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

  1. Aug 19, 2017 #26

    ChrisVer

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    However a neutral pion is either uubar or ddbar...
     
  2. Aug 19, 2017 #27

    mfb

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    No, it is the linear combination, as Orodruin said already.
     
  3. Aug 19, 2017 #28

    ChrisVer

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    OK, draw the Feynman Diagram of the linear combination for me and I would really like to see how you'd represent the (-) and where the /sqrt(2) appears, try the decay [itex]\pi \rightarrow \pi^0 e \nu[/itex]... or diagrams like in Fig.2(a,b) and the blue neutral pion (for I don't see a uubar-ddbar), neither do I know a Z decaying to both a uubar and ddbar to give a neutral pion (d) http://www2.kek.jp/en/press/2008/BellePress12e.html
    The combination appears only when you want to write the neutral pion's wavefunction in flavor space... afterall superpositions aren't natural.
     
    Last edited: Aug 19, 2017
  4. Aug 19, 2017 #29

    vanhees71

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    The mesons are grouped according to isospin (or its SO(3) generalization if you wish). Take ##\psi=\binom{u}{d}## for the first-generation quarks, forming an iso-doublet. Out of those you can build the pions by
    $$\vec{\pi} = \bar{\psi} \vec{\tau} \gamma_5 \psi,$$
    making them an isovector (isospin 1) triplet, the charged pions are given by ##\pi^{\pm}=\pi_1 \pm \mathrm{i} \pi_2## and the neutral pion is ##\pi^0=\pi_3##. The ##\vec{\tau}## are the isospin-Pauli matrices and thus indeed ##\pi^0=\bar{u} \gamma_5 u-\bar{d} \gamma_5 d## (without caring about normalizations here). The iso-singlet is the scalar ##\sigma## meson (or ##f^0##): ##\sigma=\bar{\psi} \psi=\bar{u} u + \bar{d} d##, forming the other neutral combination of light quarks.

    Whether or not you put factore ##1/\sqrt{2}## or ##1/2## there is a matter of convention, and there are at least three conventions for the pion-decay constant in the literature all different by factors ##\sqrt{2}## or ##2## from each other. For more details with all the factors (hopefully ;-)) in place, see theory lecture I under

    http://th.physik.uni-frankfurt.de/~hees/hgs-hire-lectweek17/
     
  5. Aug 19, 2017 #30

    ChrisVer

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    Still I think this works only for effective field theories, since you mention the [itex]f_\pi[/itex] (pion's decay constant). In that case you wouldn't represent the pion as a combination of quarks but as a single line (scalar field) in a Feynman Diagram (see eg some neutral pion decay diagrams), which is exactly what you do when you try to define it as a "field".
     
  6. Aug 19, 2017 #31

    vanhees71

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    Sure, and the effective models are governed by chiral symmetry and thus take the form they take. Pions are a bound state with the specific quark content. Already these quarks are constituent quarks of course!
     
  7. Aug 19, 2017 #32

    mfb

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    You would draw both diagrams, and you have to calculate both diagrams for the decay and add their amplitudes.
    Everything is a superposition - in most bases. Saying "this is not a superposition" without specifying a basis is meaningless. Superpositions are much more natural than the special case of pure states.
     
  8. Aug 19, 2017 #33

    ChrisVer

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    in any case in the 1st diagram you would end up with a ddbar or with a uubar... so according to the OP, this wouldn't be a neutral pion.
     
  9. Aug 19, 2017 #34

    Orodruin

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    The point is that you have to draw all the other diagrams as well and take the appropriate linear combination when you actually compute the decay amplitude. Indeed the uubar state is not a neutral pion, but the neutral pion projection onto the uubar direction (or vice versa) is non-zero so you have to include it when you compute the amplitude.
     
  10. Aug 19, 2017 #35

    Orodruin

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    Somebody should tell the ##K-\bar K## oscillations ... :rolleyes:
     
  11. Aug 19, 2017 #36

    ChrisVer

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    I disagree...
    The fact that you can have an electron with spin up or spin down, doesn't mean that a single electron would have both... A collection of electrons would of course have 50% up and 50% down (given that the orientation symmetry is preserved), the single electron would have either up or down (you'd need to measure it).
     
  12. Aug 19, 2017 #37

    Orodruin

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    Before the measurement an electron may very well be in a superposition of spin up and down. It is your measurement process that forces it to select either by singling out a basis. After your measurement each individual electron is in an eigenstate - in your chosen direction. If you change from the z to the x direction each electron is in an equal parts linear conbination so whether it is a linear combination of seceral states really is a question of your choice of basis.
     
  13. Aug 19, 2017 #38

    vanhees71

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    A single electron has the spin it prepared in. Usually it will be in a mixed state (aka unpolarized electron).
     
  14. Aug 19, 2017 #39

    mfb

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    Sorry Chris, but you are wrong, and repeating wrong statements won't improve them.
    Up and down in which direction? Your perfectly polarized beam of electrons "up" in the z-direction will be in a superposition of up and down in all other directions. And yes, the electrons do have both up and down in these directions. It is important to consider both directions and their relative phase to describe the electron correctly in such a basis. If you want to assign them a specific orientation in a different basis, then you don't get your perfectly polarized beam in z-direction any more.

    A pure (edit) measurement eigenstate is a very special case, in general you have superpositions.
     
    Last edited: Aug 19, 2017
  15. Aug 19, 2017 #40

    Orodruin

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    A superposition can be a pure state as well. I believe you mean "a measurement eigenstate".
     
  16. 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)?
     
  17. Aug 19, 2017 #42

    vanhees71

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  18. 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.
     
  19. 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?
     
  20. 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.
     
  21. 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?
     
  22. 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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