Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Perturbation theory presumes we have a valid family of models over some continuous (infinitely differentiable, in fact) range for some parameters, i.e. coupling constants. We have some special values for the coupling constants characterizing the unperturbed model, which presumably, is relatively easy to solve. We also assume the family of models transform smoothly under the coupling constants. Then, we perform a Taylor series analysis.

But what if the landscape of valid quantum gravity models is discrete? Even though superstring theory admits a dilaton modulus over 10 uncompactified dimensions, what about the landscape of models we get after compactifying 6 spatial dimensions with nonzero fluxes and some branes, and maybe some orbifolding? We still have moduli when supersymmetry remains unbroken, but what about the metastable states where SUSY is broken? What is the Taylor series of a Dirac delta function?

What about perturbation theory from the perspective of path integrals? With path integrals, the Wheeler-DeWitt constraint shows up in a different guise as a projection operator. We start off with some wavefunctionals, and then take the functional integral over some finite time interval T. In the limit as T goes to infinity, we are left with a projection operator singling out WDW solutions. But what happens when we interchange the order in which we take the limit of the coupling constant going to zero, and T going to infinity? If the spectrum of the Hamiltonian constraint is discrete, and varies with the coupling constant, such an interchange won't be valid! This is a fancy way of saying for most choices of coupling constants, the projection operator is zero.

share|improve this question
You might be interested in the following thread motls.blogspot.com/2011/01/twistor-minirevolution-goes-on.html .If you listen to the lecture of Nima Arkani-Hamed, linked there, you will see that people are moving beyond perturbation theory. After all, expansions in series are tools when one cannot find analytic solutions. Also one expansion of a true solution might break down because of a wrong choice in expansion, and another might be quite good and converge fast. Let alone the QCD example. –  anna v Feb 6 '11 at 8:45

2 Answers 2

The perturbative expansion of general relativity breaks at the two-loop level: one produces a term in the effective action which is cubic in the Weyl tensor, as showed by Goroff and Sagnotti, which has a UV-divergent coefficient, and which has to be canceled by a counterterm. This introduces a new unknown finite coupling to the "no longer Einstein-Hilbert action", too. Because of that, one ultimately gets an infinite number of unknown couplings and loses predictivity.

We say that general relativity is perturbatively non-renormalizable.

This shows that there has to be new physics that determines all the unknown parameters of the low-energy physics (and, typically, adds qualitatively new phenomena at high energies, too). The spectrum of possibilities is given by the "landscape" of string/M-theory. All versions of the Minkowski or anti de Sitter or de Sitter space span a complicated set with many components. Some of these components are discrete; we call them the stabilized vacua. Some of them have residual parameters - the moduli - and those are very interesting mathematically (and they are usually calculable, and often have some unbroken supersymmetry) but they are unacceptable phenomenologically.

Only the stabilized vacua are realistic candidates for a theory of the real world. The unstabilized ones violate the equivalence principle, allow the fine-structure constant and similar constants to vary, and lead to new, unobserved long-range forces.

However, around each vacuum, it's still true that the amplitudes may be Taylor-expanded in any coupling constant that happens to be weak. The fact that only one value of a coupling constant - or a scalar field, if we're in the gravitational context - is the right one is seen as the existence of a potential for this scalar field. If we find ourselves away from the minimum (or extremum) of this potential, there will exist non-zero one-point functions for this scalar field that drive the Universe back to the minimum.

When the computation is done properly, the scattering amplitudes are analytic functions of the energy-momentum vectors "almost everywhere". This fact is guaranteed by the locality (or even approximately locality) of the physical phenomena in spacetime. So there can't be any delta-functions here - except for those that impose the conservation laws.

So it is indeed true that the Minkowski-like or anti-de-Sitter-like "empty space" solutions only exist for the right values of the couplings that minimize the potential; the physical spectrum of "empty space" states is strictly vanishing away from the right stabilized value of the moduli. But this fact shouldn't be viewed as a discontinuity in the underlying maths. Instead, you should imagine that for wrong values of the coupling constants, there exist "analogous" solutions which are not "empty space". Instead, in these solutions, the scalar fields oscillate around their preferred value for which the potential is minimized.

The states "don't disappear" as you move to wrong values of the coupling constant. Instead, they just fail to be translationally symmetric in time. In this sense, there's no discontinuity and the calculations of physical amplitudes and other observables never involve delta-functions of the scalar fields.

Much more generally, however, it is of course conceivable that the perturbative expansions break down for many known and unknown reasons. It's been known for a long time that the perturbative expansions ultimately diverge; and they don't capture all the physical phenomena, anyway. If we resum the terms up to the minimal one (before they start to blow up again) in a divergent expansion in a coupling constant, the minimum term - an uncertainty of the sum - is of the same order as the first non-perturbative contributions to the amplitudes (various instantons). But attempts to "resum the perturbative expansions more properly" are not the only methods to approach the non-perturbative physics.

On the other hand, if a perturbative expansion were "qualitatively" failing even for a very small value of the coupling constant, it would probably prove that the theory is inconsistent. This shouldn't happen. Even if it happened in a hypothetical theory, you would have to think how this theory allows to calculate something, at least in principle - without that, you shouldn't talk about a theory at all.

share|improve this answer

The perturbation theory for QG breaks down as well as for any other QFT: one obtains infinite perturbative corrections. The only difference is that the UV infinities in QG cannot be discarded in a "consistent" way. They say the QG is "non renormalizable". That means it is not normal initially and is not normal after "reparations" (renormalizations).

There is a simple criterion to check if a theory physically and mathematically reasonable. This criterion is not very popular but it reveals a fallacious theory in the first Born approximation, i. e., even before encountering infinities. This criterion is the following: if the first Born approximation cannot catch the processes whose probability is unity (soft radiation), then you will obtain an explosion of perturbative corrections. If one misses the processes that happen always, one is doing something rather wrong, and no wonder the perturbative corrections will try to "violently correct" that bad start: the perturbation theory initial approximation is too far from the exact solution and needs infinite corrections and a their non linear summation to obtain a finite result. This is a simple physical and mathematical explanation of origin of those "difficulties" that are usually hided later on under rug.

Downvoters, if you disagree, give your disagreement statements, please.

share|improve this answer
You don't know nothing about Renormalization Group? here –  Nogueira Jan 22 at 15:53
@Nogueira: Like here youtube.com/watch?v=3A252xY-a0o –  Vladimir Kalitvianski Jan 22 at 20:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.