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Unitarity of CKM matrix

  1. May 24, 2008 #1
    This is a known fact that CKM matrix, a matrix that is used to connect the weak interaction eigenstates to mass eigenstates is unitary. I have studied that this is due to the conservation of probability. i.e. an up type quark will decay into exactly three type of d quarks, nothing more.
     
  2. jcsd
  3. May 24, 2008 #2
    I'm not sure what exactly you're trying to say --- it's all correct and true, but why did you decide to post that as a message?
     
  4. May 26, 2008 #3
    I just want to express the fact that an up quark can only decay to three down types of quarks only and thus the unitarity of ckm matrix mathematically.
     
  5. May 26, 2008 #4

    malawi_glenn

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    What kind of respons do you want plasmon?
     
  6. May 26, 2008 #5
    Hello, I dont understand the aim of the post too..
    but this has raised a question in my mind..
    (thanks Plasmon !)

    in the leptonic sector, the neutrino also have a CKM-analogue matrix right?
    but we dont know yet if it is unitary or not..
    if it is just related to probabilites.. where's the problem ?
     
  7. May 26, 2008 #6

    Vanadium 50

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    There is a lepton analogue called the MNS matrix, for Maki, Nakagawa and Sakata.

    The CKM (and MNS) matrices are unitary by construction. Essentially they are a change in basis - a rotation, if you will - so must be unitary. A separate question is experimental: is the 3x3 matrix that we can measure unitary? If the answer is "no", that's proof for new physics: e.g. a 4th family.

    For quarks, the answer is that we cannot tell. Partly this is because the effect of new physics can be made arbitrarily small, but there's a little more to it than that. For example, given only the upper 2x2 CKM matrix, one cannot infer the existence of a 3rd family. Additionally, the numbers don't work out nicely: Vtb is so close to 1 that one needs a measurement of single top production good to better than 0.1% to see the effect of other quarks.
     
  8. May 27, 2008 #7
    Dear All

    I guess we can express it as

    W-|u>=Vud |d>+Vus |s>+Vub |b>
    W-|c>=Vcd |d>+Vcs |s>+Vcb |b>
    W-|t>=Vtd |d>+Vts |s>+Vtb |b>

    and |d> , |s>, |b> are orthogonal states (not neccessary due to GIM mechanism)

    <u|W-(adj) W-|u>=1, <c,t|W-(adj) W-|u>=0

    => V as a unitary matrix

    Am i right in making above assumptions.
     
  9. May 27, 2008 #8
    Thanks Vanadium for your enriching contribution.
    another question, I may be off topic but it's in my mind, so I hope this wont bother you too much :)

    Is the eventual non-unitarity of the MNS matrix related to the CP violation ?
     
  10. May 27, 2008 #9

    Vanadium 50

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    There is no "eventual non-unitarity of the MNS matrix". If there are only three families of neutrinos and no new physics, you will get a unitary matrix.
     
  11. May 28, 2008 #10
    Vanadium 50,

    the eventuality is as you stated it : "If there are only three families of neutrinos and no new physics"

    ..
     
  12. May 29, 2008 #11
    Atakor

    The word "if" does not imply an eventuality.
     
  13. May 30, 2008 #12
    Really ?

    ==
    if we say 'there are only 3 families' THEN MNS is unitary,
    otherwise (if you dont like the word 'eventuality') MNS doesnt have to be Unitary.

    quiet clear for me...
     
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