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A The reaction π0 + π+ -> π+ + γ

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  1. Apr 20, 2016 #1

    Garlic

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    Hello everyone,
    Is this reaction allowed?
    π0 + π+ -> π+ + γ (just a single photon)
    I mean the π+ contributes with the reaction not just by making it possible that two particles decay into a single photon (by changing it's momentum), but also the π0 reacts with the π+, where the particle-antiparticle pairs in both mesons annihilate themselves.
    If this reaction is allowed, how would the feynman diagram of it look like? In this case we can't make the neutral pion look like a circle or triangle, because it has to be a specific state, not the uubar+ddbar/√2 one. The interaction has to force the π0 to be uubar or ddbar.
     
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  3. Apr 20, 2016 #2

    ChrisVer

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    at first glance I'd say it is possible...
    it looks pretty much like a reversed photoproduction of neutral pion, where you have replaced the more common proton with a pion+.
    I don't understand your last comment about the mixture... a neutral pion is either uubar or ddbar, and that's why it's written as a superposition. Both combinations as you mentioned make the interaction possible...
    One possible way is to let the uubar or ddbar of the pion annihilate and also get some momentum with a photon or gluon from the charged pion....
     

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  4. Apr 20, 2016 #3

    Garlic

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    Doesn't this interaction forcefully collapse the superposition of the neutral pion? So, is it allowed to represent the neutral pion as purely uubar (or ddbar) in the diagram, and not as the weird looking loop-thing (where the neutral pion is in it's superposition state)?
     
  5. Apr 20, 2016 #4

    ChrisVer

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    loops are not that strange I'd say...Especially if the gluon is a possible choice it's pretty much more probable to happen because there is a plenty of gluons in these systems (more possible than finding the quark of the charged pion).
    Again I am saying that the neutral pion is always reacting in FD as a uubar or ddbar... you never use superpositions of states. I wouldn't call it a "collapse" because as a uubar could do the job, so could a ddbar... you can't determine which was which.
     
  6. Apr 20, 2016 #5

    Garlic

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    Wouldn't the resulting photons be at different energies, and wouldn't the momentum difference of the π+ be different. Maybe we can't detect energy differences at a few electronvolts with our current technology, but it is still theoretically detectable, right? Wouldn't this possibility collapse the superposition?
     
  7. Apr 20, 2016 #6

    ChrisVer

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    For example let's take the simpler case of the neutral pion decay:
    [itex] \pi^0 \rightarrow \gamma \gamma[/itex]
    Well I don't want to get into too much depth right now...
    The key point is that what you observe (like the cross sections etc) does not depend on just one FD, but it takes into account all possible FDs and so it's proportional to [itex] | FD_1 + FD_2 + ... |^2 [/itex]... As you already know, FDs don't show the real physical process but they are there representing amplitudes (or mathematical formulas).
    So when you want to do something that concerns the neutral pion you have to take the FDs that have the uubar or ddbar of the pion, add them and take the magnitude squared.
    Check for example here:
    https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QCD/QuarksAndColour_3.html
     
  8. Apr 20, 2016 #7

    Garlic

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    Hmm.. I understand.
    Thank you very much for your explanation :biggrin:
     
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