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CKM Matrix and mass eigenstates

  1. May 21, 2007 #1
    First off, what is a mass eigenstate?! Is there a (hermitian) operator associated to mass? What should I picture when discussing a non-mass-eigenstate?! The same goes for a 'weak eigenstate' as the CKM matrix is supposed to be the basis transformation between these two... :rolleyes:

    The I read that 'a linear transformation which diagonalizes the mass terms of the u-type quarks does not necessarily diagonalize those of the d-type quarks.'

    What does this mean, and why not?! :grumpy:

    Finally, I rwill be very much helped by any insight on the meaning of this matrix and its properties, e.g. why its unitary...? :blushing:

    Sorry, the more I learn the less I seem to know...
  2. jcsd
  3. May 22, 2007 #2


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    It's more convenient to use the mass^2 operator, i.e: E^2 - p^2. A mass
    eigenstate is an eigenstate of this operator. Such a state has a determinate
    mass value.

    I'm not sure what you mean by "picture". A "non-mass-eigenstate" is a
    superposition of more than one mass eigenstate. It's mass is therefore
    indeterminate (i.e: there's some probability of measuring any of the
    mass values of the mass-eigenstates that have been superposed).

    The states that participate in the weak interaction are not mass-eigenstates
    in general, but a superposition of mass eigenstates.

    It means the operator (i.e: observable) corresponding to the flavor property
    (u,d,etc) does not commute with the operator corresponding to the mass
    property. So if you choose a Hilbert space basis corresponding to the
    flavor eigenstates, they are in general a non-trivial superposition of mass

    Such a transformation matrix is an operator in Hilbert space. It must be
    unitary to preserve inner products between states in the Hilbert space.

    I know the feeling.

    - strangerep
  4. May 22, 2007 #3
    Thanks very much for that! I guess that for any observable there is an operator, so also for (quark) mass and flavour. I just never came across (explicit representations of) such operators, so find them weird, rather than just an extension of what I know within the quantum formalism. Thanks very much again, for just showing that even in the standard model things are just like the hermitian operators and unitary transformations of ordinary quantum mechanics!
  5. May 23, 2007 #4


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    I also needed much self-teaching before I finally realized that it's all about
    Hilbert space operators. Or more precisely, it's all about relativistic QFT: unitary
    operators in a multi-particle Hilbert space (aka Fock space) which has been
    explicitly constructed so as to carry a (tensor product of) irreducible
    representations of the Poincare group and the various internal symmetry
    groups of the standard model.

    - strangerep.
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