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I Pion+/- decay - what *Exactly* is happening?

  1. Aug 19, 2017 #41

    ChrisVer

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    OK people, I will agree with you and say that the neutral pion is only the [itex]u\bar{u}-d\bar{d}[/itex] state... In that manner the decay [itex]\pi \rightarrow \pi^0 e \nu[/itex] can't happen since there is no meson with the [itex]d\bar{d}[/itex] (or [itex]u\bar{u}[/itex]) alone quark content...
    Also what's the meson that has the state: [itex] u\bar{d} \pm u\bar{s} \pm u \bar{b} \pm c \bar{s}\pm c \bar{d} \pm c \bar{b}[/itex] (that would come from the W hadronic decay addition of Feynman Diagrams)?
     
  2. Aug 19, 2017 #42

    vanhees71

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  3. Aug 19, 2017 #43

    ChrisVer

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    because I couldn't produce a neutral pion according to the definition I am getting: I either end up with processes resulting to [itex]u\bar{u}[/itex] or to [itex]d\bar{d}[/itex] (so projections of the neutral pion, as well as the sigma mesons)... I originally said that the decay was possible (like a semileptonic meson decay), since those were actually neutral pions but I was wrong.
     
  4. Aug 19, 2017 #44

    vanhees71

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    But it's not wrong, as you can see from the particle data booklet (the branching ratio is only about ##10^{-8}##, but it's a measured decay of the charged pions). I still don't understand, why you come to the conclusion that this decay cannot exist, only because the pion is the given superposition?
     
  5. Aug 19, 2017 #45

    mfb

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    They have a neutral pion component.
    There is no such thing, and a W cannot decay to a single meson anyway.

    We had a great question about partial widths a while ago, where you can see the effect of the superpositions of quark states directly in the branching fractions.
     
  6. Aug 21, 2017 #46

    ChrisVer

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    Because that decay ends up with [itex]u\bar{u}[/itex] or [itex]d\bar{d}[/itex] and not in a superposition of the two... and that is not a neutral pion (as I was repeatedly told in this discussion)...
    On the other hand, if we are to "add" every possible outcome (and so say we can have both the [itex]u\bar{u}[/itex] and [itex]d\bar{d}[/itex] in a combination, al), then I would ask on this counter-example (a possible W decay to a meson):
    https://arxiv.org/pdf/1001.3317.pdf (Fig2)
    [itex] D^0 \rightarrow \rho^+ \pi^-[/itex] : [itex] c \bar{u} \rightarrow W (d\bar{u}) \rightarrow (u \bar{d}) (d \bar{u})[/itex]
    my question would then be why do we get a [itex]\rho^+[/itex] (with the [itex]u\bar{d}[/itex] decay of a W ) and not a [itex] \alpha u\bar{d} + \beta u \bar{s} [/itex] (forgetting the bottoms due to energy). Instead we call the 1st a rho, and the second a possible Kaon (2 different outcomes)... Is it related to a flavour symmetry or something?
     
  7. Aug 21, 2017 #47

    Orodruin

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    This is self contradictory. If it can end up in uubar and ddbar, clearly it can end up in a linear combination of those states. It is a question of what the amplitude is.

    What makes you think it doesn't decay to both linear combinations? The only difference is that the s quark is heavy enough for the mass eigenstates to be essentially equivalent to the flavour states, which does not happen in the pion system.
     
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