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A Does QFT have problems

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  1. Apr 28, 2017 #1

    ftr

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    QFT seems to be a bit sick with cluster decomposition assumption ..etc. So here comes Haag's theorem and Wightman axioms to the rescue, or do they? So what do these cures actually say differently than the generic QFT . Do they solve any practical problems, if not why the fuss, millennium prize and all.
     
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  3. Apr 29, 2017 #2

    A. Neumaier

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    Only relativistic QFT in 4 space-time dimensions is sick because interacting fields are not well-defined in any logically coherent sense.

    Theoretical physicists predicting particle properties work around using either finite lattice approximations, which is a quantum field analogue of solving partial differential equations using finite difference methods breaking all spacetime symmetries, or low order formal renormalized perturbation theory, which is to a logically precise specification like Euler's treatment of functions through formal power series a few hundred years ago. These methods seem to work and give sensible physical predictions, but they have no proper logical foundation.

    The Wightman axioms spell out how the vacuum sector of such a theory should look like according to our present best guess, but (unlike in lower dimensions where there are many constructions) no interacting relativistic QFT in 4 space-time dimensions has been constructed in the more than 50 years of existence of the Wightman axioms. Neither are there nonexistence theorems. The simplest tractable case is widely held to be 4D Yang-Mills theory, which is the reason why there is a millennium prize on solving it. See https://www.physicsoverflow.org/217...osons-infinite-and-discrete?show=21846#a21846 for more comments on the latter.

    Cluster decomposition is an important principle without which physics would be impossible. It basically says that far away from other particles particles behave independently of each other. This is the basic reason why we can look at (properly prepared) small objects independently of other objects. For its physical basis see on the informal level Weinberg's QFT book, Vol. I, Chapter 5. On the rigorous level, it guarantees the uniqueness of the vacuum state; see any book on algebraic quantum field theory.

    Haag's theorem is hardly related to that; it only says that the CCR representation of the free field cannot extend to interacting fields, the interacting Hilbert space must look quite different from a Fock space. More precisely, while they have to be isomorphic as Hilbert spaces (all infinite-dimensional separable Hilbert spaces are isomorphic), the additional structure needed to set up quantum fields cannot be respected by such an isomorphism.
     
    Last edited: Apr 29, 2017
  4. Apr 29, 2017 #3

    vanhees71

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    Recently a colleague after a semester-long discussion on the foundations of QFT pointed me to this very nice review on Haag's theorem and related works:

    https://link.springer.com/article/10.1007/s10670-005-5814-y
    http://philsci-archive.pitt.edu/2673/1/earmanfraserfinalrevd.pdf

    Don't worry! Although it says, it's about philosophy, it's in fact physics.

    Whenever a serious teacher of QFT comes to explain asymptotic states of interacting particles, the S-matrix, the LSZ reduction formalism s/he gets doubts and discusses it with colleagues, of course, with not too much success; it's indeed an unsolved problem to find a mathematically rigorous formulation of 4D relativistic QFT with interacting particles.

    FAPP QFT, as used by HEP phenomenologists, works with great success (even too much since the Standard Model stands up against all experimental efforts to disprove it; although recently LHCb came up with the next attack, but only at a bit over 2% confidence level yet ;-)). The way out is a practical approach: Just put the system in a box with periodic boundary conditions, use perturbation theory and renormalize. Then you get results with up to 12 significant digits accuracy. As Arnold said the other way out is the lattice approach and high computing power (at least in QCD, where you get, e.g., the hadron mass spectrum pretty nicely out too, and that's clearly beyond perturbation theory).

    That's the approach Weinberg chooses in his book: Simply ignore Haag et al, which however is a pity since I think one should be aware of it!
     
  5. Apr 29, 2017 #4

    A. Neumaier

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    This works for kinetic or hydrodynamic questions. But the resulting compactification destroys scattering theory since under these conditions the spectrum becomes wholly discrete.

    This ultra-high accuracy is not obtained in this way but through NRQED: One uses a nonrelativistic $1/c^2$ expansion to get rid of the problems with the relativistic theory and then works essentially with nonrelativistic quantum mechanics with relativistic corrections.
     
  6. Apr 29, 2017 #5

    ftr

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    Thanks both for your replies.

    Since I am not a professional theorist, my opinion probably don't add to much. But I feel that "mathematically rigorous" should be physically rigorous. I am well aware of the successes, but I think it is mainly due to correctness of QM itself. The QFT formulation is heroic to be sure , but I have read the story of the development. A very messy process of experimentation(with it own problems, bumps.etc) and a theoretician trying to score.

    But let me ask this ,I think the main issue with QFT is removing infinities and renormalization of the mass and charge, I think these are PHYSICAL problems and not "mathematical", do you agree.
     
  7. Apr 29, 2017 #6

    ftr

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    WE wrote this within minutes. Although I am no expert, I have deduced that.
     
  8. Apr 29, 2017 #7

    vanhees71

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    Where do you need the non-relativistic approximation to evaluate the anomalous magnetic moment of the electron?
     
  9. Apr 29, 2017 #8

    vanhees71

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    Renormalization has nothing to do with infinite radiation corrections per se. It comes to the rescue, but even if everything were perfectly finite, you'd need to renormalize to adjust the free parameters of the theory (wave-function normalization, masses, coupling constants) to observations.
     
  10. Apr 29, 2017 #9

    atyy

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    Practical QFT is eg. the standard model of particle physics. Most physicists nowadays understand it to be a low energy theory, ie. to make good predictions at the energies we use in experiments, it is not necessary for the theory to exist at all energies. This is the Wilsonian viewpoint, and it is a great advance for understanding why renormalization is ok, and has nothing to do with removing infinities etc. The main problem in this practical way of thinking is that it needs a non-perturbatively defined quantum theory, eg. lattice gauge theory, but while lattice QED is thought to be ok, chiral fermions on the latttice are still problematic.

    The Millenium prize has to do with 4D relativistic QFTs that exist at all energies (and not just at low energies).
     
  11. Apr 29, 2017 #10

    A. Neumaier

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    This doesn't explain the high accuracy achieved! QM without QED has not such a high predictive accuracy.

    To first loop order you don't need it but you don't get very high accuracy. For highest accuracy you cannot work with the standard Feynman approach; it is too daunting a computational task.
    Because of Poincare invariance, any relativistic QFT (in flat spacetime) will exist at all energies and (and not just at low energies).
     
  12. Apr 29, 2017 #11

    vanhees71

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    I thought Kinoshita has calculated (g-2)/2 to order ##\alpha_{\text{em}}^5##. Are there non-relativistic approximations involved? I don't see anything mentioned like this, e.g., here

    https://arxiv.org/abs/1205.5368

    But the trouble with some QFTs (including QED) at high energies, at least perturbatively (Landau pole). The modern consensus is that the Standard Model is a low-energy approximation of some other theory, of which we have no clue (not even if it exists at all).
     
  13. Apr 29, 2017 #12

    ftr

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    I know you need QED for precision, I was just thinking that QT is real, QFT is just a technique. May one day be united:smile:
     
  14. Apr 29, 2017 #13

    PeterDonis

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    What are you basing this on?
     
  15. Apr 29, 2017 #14

    ftr

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    It just looks to me that QFT takes QM massages it and turns it into a calculational tool.
     
  16. Apr 29, 2017 #15

    PeterDonis

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    You're leaving out at least three key physical phenomena that QFT can predict and ordinary non-relativistic QM can't:

    (1) The existence of processes where particles are created or destroyed;

    (2) The existence of antiparticles;

    (3) The connection between spin and statistics.
     
  17. Apr 29, 2017 #16

    atyy

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    QFT is a type of QT.

    Condensed matter uses non-relativistic QFT. Non-relativistic QFT can be derived from the non-relativistic Schroedinger equation for many identical particles.
     
  18. Apr 30, 2017 #17

    A. Neumaier

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    No. The paper you cited calculates some contributions up to the 10th order and even includes hadronic contributions (phenomenologically included into QED) since it already affects the 10th decimal. Indeed, I checked the references with the actual calculation details that no expansion in inverse powers of c is made. Thus the results are covariant.

    I had mixed up the computation of ##g-2## with the computation of the hyperfine structure (Lamb shift), where NRQED is essential to get high precision.
     
  19. Apr 30, 2017 #18

    A. Neumaier

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    There is no such trouble in formulations such as causal perturbation theory, that preserve covariance throughout. (The Landau pole is problematic only if a cutoff must be able to move across the pole.)
     
  20. Apr 30, 2017 #19

    vanhees71

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    Well, if you need a cutoff to define the theory then you admit that it's an effective theory valid up to the cutoff scale at most. The Epstein-Glaser approach avoids UV divergences, because it's more careful in dealing with products of distribution valued operators, and thus it solves a mathematical problem, but not the physical problem of the Landau pole. As far as I know, QED is an example for a QFT more likely not to exist in a strict sense than, e.g., QCD, which is asymptotic free, but for no realistic (i.e., (1+3)-dim. theory of interacting quantum fields) there has been a proof for it to exist in a strict sense. On the other hand FAPP this doesn't matter much, because we have anyway only a limited energy available, and it's quite likely that our contemporary Standard Model will fail at high enough energies somewhere. Today, nobody known where that scale might be. Maybe there's really a dessert up to the Planck scale, where one can definitely expect something should happen concerning quantum effects of gravity. Then HEP with accelerators is doomed, funding wise :-(.
     
  21. Apr 30, 2017 #20
    No. A lattice approximation with periodic boundary conditions defines a conceptually and mathematically valid and unproblematic theory. It does not have infinities, and does not need renormalization (it is, instead, possibly part of the process of renormalization - namely a regularized theory).

    Such a lattice theory can be interpreted as some approximation of the field theory, and gives all the observable (physical) results, with some accuracy. So, the lattice theory is mathematically fine, and it is (at least if the computing power is sufficient, the lattice fine enough and the volume big enough) also physically fine. In principle, a lattice theory could be even a candidate for a more fundamental theory, given that it has neither physical nor mathematical problems, and QFT would be simply the large distance approximation of such a fundamental lattice theory.

    What is not fine is the metaphysics. The lattice theory breaks Lorentz covariance, and some people consider Lorentz covariance as something obligatory, for some metaphysical reasons - relativistic symmetry being a fundamental insight or so.
     
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