# Are effective field theories mathematically well-defined?

1. Apr 7, 2010

### petergreat

The field theories in the standard model haven't been shown to be mathematically consistent. Do "effective field theories" bypass this difficulty? Are they completely mathematically well-defined as an approximation? Of course, if the deeper underlying theory is another QFT or string theory, you still can't prove mathematical consistency. But suppose the underlying theory is a lattice model with a large but finite degree of freedom, do all the problems disappear?

2. Apr 7, 2010

### Haelfix

I consider the lattice models very well defined. The problem is they are not necessarily the same thing as what they try to approximate (you have to do some work to motivate that), nor is the approximation for finite spacing necessarily good (again you need to show this depending on the context and example in mind). Relatedly, they don't necessarily work for every type of lagrangian --it might be difficult to simulate chiral fermions, or say supersymmetric theories for instance.

3. Apr 7, 2010

### ansgar

what is not mathematically consistent?

4. Apr 7, 2010

### petergreat

Renormalization is not mathematically consistent. And the series doesn't converge even after renormalization, e.g. in QED you need to sum up the first ~137 leading orders to get a result, after which the series diverges.

A problem is that the presence of a lattice explicitly breaks Lorentz invariance. Is there a way around this? I don't know much about this subject, but I'm getting an impression (am I right?) that a mathematically consistent, Lorentz invariant, interacting quantum theory for D>=4 simply doesn't exist, since at least 3 approaches have failed the criterion: i) Point-like particle theories are ruled out by no-go theorems. ii) QFT has inconsistencies. iii) String theory doesn't even have a defining equation.

5. Apr 7, 2010

### ansgar

can you tell me a book which states this?

6. Apr 7, 2010

### Haelfix

"A problem is that the presence of a lattice explicitly breaks Lorentz invariance. Is there a way around this"

No! However its not that bad, considering you can often show that the symmetry is restored in the continuum limit (similar to the case of dimensional regularization).

Technically the discrete residual subgroup is usually large enough to guarentee that all the lorentz violating operators are irrelevant and hence dissappear when you set the spacing to zero and the volume to infinity.

However, it can happen that this is not the case, and an example of the lattice operation failing to be commutative.

7. Apr 7, 2010

### George Jones

Staff Emeritus
This was demonstrated by Dyson [1,2,3] (but not proved rigourously) more than 50 years ago.

From [4], page 451: "Thus, QED may have a zero radius of convergence in $\alpha$ space."

From [5], page 253 "Nobody knows for sure whether the series converges, but nobody seriously believes that it does."

From [6], page 259: "The belief is that the perturbation series is an asymptotic series for real e at e = 0."

[1] Phys. Rev. 85, 631 (1952): Divergence of Perturbation Theory in Quantum Electrodynamics,
http://prola.aps.org/abstract/PR/v85/i4/p631_1

[2] QED and the Men Who Made It, S. Schweber, 9.17 Divergence of the Perturbation Series

[3] Quantum Field Theory I Basics in Mathematics and Physics: A Bridge Between Mathematicians and Physicists, 15.5 The Convergence Problem in Quantum Field Theory

[4] Quantum Field Theory, M. Kaku, 13.5 Does Quantum Field Theory Really Exist

[5] Quantum Field Theory: A Tourist Guide for Mathematicians by Gerald Folland

[6] Quantum Field Theory for Mathematicians, R.Ticciati, Remark 9.4.12

Regards,
George

Last edited: Apr 7, 2010
8. Apr 8, 2010

### genneth

To return to the OP: usually effective field theories are not postulated to exist on some lattice, but simply to have some cutoff. Formally, all integrals in k-space have both upper and lower limits, and it may be the case that neither of these can be taken to infinity, whilst keeping physically relevant observables, well, physical. But you are correct that lattice models do not suffer from the usual problems of QFT, though with the caveats that people have already pointed out for whether in the limit they approximate the theories you want.

A point for thought however: if the "theory you want" is actually ill-defined, then surely not managing to converge to it is in fact the right thing to do? Just a thought.

9. Apr 8, 2010

### Parlyne

Renormalization may be weird; but, there's nothing fundamentally mathematically inconsistent about it.

And, the fact the the perturbation series diverges is not an indictment of the underlying field theory. The problem here (which is generic) is that treating interactions as a small correction to the behavior of free fields only works consistently when the couplings have particular signs. Since the expansion is around the point where all couplings are set to 0, the radius of convergence for the series can only be 0. This is independent of the consistency of the underlying theory.

10. Apr 8, 2010

### George Jones

Staff Emeritus
What underlying theory?

11. Apr 8, 2010

### genneth

The one where you can get the answers without perturbation theory, with only convergent sums.

If only we had even one of those...

Then again, I actually regard the asymptotic series to be a non-problem, but I'm a condensed matter theorist, so what do I know about rigour...

12. Apr 9, 2010

### Parlyne

My point was simply that using an approximation that formally diverges doesn't mean that the quantity you're trying to calculate actually diverges. This is not in any way specific to QFT. The added wrinkle in QFT is that the only things we know how to calculate exactly (free field results) exist at a point in parameter space where the radius of convergence of any series expansion in coupling constants is strictly 0.

13. Apr 9, 2010

### humanino

We know for a fact that the non-justified algebraic manipulations physicists test everyday provide actual results. Although we do not know how to justify them, we have all good reasons to believe that such a justification must exist. By the same token, it took roughly a century to fully justify the manipulations Joseph Fourier initiated in the beginning of the nineteenth century.

A lot of progress has been made in half a century already.
Algebra for quantum fields

14. Apr 9, 2010

### petergreat

Is there a proof showing that the actual quantity doesn't diverge? Or is there a proof showing that the actual quantity does diverge, so there is no escape from renormalization, irrespective of whether you use perturbative methods? Or is there no definitive proof on either sides?

15. Apr 9, 2010

### petergreat

My point is that if the QFTs in the SM are actually effective field theories (which may or may not be true), and if the effective field theories themselves are mathematically well-defined, then we can bypass the mathematical difficulties of QFTs. Therefore these mathematical endeavours trying to justify QFT may be unnecessary after all.

16. Apr 9, 2010

### humanino

Even being an effective theory at low energy, the SM has been tested beyond tree level, in fact the standard model without quantum corrections is sometimes ruled out by more than 100 standard deviations compared to measurements (and in agreement with measurements after quantum corrections are taken into account). The paper by Kreimer is not an attempt to remove the Landau pole at insanely high energy. It addresses the mathematical justification of order-by-order cancellations of divergences, without a non-perturbative definition of the amplitude to start with. It addresses the non-definiteness of the path integral itself, the fact that the QFT we need for the SM are von Neumann algebras of type III. It is an attempt to define QFT directly from the Feynman rules. My understanding, the ultimate goal is to have rigorous definitions including for what a path integral is, from the constructions obtained in the Feynman graph algebras.

17. Apr 9, 2010

### genneth

But why should we expect that the SM holds to arbitrarily high energies? If we impose any cut off, then there's no problem. My personal flavour is to have a fundamental theory with only a countable number of degrees of freedom.

18. Apr 9, 2010

### humanino

I do not care about arbitrarily high energy. I do not care that QED goes crazy at 137 loops.

In the finite energy where we have already tested the standard model, the classical or tree level is ruled out. You need to use the quantum version of the standard model including higher orders, loops, radiative corrections... Those are plagued by infinities canceling each other. But the quantum path integral one starts with to define the amplitude, and then expand perturbatively, this path integral is ill defined.

Renormalization is purely algebraic anyway, it is matter of counting legs in Feynman graphs. The evolution coefficients depend only one algebraic properties of the gauge groups. So there is hope to entirely redefine the amplitude without path integrals.

19. Apr 9, 2010

### genneth

Let's make sure we're not talking at cross purposes. I think you're trying to say that it would be nice to have a rigorous relativistic QFT, which can be the basis of the SM. I agree completely. I also think it is unnecessary, since I'm happy to put a cut off in, and then my life is easy.

If you mean that perturbation expansions around a free theory cannot be considered a proper theory, then I'm not sure how to proceed.

20. Apr 10, 2010

### Haelfix

I think the point is less the final result which is meant to be approximately valid at some well motivated energy scale, and moreso the derivation thereof that is problematic on a mathematical level.