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I found this post about renormalization very helpful, since I don't really know QFT. Now I'm curious, is this fundamentally all renormalization means in QFT as well, or is it just an analogy? That is, when a QFT gets renormalized, are we essentially just averaging/aggregating the quantum field over bigger and bigger chunks of spacetime, and then adjusting some parameter(s) such that the dynamics is invariant?

And if so, what quantities exactly have to be invariant under renormalization? Just the Lagrangian of the field configuration?

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  • $\begingroup$ The example in the post is "real space" renormalization (applied to the simplest case, the Ising model). In QFT the scale is typically energy, not length (even though the two things are conceptually related). $\endgroup$
    – Quillo
    Commented Feb 22, 2022 at 21:43

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The cruical detail is buried in the post you link, where the author says:

"This renormalisation analysis that explained the phenomenon of “critical universality” was introduced by Kadanoff and Wilson in the 1960s, and is most powerful not in the real-space version introduced above, but in momentum space. (These techniques differ in important ways, but I can’t work out a way of explaining the differences concisely in this answer.)" [emphasis mine]

We're not "averaging fields over spacetime" when we do the actual momentum space version of renormalization. The infinities that renormalization is designed to counter in QFT come in two flavours - "infrared", because momenta go continuously down to zero and "ultraviolet" because momenta go arbitrarily high. The infrared issues are comparatively "easy" and rarely the focus of what we do because they can essentially be solved just by putting the theory into an arbitrarily large but finite box. The "ultraviolet" issues are the ones for which we really use renormalization.

The Wilsonian view is that we introduce a momentum cutoff $\Lambda$ above which some magic prevents momenta from existing. "Renormalizing" then means choosing a different cutoff $\Lambda'$. I guess you can view this as "averaging over" the momenta between $\Lambda'$ and $\Lambda$, but what we're averaging is the integrand of the path integral that gives us the partition function of our theory, not "the quantum fields". Explicitly, we view the partition function as a (Euclidean) path integral $$ Z[J] = \int \mathrm{e}^{-L + J\phi}\mathcal{D}\phi$$ where $\mathcal{D}\phi = \Pi_{\lvert k\rvert < \Lambda} \mathrm{d}\phi_k$ with $\phi_k$ the Fourier modes of the fields (in a box) whose momentum is lower than our cutoff. Renormalizing to $\Lambda'$ means performing all the integrals over $\Lambda' < \lvert k \rvert < \Lambda$. This gives us some factor that we can write as $\mathrm{e}^{-L_\text{eff}}$ (this isn't trivial, but it's true at least perturbatively), and so the Lagrangian of our theory at $\Lambda$ effectively changes to some $L+L_\text{eff}$. The renormalization flow is now moving $\Lambda'$ around and looking at the different effective theories we get this way.

Generically, the effective Lagrangian contains all possible couplings of the fields to each other that aren't forbidden by a symmetry of the Lagrangian we started with, and so the "adjustable parameters" that the post you link talks about become a list of coupling constants - e.g. for $\phi^4$ theory, we get a coupling $\lambda_{2n}(\Lambda')$ for each even number of fields. The choice of $\lambda_{2n}(\Lambda')$ that produces the same scattering amplitudes as our original theory are the correct choice (because a theory whose scattering amplitudes differed would obviously be a different theory), so the "quantity invariant under renormalization" is "just" the S-matrix, not some high-level observable property. There are interesting observations to make that of all the possible $\lambda_{2n}(\Lambda')$, only a few - namely the "renormalizable" and "super-renormalizable" - are relevant at $\Lambda'\ll \Lambda$ (but this is another technical argument I don't want to add here), so if we suppose that the "natural" $\Lambda$ is very high, this can explain why only renormalizable QFTs seem to be relevant for describing our universe at the "low" energy scales $\Lambda'$ at which we're looking at it.

Lastly, let me remark that this Wilsonian view on renormalization is not universal - this is a nice conceptual view, but in practice "integrating out Fourier modes in the path integral" might be a bit of a hassle. Other renormalization schemes exist, and while they will also have a "cutoff" (more generically call "renormalization scale") $\Lambda$, their running couplings $\lambda_{2n}(\Lambda')$ will not agree with Wilsonian renormalization beyond leading order in perturbation theory.

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  • $\begingroup$ How would the interpretation of renormalizing change in other schemes? For example, dimensional regularization and nonlocal regularization? $\endgroup$
    – Dan
    Commented Feb 22, 2022 at 21:01
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    $\begingroup$ @Dan The $\Lambda$ often isn't a cutoff there, but more "the typical energy scale of a process". See also this answer of mine. $\endgroup$
    – ACuriousMind
    Commented Feb 22, 2022 at 21:03
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    $\begingroup$ Never said it was a cutoff. However, $\Lambda$ cannot be a physical parameter. It has to go to infinity in the final calc. So it has to at least get pushed to the next-up-level of your perturbation scheme, else you lose just masses of things you need, such as unitarity. So, my question remains, how do you interpret renormalization when you don't have any $\Lambda$? $\endgroup$
    – Dan
    Commented Feb 22, 2022 at 21:10
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    $\begingroup$ @Dan 1. Some people would disagree, and say the "natural" Wilsonian cutoff as the point where non-QFT physics becomes relevant is a very physical parameter (of course assuming there is non-QFT high-energy physics). 2. I don't think there's a lot of "physical" interpretation of renormalization when it's not phrased in terms of physical momentum scales. E.g. a very different technical viewpoint is that renormalization is essentially resolving the ill-definedness of point-wise products of operator-valued distributions (buzzword Epstein-Glaser renormalization). $\endgroup$
    – ACuriousMind
    Commented Feb 22, 2022 at 21:24
  • $\begingroup$ But that's not what this question was about - please ask a new question if you have a different question. $\endgroup$
    – ACuriousMind
    Commented Feb 22, 2022 at 21:24

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