Are non-renormalizable theories less predictive than renormalizable theories?

Question

Non-renormalizable field theories contain non-renormlizable operators whose couplings have negative mass dimension (for example, Fermi coupling in the Fermi theory of weak interaction). These couplings provide an energy scale $\Lambda$ built into the theory, and it is said that the predictions above energy $E\geq \Lambda$, is not reliable.

How does one understand whether the predictions of a nonrenormalizable theory, below $\Lambda$ are reliable but bound to fail above $\Lambda$?
Consider a renormalizable theory such as the Standard model. There is no built-in length scale. Therefore, if its predictions are tested at energy $E=E_1$, can we not claim that it's predictions, will be perfectly reliable to arbitrarily high energies (for example, $E=10^{16}\times E_1$), if there is no new physics that enter in between.
Such a claim may be (or must be?) false for a nonrenormalizable theory. Isn't it?

@ACuriousMind- 1. For concreteness, I want to understand it for the Fermi theory. 2. SM was just an example. My question was meant for a comparison between a renormalizable theory and a non-renormalizable one. That is why I carefully stated, if there are no new physics. I think that is an advantage of a renormalizable theory. If we find deviation from its predictions in high energy experiments, we are sure that it is due to new physics. — SRS, Nov 29 '16 at 14:00
@ACuriousMind- But if Standard Model were non-renormalizable, then the deviation above a characteristic energy scale would be expected even theoretically. And need not be a signature of new physics. Isn't it? — SRS, Nov 29 '16 at 14:09
About the first question: in an effective field theory you usually work with power series expansions of $E/\Lambda$, so you can't expect your results to be valid at $E\geq \Lambda$. — coconut, Nov 29 '16 at 14:51
The Standard Model does have a built-in length scale of $1\times10^{-18}$ m set by the Higgs mass. The coefficient of the Higgs mass term in the (pre-symmetry-breaking) Standard Model Lagrangian is dimensionful. — tparker, Dec 05 '16 at 07:41

coconut · Answer 1 · 2016-12-06T17:06:57.723

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The effective lagrangian is usually written as an expansion in inverse powers of the energy cut-off $\Lambda$. That means that observables at some energy $E$ will be computed as the first terms of an expansion in powers of $E/\Lambda$. When $E\gtrsim\Lambda$, higher powers contribute more than lower ones and the calculations using the effective theory are no longer valid.
The Standard Model does indeed have a built-in length scale, but it doesn't matter for this discussion. In the case of the Standard Model (as in any other renormalizable theory) we can of course claim that it is reliable to arbitrarily high energies if new physics don't arise in between. But notice that this is almost by definition of new physics! The statement "there is no new physics" means that "old physics" remain valid.
For a non-renormalizable theory the logic is the same, so the same statement can be made, with the extra advantage that the theory itself tells you at which maximum energy scale the new physics are expected to appear.

edited Dec 06 '16 at 17:06

answered Dec 06 '16 at 12:48

coconut

4,653

No one theory tells you at which scale the new physics is expected to appear. It is the experiment who tells you that. – Vladimir Kalitvianski Dec 06 '16 at 13:11
Effective field theories usually tell you the energies around which where they will fail. That means that a new theory (new physics) has to replace them at that energy. Of course, we can only know what the new physics are by experiment. – coconut Dec 06 '16 at 13:32
Consider Classical Electrodynamics and tell me what it tells you, please. – Vladimir Kalitvianski Dec 06 '16 at 14:23
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We are talking here about quantum non-renormalizable effective field theories. Classical electrodynamics is not an example – coconut Dec 06 '16 at 14:33
Yes, CED is a perfect example of a non-renormalizable theory that shows the necessity of experimental data in order to determine its region of validity. – Vladimir Kalitvianski Dec 06 '16 at 16:59
Classical electrodynamics is not a quantum theory. – coconut Dec 06 '16 at 17:06
Yes, CED can be "quantum" if one considers the fields in a cavity with a discrete EMF spectrum. CED fails at low intensity fields then. – Vladimir Kalitvianski Dec 06 '16 at 17:42

score 2 · Accepted Answer · answered Dec 06 '16 at 16:41

A renormalizable theory is determined by a fixed number of parameters; once these are determined, all of its predictions are determined (though not necessarily easily calculable) at every energy.

A nonrenormalizable theory requires at higher and higher energies more and more counterterms and corresponding parameters that must be determined to make definite predictions at a fixed accuracy, and depending on the behavior of the resulting higher order terms, the asymptotic expansion may break down completely above some energy. Given a fixed number of terms, we still have an infinite family of theories with these low order terms, which have different behavior at higher energies. Thus the theory is not predictive enough at high energies.

The above holds for any quantum field theory, irrespective of its realization in Nature. Whether the predictions of a specific (renormalizable or unrenormalizable) theory are matched by Nature is a completely different matter and can be decided only by comparing the predictions with experiment. The latter is also needed to adjust the constants on which the predictions depend.

Your answer is quite helpful except I want to be sure at one point. You explain, "Given a fixed number of terms, we still have an infinite family of theories with these low order terms, which have different behavior at higher energies. Thus the theory is not predictive enough at high energies." By this, Do you mean, for example, the low energy non-renormalizable Fermi theory could have come from more than one high energy renormalizable theories other than standard model? And therefore, it is not predictive enough at high energy? — SRS, Dec 28 '16 at 10:09
Yes. It is like approximating a function by a low order polynomial. The approximation could have comte from many functions. — Arnold Neumaier, Dec 31 '16 at 11:25

Are non-renormalizable theories less predictive than renormalizable theories?

2 Answers2