This post relates to this previous one. My question is, what is the actual meaning of a theory being renormalizable?

There might be at-least two possibilities (correct me if I am wrong)

1.Power-counting renormalizable

If I understood correctly, the BHPZ theorem guarantees that all superficial divergences for a power-counting-renormalizable theory can be absorbed by counter terms. Formally, it might be possible that a counter term is outside the form of the Lagrangian.

2.Stronger than power-counting renormalizable

The counter term be produced from the original Lagrangian, such as QED. We need to look at all possible divergence explicitly and show how to regularize and renormalize them.

(3. Any further possibility from non-perturbative aspect (I don't know))

This problem also relates to 't Hooft and Veltman's achievement of proving Yang-Mills theory with Higgs mechanism is renormalizable. It cannot simply be power counting.

When people say a theory is renormalizable, what does it mean? Meaning 1 or 2? Is this distinction important?


2 Answers 2


The correct option is really option 3. Most of the time when a physicist says a theory is renormalizable, they mean that the theory is a relevant deformation of some conformal field theory.

This is a non-perturbative definition. It contains the physically meaningful content that the other more technical definitions about counterterms in perturbation theory are attempting to capture. Indeed, it implies them. (However, the reverse implication is not always true. For example, perturbative QED is renormalizable in the sense of Option 2, but there is no underlying non-perturbative QED, so one can't even ask about Option 3.)

It's good practice to always try to think about the non-perturbative meaning of the physical formalism you are studying. Perturbation theory is a sometimes useful tool for computations, but it can obscure the physics in a cloud of virtual technicalities.

So what does it mean for a theory to be a relevant deformation of a CFT?

It means that there's a CFT whose observables are essentially the same as the observables in your QFT, and that you can compute any correlation function in the QFT as

$\langle \mathcal{O} \mathcal{O}' ...\rangle_{QFT} = \langle \mathcal{O} \mathcal{O}' ... e^{\sum_i g_i \int\mathcal{O}_i} \rangle_{CFT}$

where the $\mathcal{O}_i$ are relevant operators in the CFT and $g_i$ are (dimensionful) coupling constants. Knowing that your QFT is near a CFT in this sense is what allows you to study the behavior of the QFT's expectation values under changes of scale, which is the heart of the renormalization group analysis.


First, an easy example: The free scalar field theory is a conformal field theory. This theory is basically described by $\langle \mathcal{O} \rangle_{CFT} = \int \mathcal{O}(\phi) e^{i\int |d\phi|^2} \mathcal{D}\phi$. In this theory, in dim >= 3, the operator $\phi^2(x)$ is relevant, so we can deform with this term and get a non-conformal field theory. The expectation value in this QFT is then described by

$ \langle \mathcal{O} \rangle_{QFT} = \langle \mathcal{O} e^{i m^2 \int \phi^2(x)dx}\rangle_{CFT} = \int \mathcal{O}(\phi) e^{i\int [ |d\phi(x)|^2 + m^2\phi^2(x)]dx} \mathcal{D}\phi.$

So, not surprisingly, the theory we get by deforming the free scalar CFT with a mass term is the massive free scalar.

Second, a subtlety that I should point out. The first equality above isn't exact in most situations. The problem is the deformations we want may not be integrable with respect to the CFT's path integral measure, thanks to UV singularities. This is dealt with by regularizing. So, in most QFTs, what we get is a family of approximations

$\langle \mathcal{O} \mathcal{O}' ...\rangle_{QFT} \simeq \langle \mathcal{O} \mathcal{O}' ... e^{\sum_i g_i(\Lambda) \int\mathcal{O}_i(\Lambda)} \rangle_{CFT}$

where the relevant operators and the coupling constants depend on a cutoff scale $\Lambda$ and the errors vanish as $\Lambda \to \infty$.

  • $\begingroup$ Can you please elaborate on your statement "there is no underlying non-perturbative QED". $\endgroup$
    – user10001
    Commented Dec 4, 2013 at 6:22
  • 1
    $\begingroup$ QED is believed to have a Landau pole. This implies that, while perturbative QED may make sense as a collection of statements about formal power series, non-perturbative QED must be a non-interacting theory. The perturbative computations are not an approximation to non-perturbative computations made using the same variables. $\endgroup$
    – user1504
    Commented Dec 4, 2013 at 12:47
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    $\begingroup$ @user10001: Perhaps I should have underlined more strongly the condition "using the same variables". Perturbative QED calcuations are a useful approximation to computations which make sense in the context of some effective field theory. But they are not an approximation to computations done in a continuum non-perturbative QFT whose basic fields are a U(1) gauge field and a Dirac spinor. It's exceedingly unlikely that such a theory exists. $\endgroup$
    – user1504
    Commented Dec 4, 2013 at 18:45
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    $\begingroup$ Yeah, the point is that you can't treat QED as anything other than an effective field theory. What you're suggesting is possible, but seems very improbable. The lattice studies I know of point towards triviality when the theory is weakly coupled at long distances. $\endgroup$
    – user1504
    Commented Dec 4, 2013 at 19:45
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    $\begingroup$ I'm just doubtful about the statistics when you say "Most of the time when a physicist says a theory is renormalizable... ", most of the physicist I've met would take option (2), e.g. they would say QED is renormalizable without further qualifications. I hope this is not nitpicking since the question is about the conventional usage of terminology. $\endgroup$
    – Jia Yiyang
    Commented Feb 26, 2014 at 12:03

Usually one refers to option 2) when talking about the renormalizability of a theory. Often power counting is used to determine at a glance whether a theory has a chance of being renormalizable or not.

  • $\begingroup$ In practice, if I compute some loop diagram and I found I need to add some counter term outside the Lagrangian (the power-counting renormalizable but not option 2 theory), is this issue causing any arbitariness in prediction? Does it regard as a serious problem? $\endgroup$
    – user26143
    Commented Dec 4, 2013 at 0:14

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