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Baryon Octet/Decuplets

  1. May 2, 2010 #1
    Can anyone explain to me why you can't have uuu and ddd on the first (spinhalf.gif) diagrams I have uploaded please?

    Attached Files:

  2. jcsd
  3. May 2, 2010 #2
  4. May 2, 2010 #3
    The answers in that thread were completely beyond me. Can somebody explain in a single sentence for me? I'm only studying very elementary particle physics..
  5. May 2, 2010 #4


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    The short answer is the exclusion principle. But to see that you need to know a bit about groups and their representations. Specifically that the direct product of 3 fundamental '3's of SU(3) decomposes like 3 * 3 * 3 = 10 + 8 + 8 + 1. The decuplet (which contains uuu) is necessarily spin 3/2 and completely symmetric.
  6. May 3, 2010 #5


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    The attempt to construct a spin 1/2 state |uuu> results in a violation of the Pauli exclusion principle.

    As far as I remember from my QCD lectures (~ two decades ago!) the explanation in the other thread misses the spatial part of the wave function. One must take into account
    |color> * |spin> * |isospin> * |space>
    Ofcourse this doesn't make things easier ...
  7. May 3, 2010 #6
    yes the most basic answer is to combine the two of haelfix and tom.stoer
  8. May 3, 2010 #7
    The ground state L=0 is space-symmetric in the constituent quark model.
  9. May 3, 2010 #8


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    The answer is simpler than group theory, but is based on the Pauli principle (Fermi-Dirac statistics) and the addition of spins. The combined state of (color)(space)(spin) must be completely antisymmetric for three identical u quarks. The wave function is antisymmetric in color. The spin addition 1/2+1/2+1/2=1/2 is of mixed symmetry, and so cannot combine with the presumed symmetry spatial ground state and antisymmetric color state to form a completely antisymmetric state. Spin 1/2 for three quarks is only possible if they are not all identical, for instance uud.
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