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I On the decay of the neutral Sigma particle

  1. Oct 15, 2017 #1
    Hello!
    I wanted to ask why does the neutral Σ decay almost always to one γ(or two) and a neutral Λ.
    Why can't it decay to anything else?
    Thanks!
     
  2. jcsd
  3. Oct 15, 2017 #2
    A possible decay for the neutral Σ particle that I have found is the one which it decays to a proton and a negative pion. But, I don't know why this is a forbidden decay. A friend pointed out that this decay does not conserve strangeness, although I have found that this decay could happen(possibly) through weak interactions which do not conserve strangeness. So, I don't know why this is forbidden.
     
  4. Oct 15, 2017 #3
    It could.
    Anything else is by 9 or more orders of magnitude slower.
     
  5. Oct 15, 2017 #4

    mfb

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    I merged your two threads as they are about the same question.

    If a particle can decay via the strong or electromagnetic interaction, then these decays will dominate. The weak interaction is weak - these decays are not impossible, but they are only relevant if there is no alternative to it (and typically the particles are long-living in this case).
     
  6. Oct 15, 2017 #5
    Thank you!
     
  7. Oct 16, 2017 #6
    Not quite "will", "no alternative". Compare with internal transition of nuclear isomers. There are excited states of nuclei for which electromagnetic decay to a lower lying state is possible but difficult enough that decay via weak interaction (beta decay) is a relevant competing process, or the prevalent ones.
    Elementary particles are simpler than nuclei and such cases are rare. Neutral sigma behaves as an excited state of uds, and electromagnetic decay to the lowest uds state, which is lambda, prevails by 9 orders of magnitude over weak decay straight from the excited state.
     
  8. Oct 16, 2017 #7

    Orodruin

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    I think it is worth pointing out that there theoretically should be a branching ratio of order ##10^{-3}## to ##\Lambda e^+e^-## mediated by an off shell photon at tree level. I do not think this has been verified experimentally.
     
  9. Oct 16, 2017 #8

    mfb

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    I‘m not aware of measurements of the branching fraction but the decay has been used to determine the parity. Example - they found 314 events with a ##\pi^0##!

    @snorkack: As this thread is about particle physics I didn’t consider nuclear isomers.
    I don’t know any case in particle physics where possible electromagnetic decays are suppressed sufficiently to make weak decays compete with them.
     
  10. Oct 16, 2017 #9

    Vanadium 50

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    They don't exist. The Dalitz decay experiments were done in the 1960's using bubble chambers. While they can see the Dalitz decays, they have no means of normalizing to any measured decay.
     
  11. Oct 16, 2017 #10

    Vanadium 50

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    We are about an order of magnitude away, perhaps two. The [itex]J/\psi[/itex] is probably the best candidate, since it can be copiously produced, easily identified, and has a relatively small non-weak decay width of about 90 keV. Decays that are promising are [itex]J/\psi \rightarrow D +[/itex] anything and [itex]J/\psi \rightarrow [/itex] neutrinos.
     
  12. Oct 16, 2017 #11

    mfb

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    BES sets some 10-5 limits for D and 10-6 limits for Ds and quotes 10-8 "or smaller" predictions in the introductions. The fully hadronic modes have worse upper limits. Reaching 10-8 sounds very challenging.
     
  13. Oct 17, 2017 #12

    Vanadium 50

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    If it were easy, they would already have done it!
     
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