Non-renormalizable theories, when regarded as an effective field theory below a cut-off $\Lambda$, is perfectly meaningful field theory. This is because non-renormalizable operators can be induced in the effective Lagrangian while integrating out high energy degrees of freedom.

But as far as modern interpretation is concerned, renormalizable theories are also effective field theories. Then why the renormalizability of field theories is still an important demand? For example, QED, QCD or the standard model is renormalizable. What would be wrong if they were not?

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    $\begingroup$ Related: physics.stackexchange.com/q/4184/2451 $\endgroup$
    – Qmechanic
    Commented Nov 28, 2016 at 14:36
  • $\begingroup$ Notice that in the standard literature non-renormalisable refers to non-renormalisable perturbatively; this does not however exclude any other renormalisation procedure to be in place (e. g. all gravity models with spin networks and discrete space-time). $\endgroup$
    – gented
    Commented Nov 28, 2016 at 14:40

2 Answers 2


In the modern effective field theory point of view, there's nothing wrong with non-renormalizable theories. In fact, one may prefer a non-renormalizable theory inasmuch they tell you the point at which they fail (the energy cut-off).

To be concrete, consider an effective lagrangian expanded in inverse powers of the energy cut-off $\Lambda$:

\begin{equation} \mathcal{L}_\mathrm{eff}(\Lambda)=\mathcal{L}_\mathrm{renorm}+ \sum_\mathcal{\alpha}\frac{g_\alpha}{\Lambda^{ \operatorname{dim}\mathcal{O}_\alpha-4}}\mathcal{O}_\alpha \end{equation}

where $\mathcal{L}_\mathrm{renorm}$ doesn't depend on $\Lambda$, $\mathcal{O}_\alpha$ are non-renormalizable operators (dim. > 4) and $g_\alpha$ are the corresponding coupling constants. So at very low energies $E\ll \Lambda$ the contributions from the non-renormalizable operators will be supressed by powers of $E/\Lambda$.

That's why the Standard Model is renormalizable, we're just unable to see the non-renormalizable terms because we're looking at too low energies.

Notice also that as we increase the energy, the first operators to become important will be the ones with the lower dimension. In general, contributions from non-renormalizable operators will become important in order given by their dimension. So you can see that, although there are infinite possible non-renormalizable coupling constants, you can make the approximation of cutting the expansion of the effective lagrangian at some power of the cut-off and get a finite number of parameters.


Renormalizability allows one to fix all the parameters in the theory by measuring just a few amplitudes (or cross sections, or decay rates), because there is a finite number of divergent loop integrals, and then compute all the observables in terms of these "physical" parameters, while for the non-renormalizable theory it is impossible, and one needs an infinite number of measurements to fix an infinite number of couplings which must be renormalized.

There is a beautiful discussion of the subject in the first volume of Weinberg's course.

  • $\begingroup$ @ Andrew Feldman- Can you give the reference to that passage or page number of Weinberg? $\endgroup$
    – SRS
    Commented Nov 28, 2016 at 14:51
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    $\begingroup$ @SRS Read the subsection 12.1. More concretelely, you can read the discussion around the formula (12.1.10). $\endgroup$ Commented Nov 28, 2016 at 15:32
  • $\begingroup$ The downvoter should point out what's wrong with this answer. $\endgroup$
    – SRS
    Commented Apr 9, 2019 at 9:04
  • $\begingroup$ But should not the couplings follow certain pattern? $\endgroup$
    – Anixx
    Commented Feb 11, 2021 at 20:54
  • $\begingroup$ Yes. If one defines a coupling as an amplitude at certain momenta $\Lambda$, then $g(\Lambda)$ is of course change as we change $\Lambda$. $\endgroup$ Commented Feb 12, 2021 at 7:43

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