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?

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    $\begingroup$ My guess is that he means that even if you start with a QFT with a certain U.V. cutoff or maybe something else implying that all loop contributions are finite you could still compute the running of the coupling constants and their dependence on the energy scale you are probing your system, thus renormalization is still relevant although not motivated by the removal of infinities. I have no examples to present so please take this with a grain of salt ;) $\endgroup$ – Learning is a mess Aug 17 '13 at 21:27
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    $\begingroup$ By "finite" you mean usual quantum mechanics? $\endgroup$ – user10001 Aug 17 '13 at 21:28
  • $\begingroup$ I missed this question. joshphysics' answer is an accurate reflection of what I meant. Regularization and renormalization are conceptually distinct processes. $\endgroup$ – Michael Sep 18 '13 at 3:58

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|<k_b}d\tilde\phi(k) \end{align} schematically denotes integration over field degrees of freedom corresponding to momenta between $k_a$ and $k_b$.

Now, let any real number $0<s\leq 1$ be given. We will call $s$ the "scale." Then we note that we can split the functional integration measure into an integration over momenta in the range $(0,s\Lambda)$ and those in the range $(s\Lambda, \Lambda)$, and we imagine defining a rescaled action $S_s[\mathbf u, \Lambda,\phi]$ by \begin{align} e^{-S_s[\mathbf u, \Lambda, \phi]} = \int [d\tilde\phi]_{s\Lambda}^\Lambda e^{-S[\mathbf u, \Lambda,\tilde\phi]} \end{align} We can now write the original partition function as an integral over only the momentum modes in the range $(0, s\Lambda)$ provided we use the action $S_s$; \begin{align} Z(\mathbf u, \Lambda) = \int [d\tilde\phi]_0^{s\Lambda} e^{-S_s[\mathbf u, \Lambda,\tilde\phi]} \end{align} and we say that we have "integrated out" the higher momentum modes. The action $S_s$ can now be regarded as the action that effectively governs the physics for momentum scales below $s\Lambda$.

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.

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    $\begingroup$ Why is this necessary, though? In a finite theory, why must I assume that it's effective and do renormalization etc? $\endgroup$ – innisfree Apr 9 '16 at 13:17
  • $\begingroup$ The partition function $Z(\mathbf u, \Lambda)$ is for a finite theory (one that perhaps happened to arise from regularization of an ill-defined one), and the ensuing renormalization analysis (which has nothing to do with infinities and everything to do with scaling) is illuminating. Another example would be some theory on a lattice in condensed matter. The theory may be finite, but a renormalization analysis is useful to understand how it behaves at different scales. en.wikipedia.org/wiki/Renormalization_group $\endgroup$ – joshphysics Apr 9 '16 at 20:48
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    $\begingroup$ Useful, yes, illuminating, yes, but is it necessary? In a finite theory, I wouldn't have to introduce any dimensionful regulator like a cut-off $\Lambda$ and worry about the sensitivity of my amplitudes (at a finite order in perturbation theory) to that cutoff. In other words, I wouldn't have to think about the RG. $\endgroup$ – innisfree Apr 10 '16 at 0:21
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    $\begingroup$ @innisfree I think we're getting unnecessarily pedantic / concerned with semantics. RG is so useful in analyzing certain systems (e.g. studying critical phenomena), that it would be difficult to make do without it. I am not a condensed matter expert, but I would also wager that there are certain results that cannot currently be derived without RG, so in that sense, one could legitimately say that it's necessary. Perhaps something along these lines is what the authors are trying to say in the original quote. $\endgroup$ – joshphysics Apr 10 '16 at 0:31
  • $\begingroup$ @innisfree I think it's also worth noting that regularization allows us to calculate the bare parameters, e.g. the bare charge $e_0$, but those are not the parameters we measure due to screening. Renormalization allows us to calculate the physical parameters that we actually measure by re-scaling the bare parameters. Rescaling the field, for example, would amount to $\phi = \sqrt{Z}\phi_r$ where $Z$ is the field strength renormalization. This would give us new terms in the Lagrangian called counter terms which can be adjusted to uphold the renormalization conditions for e.g. $m$ and $\lambda$. $\endgroup$ – QuantumMechanic Jun 5 '18 at 9:14

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