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Can a neutral pion decay to a neutrino and an anti-neutrino?

  1. Feb 27, 2008 #1
    Is it allowed? such as a electron neutrino and an electron anti-neutrino.
    And why?
    Now, I am confused.....

    Thanks.
     
  2. jcsd
  3. Feb 27, 2008 #2
    This decay is forbidden by angular momentum conservation if neutrinos are purely massless.
     
  4. Feb 28, 2008 #3

    Vanadium 50

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    But they are not. So the decay occurs with some rate. (It's GIM suppressed, though).
     
  5. Feb 28, 2008 #4

    malawi_glenn

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    Strong- and EM-decay would dominate altough right?
     
  6. Feb 28, 2008 #5

    pam

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    There is no strong decay of the pi0.
    It is not a strong interaction.
     
  7. Feb 28, 2008 #6

    malawi_glenn

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    Ah yeah thats right! I forgot :-)
     
  8. Feb 28, 2008 #7
    Hello,

    what do you mean by GIM suppressed ?

    From PGD, existing limits are :
    Br(pi0 -> nunu) < 2.7e-7
    Br(pi0 -> nunugamma) < 6e-4
    I have no idea, how much we should expect from SM theory, do you know ?
     
  9. Feb 28, 2008 #8
    Hi
    I wanted to quote the same values as you just did. You can find the original references for those upper bounds, they describe the various existing predictions.
     
  10. Feb 28, 2008 #9
  11. Feb 28, 2008 #10

    Vanadium 50

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    Suppressed by the mechanism of Glashow, Illiopoulos and Maiani. This is a cancellation that occurs in flavor changing neutral currents: decays like K0L -> mu+ mu- and pi0 -> nu nubar. If the neutrinos were exactly degenerate the cancellation would be exact and this decay wouldn't occur. Because they have slightly different masses, the decays are suppressed.
     
  12. Feb 29, 2008 #11
    Hello Vanadium,

    GIM mechanism start to be old for me last time I studied it.
    I understand that for K0 we have FCNC, but why can't we have :
    u ubar + ... -> Z0 -> nu nubar ?
    This diagram is suppressed by mnu/MZ, but why do we need CKM and stuff like that ?
    Can you explain a bit more ?
     
  13. Feb 29, 2008 #12

    Vanadium 50

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    The pi0 carries no weak charge, so it cannot couple directly to a Z, just as it contains no electric charge and cannot couple directly to a photon.
     
  14. Mar 3, 2008 #13

    arivero

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    But suppose that the up quark were massless. Could the neutral pion decay into a pair of up, anti-up quarks?


     
  15. Mar 3, 2008 #14

    Vanadium 50

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    The pion is already composed of u-ubar quarks, so I don't know what a decay to u-ubar would mean. Since quarks are confined, you'll end up with a pi0 anyway, so that's not much of a decay.

    There is an additional subtlety - a pi0 composed of massless quarks would itself be massless (it becomes a Goldstone) and massless particles do not decay.
     
  16. Mar 3, 2008 #15
    Hello,

    Your argument looks sensible (why no weak charge ? I would say only T_3 = 0) but I'm not sure to understand it completly. Looking at http://doc.cern.ch//archive/electronic/hep-ph/0501/0501117.pdf ,
    it seems that pi0 -> Z -> nu nubar vanish only for massless neutrinos.
     
    Last edited: Mar 3, 2008
  17. Mar 3, 2008 #16

    Vanadium 50

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    That very paper points out that at tree level pi0 -> Z -> nu nubar is zero. For massive neutrinos you can have a box diagram involving two W's, a diagram that is both GIM and helicity suppressed, but there is no symmetry forcing it to be exactly zero.

    As for only T_3 being zero, that alone is enough. The SU(2) nature of the weak interaction tells you that this coupling is zero. (The (1,0)+(1,0)=(1,0) Clebsch-Gordan coefficient is zero) Now that I think about it, I think the other SU(2) symmetry, isospin, also blocks this for the same reason.
     
  18. Mar 3, 2008 #17

    arivero

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    Well, it is a subtlety but it is not an additional subtlety (to the original question). It roots on chiral aspects too, doesn't it?
     
  19. Mar 3, 2008 #18

    Vanadium 50

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    If the quarks were massless, the pion (unlike the proton) would also be massless. That would prevent it decaying by any mechanism whatsoever - not just the weak interaction.
     
  20. Mar 3, 2008 #19

    arivero

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    Yes, but why does the mass of the pion depends of the mass of the quarks? It is because of chiral symmetry breaking, isn't it?
     
  21. Mar 3, 2008 #20

    Vanadium 50

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    Yes, but chiral symmetry breaking is essentially a QCD phenomenon. The fact that the weak interaction treats states of different chirality differently isn't necessary to break chiral symmetry: the masses of the u and d quarks do it by themselves.

    It's broken by terms in the Lagrangian like u_L(m)u*_R.
     
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