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Coleman Mandula and CKM

  1. Jul 1, 2010 #1

    arivero

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    How is it that the Coleman Mandula theorem does not tell anything about the values of CKM matrix?
     
  2. jcsd
  3. Jul 1, 2010 #2
    As far as I am aware, there is currently no theoretical backing to the non-zero diagonal elements of the CKM matrix. They are entirely experimental.

    It is one of the most unsatisfying, niggling facts in particle physics today that we have no theoretical explanation of flavour mixing.


    Cheers,

    Simon
     
  4. Jul 1, 2010 #3

    blechman

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    Why would it?! CM Theorem is about what kinds of symmetries are allowed in "healthy" quantum field theories. CKM parameters are free parameters of the standard model. What does one have to do with the other?
     
  5. Jul 1, 2010 #4

    arivero

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    Point is, if flavour becomes incorporated to any gauge theory (horizontal symmetry, E8s, lot of gauge GUTS...), then it applies, mass eigenstates should be [flavour-]charge eigenstates, and CKM and MNS should be zero.
     
  6. Jul 2, 2010 #5

    tom.stoer

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    I was never aware of the fact that flavour symmetry becomes a gauge symmetry in these models. Would that mean that you have flavor gauge bosons?
     
  7. Jul 2, 2010 #6

    arivero

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    I do not know how mainstream they are, but if a GUT multiplet includes particles with different generations, it means that you have flavour-changing currents. So perhaps the horizontal "flavour" symmetry of such models is a different beast, or perhaps it is another way to bypass Coleman Mandula. The "small print" of CM mumbles something about "non trivial scattering".

    Of course this should aply to Garret's too, but his model already ignores CM in a different way, having both bosons and fermions in the same representations.

    An interesting point, perhaps related to this, is that Kaluza Klein was abandoned because the representations of the symmetry coming from the gravity group (from the isometries of the intermal manifold) were not able to reproduce the charges of the standard model.
     
  8. Jul 2, 2010 #7

    blechman

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    such proposals have been made to solve the flavor problem, the most famous of which that I know of is the "Froggatt-Nielsen Mechanism" propsed in C. D. Froggatt and H. B. Nielsen, Nucl. Phys. B 147, 277 (1979). You gauge the flavor symmetry at the GUT scale. Some work has to be done since the visible flavor symmetry is anomalous so you have to add more fermions, as well as adding new gauge bosons. When the symmetry is spontaneously broken, these fermions get GUT-scale masses along with the "flavons" (gauge bosons of flavor symmetry) and the Yukawas are fixed (just like the Higgs vev), and then CKM, etc comes from running.

    I still don't see what this has to do with Coleman-Mandula, though...
     
  9. Jul 2, 2010 #8

    blechman

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    But FCNC's have nothing to do with CM! Almost all extensions of the standard model have FCNC's, none of them violate CM. A violation of CM would be a theory that tried to group the Higgs boson with the W-bosons, for example. THAT would be a problem.

    I don't know what this is, but CM would require supersymmetry for this to work, I believe.
     
  10. Jul 2, 2010 #9

    tom.stoer

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    fancy ...

    the main problem is that one can write down the GR, QED and QCD Lagrangians in one line; for the el.-weak theory one needs approx. one page.

    I never tried to understand the MSSM in all details but it seems that already the lagrangian is as long as a one-loop Feynman diagram in QCD :-) I don't think it gets better if one adds Xons to get something right but unfortunately spoiling something else which has to be cured by Yons, mirros-As, hidden Bs, weakly interacting Cs and so on.
     
  11. Jul 2, 2010 #10

    blechman

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    Have you ever written out the GR action by expanding [itex]g=\eta+h[/itex] (what you need to do perturbative calculations in GR)? Or the QCD Lagrangian by expanding out [itex]F_{\mu\nu}^a[/itex]? The latter takes a page, the former takes a forest, and you're still not able to do it!!

    If you're a one-line-fiend, just use covariant derivatives and summation notation, and you can easily fit the EW action on one line!
     
  12. Jul 2, 2010 #11

    tom.stoer

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    I was a member of a group doing non-perturbative canonical quantization of QCD in non-covariant but physical gauges w/o ghosts; therefore I know what lengthy expressions are :-)

    The problem is not with el.-weak symmetry but that one seems to introduce many more fields, concepts etc. whereas in principle one is looking for a simple GUT. The right way to do it is to find a simple theory from which all these complicated concepts may emerge - but are not already present in the definition.
     
  13. Jul 2, 2010 #12

    Haelfix

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    "I never tried to understand the MSSM in all details but it seems that already the lagrangian is as long as a one-loop Feynman diagram in QCD "

    I would say most modern formulations of beyond the standard model particle physics try to keep things as simple as possible. For instance, writing the lagrangian for the MssM is a complete mess if you include all the supersymmetry breaking terms, however it simplifies remarkably (one or two lines, depending on some specifics) at high energies.

    One would say that the mess is emergent, and the theory itself relatively simple.
     
  14. Jul 2, 2010 #13

    arivero

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    That was my doubt, so I wonder what I am missing.

    1) CM tells that the Poincare generators conmute with the gauge generators. So they have a common set of eigenvectors, labeled by mass and charge.

    2) But CKM (and MNS) tell us that the mass eigenvectors are not eigenvectors of flavour.

    3) I would conclude that flavour can not be a charge of a GUT theory.
     
  15. Jul 2, 2010 #14

    blechman

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    I suspected that might have been your confusion.

    If your argument is correct, then the Standard Model must be wrong! The SM groups the top and bottom quark into the same SU(2) multiplet, and yet they have different masses!

    It is usually SPIN that is important for CM theorem, not mass. For example: the original CM paper was [according to Weinberg] trying to promote the SU(6) constituent quark model to a relativistic QFT. This model groups mesons of different spins (pion and rho, for example). CM tells us that such an attempt is hopeless.

    My off-hand response is that in gauged flavor models, all fermions are massless! The Higgs has not yet gotten its vev. After all, the gauged theory lives way up at the GUT or Planck mass! I have to think a little more about it to nail down a better explanation...

    Counterexamples to CM Theorem:
    (1) Heavy Quark Effective Theory (HQET) groups mesons of different spins (B, B*, for example), but sidesteps the CM theorem because they explicitly break the Lorentz invariance by introducing a preferred rest frame (the rest frame of the heavy quark).

    (2) Supersymmetry groups particles of different spin but the algebra is a GRADED Lie algebra and sidesteps the CM theorem that way.
     
  16. Jul 4, 2010 #15

    arivero

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    I am not sure if it is the same problem, SU(2) has all of the chirality, different multiplets for left and right, etc. It is part of the problem, perhaps.

    But consider only the eigenvectors of SU(3)xU(1). Each eigenvector is also a mass eigenvector, so it is not a problem even if they have a different eigenvalue.
     
  17. Jul 6, 2010 #16
    I concur with blechman (naturally)... Sans Higgs, all fermions are massless (or perhaps have the same mass? - in agreement with Coleman-Mandula), and CKM does not exist because you can always find a basis where it's diagonal. When Higgs is spontaneously broken, fermions acquire nontrivial masses, and CKM acquires nontrivial off-diagonal terms.
     
  18. Jul 6, 2010 #17

    blechman

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    Thanks for the support, hamster! :smile: Unfortunately, I myself am still a little confused now.

    I still stand by my statement that CM is irrelevant for the fermion mixing, but what really confuses me is the region right near the EW phase transition.

    At very high energies, I think this is a correct argument. And at very low energies, EW theory is broken and the top and bottom quarks are no longer grouped together, and so CM is irrelevant. My question is: what happens at 200 GeV, which is neither "high" or "low"?

    Also: what you said is actually true: the CKM matrix elements should not VANISH at high energies! Otherwise, particles would just stop decaying at the LHC!!

    I think there is more to it than what I was saying. In particular: every reference to CM has to do with *SPIN* constraints, not mass constraints. So I am starting to think that we're missing something else...

    As to arivero's comment: I agree with you that QCD and QED (but NOT hypercharge) is not a problem - these are "flavor-blind" theories with no FCNCs. But that doesn't resolve the EW "paradox".

    Still thinking...
     
  19. Jul 6, 2010 #18
    You have bubbles of different phases?

    Without CM, we could construct some bizarre symmetry that naturally gives different masses to e.g. electron and muon, maybe because it connects them into an unbroken SU(2) subgroup, and whose flavor-changing generators are either massive scalars or vectors, and therefore don't commute either with translation or rotation operators. Just like you could connect top and bottom into an unbroken group with vector gauge bosons as generators.
     
    Last edited: Jul 6, 2010
  20. Jul 6, 2010 #19

    blechman

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    I think I'm beginning to see where I went wrong...

    When talking about the S matrix (and CM Theorem is all about S-matrix theory!), you have to be a little careful about how you define you "states". The only objects that may rightly be called "states" are MASS EIGENSTATES, as they correspond to the objects that have well-defined wavepackets as [itex]T\rightarrow\pm\infty[/itex]. "Flavor states" cannot be used to compute an S-matrix element. Attempting to do so leads to some pretty weird pathologies. For example, the "flavor states" (that is: states that are annihilated by the quark operators in the flavor basis) would include a sum over various multiparticle states.

    Now Weinberg's QFT Text, Chapter 24.1 quotes the CM Theorem:

    So the flavor symmetry does not satisfy the conditions of the CM theorem (single-particle states stay single-particle states after a transformation).
     
  21. Jul 6, 2010 #20
    You can define the flavor basis for either the up-type or the down-type quarks to coincide with the mass eigenstate basis, and have a diagonal CKM matrix.

    Or you can do as conventional, to define both bases to coincide with mass eigenstates at the same time, and then CKM becomes non-diagonal.
     
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