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Quark mixing factor in CKM matrix

  1. Nov 25, 2015 #1
    I find that the quark mixing factor say for example ##V_{ub}## is the same for:
    u ##\Leftrightarrow## b
    ##u\Leftrightarrow\bar{b}##
    ##\bar{u}\Leftrightarrow## b
    ##\bar{u}\Leftrightarrow\bar{b}##
    Does this have something to do with weak interaction being unable to distinguish these from one another?
    Thanks in advance.
     
  2. jcsd
  3. Nov 25, 2015 #2

    Orodruin

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    There is no (known) interaction which turns a quark into an anti-quark as that would violate baryon number.

    For the coupling between ##u## and ##b## being equal (in magnitude) to the coupling between ##\bar u## and ##\bar b##, this is a simple consequence of the Lagrangian density being Hermitian.
     
  4. Nov 25, 2015 #3
    I can understand that the coupling between u and b is equal (in magnitude) to the coupling between ##\bar{u}## and ##\bar{b}## due to the Hermitian property but in the decay of charged Kaon,
    ##K^+ \rightarrow \mu^+ + \nu_\mu##
    the hadronic part of the current has ##V_{c\bar{s}}## which is the same coupling as ##V_{cs}##.
     
    Last edited: Nov 25, 2015
  5. Nov 25, 2015 #4

    mfb

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    It is not a transition between them, that is the point.
    You can rotate the diagram to get u -> s+W+, for example (no c involved in a kaon decay). The vertex stays the same, so you need Vus.
     
  6. Nov 25, 2015 #5
    Sorry ##V_{us}##. Ah, now I get it. Thanks.
     
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