- #1

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*finite*degree of freedom, do all the problems disappear?

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- #2

Haelfix

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what is not mathematically consistent?finitedegree of freedom, do all the problems disappear?

- #4

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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.what is not mathematically consistent?

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.I consider the lattice models very well defined.

- #5

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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.

can you tell me a book which states this?

- #6

Haelfix

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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.

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George Jones

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This was demonstrated by Dyson [1,2,3] (but not proved rigourously) more than 50 years ago.can you tell me a book which states this?

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

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- #8

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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

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

And, the fact the the

- #10

George Jones

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What underlying theory?This is independent of the consistency of the underlying theory.

- #11

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

If only we had even

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...

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- #13

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A lot of progress has been made in half a century already.

Algebra for quantum fields

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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?

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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.

A lot of progress has been made in half a century already.

Algebra for quantum fields

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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.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.

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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....the QFT we need for the SM are von Neumann algebras of type III...

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

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.

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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.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.

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

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Haelfix

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