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I Interpreting K-minus decay

  1. Nov 11, 2017 #1
    So I am trying to draw Feynman diagrams for the following reaction:

    K- → μ- + (νμ) (anti muon neutrino, not very skilled at typing symbols sorry).

    And I have the strange quark and the anti-up quark colliding to produce the muon and anti-muon neutrino via a W- boson. I'm not sure if this is right but it seems to be the most logical way to draw the diagram, could someone please verify if I am using the correct feynman diagram and potentially explain how the strange and anti-up quark produce a W-boson?
     

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  3. Nov 11, 2017 #2

    vanhees71

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  4. Nov 11, 2017 #3

    mfb

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    @vanhees71: Huh? The diagram is correct.

    This forum supports LaTeX, the neutrino can be written as ##\bar \nu_\mu## —> ##\bar \nu_\mu##.
     
  5. Nov 11, 2017 #4

    ChrisVer

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    it's the correct diagram for this case.

    This question is vague. In the quark sector, W bosons don't interact with mass eigenstates but with weak eigenstates. One can choose the down type quarks (|q|=1/3) to write them in the weak eigenstates, [itex] d'_i [/itex] (i identifies the generation), which are related to the mass eigenstates [itex]d_i[/itex] with the CKM matrix elements [itex] d'_i = V^{CKM}_{ij} d_j[/itex] . Eg the interaction of W with (u,d') in terms of quark mass eigenstates is [itex]u ( V^{CKM}_{ud} d + V^{CKM}_{us} s + V^{CKM}_{ub} b ) W[/itex] . As a result of this you can have ud,us,ub interactions with W-bosons (of course the CKM matrix is an almost diagonal unitary matrix and so the further away in generations you look, the weaker the couplings get, something you can see as "CKM suppressed").
     
  6. Nov 12, 2017 #5

    vanhees71

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    Perhaps, I misread the labels on the diagram, but a kaon doesn't consist of an s quark and an anti-neutrino!
     
  7. Nov 12, 2017 #6

    mfb

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    I'm quite sure that is a "##\overline u##" at the lower left. The ##\nu## at the lower right is not round at the bottom.
     
  8. Nov 12, 2017 #7

    vanhees71

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    Argh. Yes, then it's of course right. Sorry for the confusion.
     
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