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Why quarks' wave functions are eigenstates of I and Iz

  1. Mar 12, 2013 #1
    This might sound like a totally dumb question but anyway:

    QCD lagrangian in the limit of mu=md has flavor SU(2) symmetry with respect these two quarks. And we say that these quarks' wave functions are eigenstates of I and Iz. The question is why?

    Thanks.
     
  2. jcsd
  3. Mar 12, 2013 #2

    fzero

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    Whenever there is a symmetry of the system, the corresponding charges of the states become good quantum numbers (since the charge generators commute with the Hamiltonian). So we can always choose a basis for the states such that the basis elements are eigenstates of [itex]I^2, I_z[/itex]. In this particular case, the SU(2) flavor symmetry acts in such a way that the up and down quarks have [itex]I_z=+1[/itex] and [itex]I_z=-1[/itex].
     
  4. Mar 12, 2013 #3
    but how do you know that the eigenstate of Iz are the real u and d quark and not some linear combination of them?
     
  5. Mar 13, 2013 #4

    Bill_K

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    Well for one thing, the charge. The u and d quark states are the charge eigenstates. By convention we choose isospin to obey Q = Iz + Y/2 where Y is the hypercharge, a constant for the multiplet and equal to 1/3 for quarks.
     
  6. Mar 13, 2013 #5
    How does all these follow from qcd lagrangian?
     
  7. Mar 13, 2013 #6

    fzero

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    Electric charge and isospin are symmetries of the QCD Lagrangian. Therefore, there are no terms in the QCD Lagrangian that can change those quantum numbers. So if a state is prepared with a given set of charges, these will be conserved.
     
  8. Mar 13, 2013 #7
    Am I right on this: the direction of axes in isospin space can be chosen arbitrarily, hence we choose them in such a way as to attribute to u quark the 1/2 eigenvalue of Iz operator and to d quark - the -1/2 eigenvalue; I is the casimir operator for the representation so it gives the indexing of representation.
     
  9. Mar 13, 2013 #8

    fzero

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    This is true if we ignore electric charge. As Bill_K points out, after electroweak symmetry breaking, the physical electric charge depends on the eigenvalue of ##I_z##, so the identification of quarks is actually fixed on us by self-consistency.
     
  10. Mar 15, 2013 #9
    But isn't it strange that it's fixed? I mean, then how does that happen that eigenstates of Iz are precisely the quarks of electrodynamics interaction and not some linear combination of them? How would that follow from SM lagrangian?
     
    Last edited: Mar 15, 2013
  11. Mar 15, 2013 #10

    fzero

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    Having [itex]I_z[/itex] diagonal is a choice of basis. It's natural to choose it diagonal in the basis of states handed to us by observation. Choosing some other basis would just make things inconvenient, not any more correct.
     
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