I have heard that nonrenormalizable operators (i.e., mass dimension greater than 4) can be "induced" in the Lagrangian (that we started with) via loop effects. However, I do not understand what does it mean by a new operator or term in the Lagrangian being induced.

In QED, loops generally correspond to self-energy diagrams, vacuum polarization diagrams and vertex correction diagrams which modify the bare mass, bare charge, and bare fields to the corresponding renormalized quantities by adding counterterms (or by splitting the original Lagrangian into a renormalized part and a counterterm part). Counterterms must be included, as I understand, to get rid of various divergences (or cut-off dependence).

$\bullet$ But why should one include nonrenormalizable terms in the original Lagrangian (in which none of the terms did resemble the induced operator)?

$\bullet$ Can they be thought of as counnterterms?

$\bullet$ How many of them can/should be included?

$\bullet$ Can one clarify the concept inducing non-renormalizable operators in the context of a simple field theory?


1 Answer 1


Oddly enough, non-renormalizable operators appear when renormalizing:

In the Wilsonian viewpoint, every QFT is defined as an effective theory with an intrinsic momentum cutoff $\Lambda_0$. Renormalizing the theory corresponds to lowering this cutoff by integrating out the Fourier modes of the fields above the new cutoff $\Lambda$ in the path integral. It turns out that, if you (formally) compute this integral, that the result is a new partition function which looks like the partition function for a different action, the Wilsonian effective action, and that it includes all terms not forbidden by symmetry no matter their renormalizability. Effectively, you should think of this as the original action including all those terms with a coupling constant of 0, and when renormalizing by lowering the scale towards the infrared cutoff, they start to appear again because the coupling gets renormalized away from 0 unless protected by symmetry.

An illustrative example of how non-renormalizable terms can appear in the infrared is the Fermi theory of beta decay, which has a non-renormalizable interaction part $\propto \bar\psi_p\psi_n\bar\psi_e\psi_\nu$ where $n,p,e,\nu$ denote the spinor field of the proton, neutron, electron and neutrino, respectively. This theory is amazingly predictive in the low-energy regime, but this is just an effective vertex one gets from integrating out the weak interaction - in the full Standard model, this four-fermion vertex is resolved into the fermions interacting not directly, but via vertices involving W- and Z-bosons of the weak interaction.

  • $\begingroup$ @ACuriousMind- This answer is extremely illuminating. However, is it exactly true that every quantum field theory is effective field theory? For example, I don't know much about string theory but what I have heard is that it is super-renormalizable. It's not an effective field theory. Correct? $\endgroup$
    – SRS
    Commented Nov 17, 2016 at 15:33
  • $\begingroup$ Moreover, can you also give reason for including such terms if one is used to renormalization calculations using the method of counterterms? $\endgroup$
    – SRS
    Commented Nov 17, 2016 at 15:35
  • 1
    $\begingroup$ @SRS String theory is not an effective field theory, but it's also not a quantum field theory. It's not true that every QFT must be effective, for instance, (some) conformal field theories are fixed point in the UV, and you can effectively lift the cutoff to $\infty$. $\endgroup$
    – ACuriousMind
    Commented Nov 17, 2016 at 15:55
  • $\begingroup$ You said that I should think of the non-renormalizable terms to be absent in the original Lagrangian (or present with zero coupling), and they appear when we integrate high-momentum modes (or equivalently, lower the cut-off) i.e., in the low-energy approximation of the theory. But isn't it odd that non-renormalizable (irrelevant) operators start contributing at low-energy Lagrangian rather than the original high-energy Lagrangian? This behaviour is opposite to how irrelevant operators behave. Do I misunderstand something? @ACuriousMind $\endgroup$
    – SRS
    Commented Jan 3, 2018 at 9:48

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