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A quantum field theory can be classified as superrenormalizable, renormalizable, or nonrenormalizable and in the renormalization group an operator can be classified as relevant, marginal, or irrelevant depending on how it grows/shrinks in the IR. What is the relationship, if any, between these two classifications?

As a concrete example let us consider a scalar field with a $\phi^4$ interaction. In three dimensions this theory is superrenormalizable and $\phi^0$, $\phi^2$, and $\phi^4$ are all relevant, $\phi^6$ is marginal, and $\phi^n$ is irrelevant for $n > 6$. Here I am only considering even powers due to symmetry. It is known that this theory does not need a coupling constant renormalization.

In four dimensions $\phi^0$ and $\phi^2$ are relevant, $\phi^4$ is marginal, and $\phi^n$ for $n \geq 6$ is irrelevant. This theory also needs a coupling constant renormalization.

Is there any reason why $\phi^4$ and $\phi^6$ don't need renormalization despite being marginal/relevant in three dimensions? Is the reason $\phi^4$ is only renormalizable in four dimensions because the bare Lagrangian contains terms that are marginal? More generally, given a Lagrangian is there a way to tell which terms will need to be renormalized? It seems like their classification under the renormalization group is not enough.

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  1. If you take a free Lagrangian and add a relevant interaction, you get a super-renormalizable theory.
  2. If you take a free Lagrangian and add a marginal interaction, you get a renormalizable theory.
  3. If you take a free Lagrangian and add an irrelevant interaction, you get a non-renormalizable theory.

To convince yourself of point 1, you can show that the leading term of the beta function for the coupling you added looks like $\beta(\lambda) = (\Delta - d)\lambda + O(\lambda^2)$. Comparing this with the familiar $-\epsilon \lambda$ from marginal operators tells you that your Feynman diagrams are going to have inverse powers of $\Delta - d$ instead of $\epsilon$. This is not a divergence because $d$ going to an integer does not make it zero. Therefore, counterterms are not needed to get finite answers.

To convince yourself of point 3, you can use the interaction vertex with "too many legs" to build diagrams that have "even more legs" and show that some of them have a positive degree of divergence. In order to cancel this divergence with counterterms, interactions with these extra fields need to be added in a runaway process.

The evidence you give about this correspondence not being simple seems to be an error.

Is there any reason why $\phi^4$ and $\phi^6$ don't need renormalization despite being marginal/relevant in three dimensions?

When you add the marginal operator $\phi^6$ in three dimensions, you get divergent diagrams so you do need renormalization.

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  • $\begingroup$ Thank you. I have two followup questions. Could we use these RG classifications to determine whether a term needs to be renormalized or not? For example, with $\phi^4$ in three dimensions we need counter-terms for $\phi^0$ and $\phi^2$, but not for $\phi^4$. Would we be able to predict this without doing any perturbative calculations and just from the RG analysis? Second, does the argument you give also imply that no relevant or marginal terms can appear in the RG flow if none are present in the original (effective) Lagrangian (i.e. only irrelevant terms can appear)? $\endgroup$ Commented 2 hours ago
  • $\begingroup$ The circle with a line going through it has two 3d momenta to integrate and three quadratic propagators. So that is 6 powers is the numerator and another 6 in the denominator leading to 0 in total which is a logarithmic divergence. Do you consider this power counting to be a perturbative calculation? My way of predicting would go along those lines since I'm not sure if there's an easy rule. $\endgroup$ Commented 2 hours ago
  • $\begingroup$ Whether marginal and relevant operators appear along the flow is a question of scheme. Often, there will be papers starting off with a massless scalar in the UV and then saying mass is never generated. But that is just because they are using the MS-bar scheme. With chiral fermions however, the massless limit is protected by symmetry. $\endgroup$ Commented 2 hours ago

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