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In the physics literature a quantum field theory is qualitatively classified as renormalizable, super-renormalizable, or non-renormalizable. This heuristic is based on how many Feynman diagrams converge. In more rigorous approaches to quantum field theory, especially in the stochastic quantization literature, they instead use the classifications critical. sub-critical, and super-critical, as is commonly done in the PDE community.

Since these classifications are used in a seemingly interchangeably way there should be a connection between the two but I haven't been able to see it. Criticality has to do with how conserved quantities or norms behave at various scales and is an analysis done directly on the PDE. In contrast, renormalizability, as described by the divergence of Feynman diagrams and the inclusion of counterterms, is a perturbative description which is done on the Lagrangian and the PDE itself is never considered.

What is the connection between these two points of view?

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The terms describe how the coupling terms of the theory change as one increases the energy. A theory is renormalizable = critical if the coupling terms remain unchanged, super-renormalizable = sub-critical if the coupling terms decrease with increasing energy, and non-renormalizable = super-critical if the coupling terms increase (and eventually diverge in the limit that the energy tends to infinity).

These lecture notes by Witten describe the classification in some detail, with examples, from the perspective of perturbation theory (Feynman diagrams). For a presentation in terms of PDE's (stochastic heat equation), see Stochastic PDEs, regularity structures, and interacting particle systems by Chandra & Weber.

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  • $\begingroup$ I'm not sure what is meant to be unchanged/decreasing/increasing here. In a renormalizable theory, the values of the couplings change with energy. The point of renormalizability is that there are only a fixed number of such couplings that need to be adjusted. Non-renormalizability means that the number of such couplings proliferates. Super-renormalizability means that the couplings don't run (just a few infinite constants might have to be subtracted). $\endgroup$
    – Michael Engelhardt
    Commented Nov 23 at 20:04
  • $\begingroup$ I was just referring to the running coupling of a QFT, which scales to 0, $\infty$, or is scale independent; definition 1.3 in the PDE paper I linked to gives a formal description. $\endgroup$
    – Carlo Beenakker
    Commented Nov 23 at 20:28
  • $\begingroup$ Yes, and that is what confuses me - in a renormalizable theory, the coupling runs. So, what is the meaning of "the coupling terms remain unchanged"? The coupling of a renormalizable theory can run to 0 (asymptotic freedom), or to $\infty $ (at least perturbatively, i.e., a Landau pole) ... $\endgroup$
    – Michael Engelhardt
    Commented Nov 23 at 21:30
  • $\begingroup$ A renormalizable theory has a fixed point where the coupling constant is scale invariant, the correspondence to a critical PDE refers to that case, as far as I understand. $\endgroup$
    – Carlo Beenakker
    Commented Nov 23 at 22:39
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    $\begingroup$ this is discussed in Witten's lecture 2: how a coupling strength $g$ scales under a scaling transformation ($x\mapsto x/t\Rightarrow g\mapsto t^\alpha g$) determines the number of divergent graphs in the Feynman diagram expansion, see theorem 2.1 $\endgroup$
    – Carlo Beenakker
    Commented Nov 24 at 11:51

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