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I Are renormalizable QFT the ''really renormalizable''?

  1. Sep 18, 2016 #1
    The renormalizable QFT is the theory with only a finite number of Feynman diagrams superficially diverge(in all order) and the non-renormalizable QFT is the theory with infinite diagrams superficial diverge.
    Then my question is in all renormalizable theories can we absorb all divergences into counter terms or not?(All renormalizable theories are ''really renormalizable'')
    Is there any case in non-renormalizable QFT we can absorb all divergences into counter terms?
     
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  3. Sep 18, 2016 #2

    vanhees71

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    A Dyson-renormalizable QFT is defined as one where you can renormalize the theory by renormalizing a finite number of parameters (in 1+3-dim spacetime usually the wave-function normalization, masses and coupling constants) loop-order by loop-order of perturbation theory. Usually you have an infinite number of divergent Feynman diagrams symbolizing proper vertex functions. E.g., in simple ##\phi^4## theory the superficial degree of divergence of a diagram is ##4-E## (where ##E## is the number of external (truncated) legs of any 1PI truncated diagram). Due to field-reflection symmetry the 1PI diagrams with an odd number of external legs vanish. Thus the only divergent diagrams are the self-energy diagrams ##E=2##, which are renormalized by wave-function and mass renormalization and the four-point proper vertex function, which is renormalized by the coupling-constant renormalization.

    It has been shown by Bogoliubov, Parasiuk, Hepp, and Zimmermann that superficially renormalizable theories are indeed renormalizable. When it comes to theories with symmetries there's, however, a bit more to it. Usually symmetries restrict the freedom of terms in the Lagrangian. E.g., in QED if you want a superficially renormalizable theory you cannot write down a four-photon interaction contribution since there is no gauge-invariant term which has a coupling constant with non-negative energy dimension, i.e., there is no such term that is superficially renormalizable. On the other hand symmetries, particularly local gauge symmetries, lead to strong constraints on the proper vertex functions themselves known as Ward-Takahashi or Slavnov-Taylor identities, and these can help to make superficially divergent contributions in fact finite. In QED the four-photon vertex is superficially logarithmically divergent, but the WTI of this function in fact tells you that it delivers only a finite contribution.

    Now another large class of important QFTs are the socalled effective QFTs. They are non-renormalizable in the sense that you need an infinite number of parameters to renormalize the theory. These theories are also based on symmetry like chiral symmetry as a guideline to build effective QFTs for hadrons, which are, as composite particles, only effectively described as elementary quantum fields for low collision energies, at scales where you don't probe their composite structure too much. Thus you have an expansion in powers of energy and momentum (usually relative to some scale, which in chiral perturbation theory is ##4\pi f_{\pi} \simeq 1 \; \mathrm{GeV}##). Then you start with a tree-level Lagrangian obeying the symmetries containing all terms not only the Dyson-renormalizable ones allowed by the symmetry. Then you start at a certain order in momentum. Usually the more loops are contained in a proper-vertex diagram the higher the momentum order of divergent pieces get, and you have to renormalize the contribution by renormalizing the corresponding coupling constants in the Lagrangian. One can show that this is possible, i.e., usually you are not forced to introduce symmetry-violating types of contribution to the diagram.

    However, there are important exceptions, known as anomalies. An anomaly occurs when a classical field theory obeys some symmetry which, however, is necessarily destroyed when quantizing the theory. A famous example is the axial anomaly. When you consider massless QCD it obeys a symmetry where the quark fields are multiplied by a factor ##\exp(-\mathrm{i} \alpha_A \gamma^5)##. Now it turns out that you cannot renormalize the theory such that both the vector currrent (due to the usual symmetry by just mutliplying the quark fields with a usual phase factor) and the axial current stay conserved. The vector current, however, must be conserved. So you necessarily have to break the axial U(1) symmetry. An that's in fact a good thing, because it explains the decay of neutral pions to two photons.
     
  4. Sep 18, 2016 #3
    In effective QFTs we need infinite parameter to renormalize,then are there infinite parameters in a Lagrangian,so are there infinite terms in a Lagrangian?Are there any meaning in such Lagrangian?Which books say about Effective QFTs?
     
  5. Sep 19, 2016 #4
    Now I have seen in the effective QFT the Lagrangian has finite number of terms.
    Why we do not accept infinite number renormalizable parameters in non-renormalizable QFTs,but we must only build renormalizable QFTs(except effective QFTs)?
     
    Last edited: Sep 19, 2016
  6. Sep 19, 2016 #5
    Effective QFTs are nonrenormalizable still OK because we only need the theories at low energy.We do not accept general nonrenormalizable QFTs because we need other parameters than wave-function,mass and interaction constants.Are these correct?
    Why do we not simply cut off the devergence parts(consider as the parts of unknown high energy physics),but we must put out the counter terms to absorb the infinities,therefore rising the renormalizable and non-renormalizable QFTs?
     
    Last edited: Sep 19, 2016
  7. Sep 21, 2016 #6
    I have been wrong in my last questions.Now I already understand what is non-renormalizable QFTs!But I wonder whether some non-renormalizable QFTs are still renomalizable?(mean that still can eliminate divergences)
     
  8. Sep 21, 2016 #7

    vanhees71

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    Yes, understood in the sense of expansion wrt. powers of energy/momentum they are renormalizable. It's only that you need more and more "low-energy constants" the higher you go in this expansion. A nice paper on this more relaxed view towards effective theories which are renormalizable in this extended sense but not renormalizable in the stricter sense (one often calls such theories "Dyson renormalizable" since it was Dyson after all who has shown that the approaches by Schwinger, Feynman, and Tomonaga on the renormalization problem of QED are equivalent and can be understood from QFT) is

    http://arxiv.org/abs/hep-th/9702027

    or Sect. 12.3 in his book "The quantum theory of fields, vol. 1, Cambridge University Press (1995)".
     
  9. Sep 22, 2016 #8
    Non-renormalizable QFTs are theories with the superficial degree of divergence D increases together the increasing of vertices.Then my question is that: Are there any non-renormalizable QFTs(coupling constant has negative mass dimension) that we need only a finite number of parameters to renormalize them?(e.g only with field,mass and interaction constants)(despite of infinite number of diagrams that are divergent)
    Can we demonstrate that non-renormalizable QFTs always need infinite number of observations(measurable quantities or parameters) to renormalize ?
     
    Last edited: Sep 23, 2016
  10. Sep 23, 2016 #9

    vanhees71

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    That's an interesting question. I don't know the answer. It could well be that you can find some theory with some pretty strong symmetries that prevent an infinite number of superficial divergences to be in fact finite because they are somehow "protected" by the symmetry.

    An example for the renormalizable case are local gauge symmetries. In both abelian (e.g., QED) and non-ablian (e.g., QCD) gauge theories the gauge-boson four-vertex is superficially logarithmically divergent, but in fact it's finite at all orders of perturbation theory since a Ward-Takahashi identity "protects" this vertex from being divergent.
     
  11. Sep 23, 2016 #10

    atyy

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  12. Oct 1, 2016 #11
    In Phi-4 and Yang-Mills theories there are not exist the Ward identities,then I wonder whether that they are renormalizable or not?
     
    Last edited: Oct 1, 2016
  13. Oct 1, 2016 #12

    vanhees71

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    Simple ##\phi^4## theory has only "field-reflection symmetry" as a symmetry, i.e., it's invariant under ##\phi \rightarrow -\phi##, which makes all proper vertex functions with an odd number of legs vanish. Then Weinberg's theorem together with Zimmermann's forest formula tells you that you need to renormalize only the self-energy (mass and wave function renormalization) and the four-point vertex (coupling-constant renormalization) which makes ##\phi^4## theory renormalizable.

    Yang-Mills theories as well as their extension with matter fields, including quantum flavor dynamics describing the electroweak interaction are local gauge symmetries and thus fulfill Ward-Takahashi (or Slavnov-Taylor) identities which are crucial for the proof of their renormalizability.
     
  14. Oct 2, 2016 #13

    A. Neumaier

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    There are theories that are nonrenormalizable according to standard perturbation theory but are renormalizable when perturbed around the large N limit. See, e.g., G. Parisi, The theory of non-renormalizable interactions: The large N expansion, Nuclear Physics B 100.2 (1975): 368-388.

    The only difference between renormalizable and nonrenormalizable theories is that the collection of the former/latter forms a finite/infinite-dimensional manifold of theories. But of course, infinite-dimensional manifolds contain lots of finite-dimensional ones; for example those that set all but finitely many of the parameters to zero. The problem is just identifying the right ones to describe given physics is much harder, especially if one aims at an infinitely accurate (''fundamental'') description of Nature.
     
  15. Oct 3, 2016 #14
    Is it possible there is in general a criterion to confirm a non-renormalizable theory is renormalizable?
     
  16. Oct 3, 2016 #15

    A. Neumaier

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    Should there be such a criterion, nobody has found it yet.
     
  17. Oct 15, 2016 #16
    The BPHZ theorem guarantees the renormalization of renormalizable QFT then why do we also need Ward identity for renormalizing gauge QFT?
     
  18. Oct 15, 2016 #17

    vanhees71

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    Gauge symmetries restrict the "allowed" terms in the Lagrangian. So if you have a proper vertex function, for which you cannot find a gauge invariant counterterm, you are seemingly in trouble with BPHZ. You'd violate the gauge symmetry, and this would make the entire model meaningless. On the other hand gauge symmetry puts also severe restrictions to the structure of the proper vertex functions in terms of Ward-Takahashi (or Slavnov-Taylor identities), which prevent the superficially divergent vertex functiont to be really divergent. An example is the four-photon scattering vertex function. In leading order (one loop) it's a set of "box diagrams". This diagrams are superficially logarithmic divergent, and there's no renormalizable counter term that's gauge invariant available. On the other hand thanks to the WTI restricting the structure of the proper vertex function you can be sure that in fact the vertex function is convergent, and you don't need a counterterm for it. So QED is renormalizable although it looks as if there may be trouble with the four-photon vertex function.
     
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