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Muon decay feynman graph

  1. Mar 22, 2012 #1
    Hello,

    The Feynman graph of muon decay is
    attachment.php?attachmentid=45391&stc=1&d=1332436463.gif
    and I asked my professor if we could also write
    attachment.php?attachmentid=45392&stc=1&d=1332436463.gif
    and he said no, because then out of nothingness an electron and an (anti)electron-neutrino would appear and send out a boson to the muon.

    However, I was not very convinced, so I wanted to double check here. Thank you!
     

    Attached Files:

  2. jcsd
  3. Mar 22, 2012 #2

    fzero

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    Those diagrams are identical. Because the [itex]W^+[/itex] is the antiparticle of the [itex]W^-[/itex], the direction of the momentum in the Feynman graph relates them. Your professor is really just choosing the option that makes the most sense from the point of view of cause and effect.
     
  4. Mar 22, 2012 #3
    But if they're identical, isn't his choice illusory?
     
  5. Mar 22, 2012 #4

    Vanadium 50

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    The W doesn't get an arrow. It's not a fermion. Simple as that.
     
  6. Mar 22, 2012 #5

    fzero

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    As a practical matter we only have to consider one of those diagrams, not both. The first makes complete sense as a decay process. The second has a stranger interpretation. There is a certain about of bias, completely reasonable in my opinion, to choose the first. From the point of view of getting a number out at the end of a computation, it doesn't matter. Feynman was quite fond of the interpretation of an antiparticle as the particle traveling backwards in time. It is probably the case that always drawing diagrams with particles traveling forwards in time could avoid some confusion at some point, but it is not necessary if you're completely consistent.
     
  7. Mar 22, 2012 #6

    blechman

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    Actually, that's not correct. The arrow on the W boson represents "charge flow". And it is fine to include it, even though it is not a fermion. Indeed, when you have charged bosons, it is useful to include arrows to make sure to distinguish charge and momentum flow in a Feynman diagram (it could lead to minus signs).

    The point is that (as fzero correctly mentioned) both of these diagrams are identical diagrams! Remember that in Feynman's perturbation theory (with the [itex]+i\epsilon[/itex] in the propagator) a single Feynman Diagram represents an infinite number of graphs where things happen in various time-order. The choice of Feynman propagator then picks out the correct time-ordered contributions. Feynman diagrams are more slick than many people give them credit for!

    A nice description of this is in the relatively older text, "Quantum Field Theory" by Mandl and Shaw (see, for example, Sec 4.4 of the "Revised Edition").
     
  8. Mar 22, 2012 #7

    Vanadium 50

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    But it's normal not to include the arrow because it's completely unnecessary - and confusing - as you point out, it's one diagram, not two. And without the arrow, it's explicitly one diagram. I don't ever think I have seen anyone stick an arrow on the W in this process. Ever.

    Fermions, however, must have arrows, or, as you say, you will drop a minus sign.
     
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