda, \phi] = S[\mathbf u_s, \Lambda, \phi_s] \end{align} This is often referred to as the "running of the couplings." In other words, the process of renormalization which leads to the running of the couplings is simply the process by which we examine how the field theory effectively behaves at different scales; this is conceptually distinct from the issue of removing infinities.