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Quark Model (Supermultiplet and States questions)

  1. Aug 24, 2011 #1
    From Ryder:
    "We can now see how a supermultiplet of ten baryons may arise. Baryons are made of three identical fermions, so the possible states may be classified according to their symmetry under interchange of quark labels. Altogether there are 27 states. One of these is totally anti-symmetric: uds+dsu+sud-usd-sdu-dus,
    and ten symmetric states:

    I am a little bit lost here, first question:
    what is a supermultiplet...? Also, what does he mean by interchange of quark labels?

    second question:
    How are these states supposed to clearly follow from the existence of 27 states...? I'm not really sure how I'd come up with these states.

    For "uud+udu+duu"
    (assuming I'm not missing something really obvious) my first thought was "2 up quarks, 1 down...proton"? I feel like I'm missing something obvious here, besides the idea that the terms don't commute (which I'm guessing is part of this). Wait, it said these were non-normalized...normalized would we have 1/sqrt 3(uud+udu+duu) and then the whole thing would just be the different states of a proton? I feel like I'm seriously over simplifying at this point.

    Thanks for any and all help,

  2. jcsd
  3. Aug 24, 2011 #2


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    Elwin, In Ryder's terminology, a multiplet is a group of particles related to each other under isospin, and a supermultiplet is a group of particles related to each other under SU(3). (Usually, I think, the term 'supermultiplet' is reserved for SU(4) and higher, but Oh well.)

    By quark labels he means the flavors u, d, s. There are 27 possible states (33 = 27.)

    Each of the 27 states has a certain charge and strangeness. The ten states he's listing form the baryon decuplet, and to form the ten, he takes the states having similar charge and strangeness and forms the symmetric combination. So starting with uus we get uus + usu + suu.

    Since the overall symmetry of a fermion state (space, spin, flavor, color) must be antisymmetric, the decuplet is symmetric under spin, i.e. spin 3/2. So uud+udu+duu is not a proton, it's a resonance usually called Δ+, or the "3,3" resonance.
  4. Aug 24, 2011 #3
    How do we know that the decuplet won't be antisymmetric under something like color...? Is it because hadrons are colorless? I know it's antisymmetric under space (Pauli exclusion principle). What does it mean to be symmetric under flavor? I know what the flavors are, is it just symmetric because of the additive nature of the states?

    Thanks for your help :) Sorry if I'm asking slightly stupid questions
  5. Aug 25, 2011 #4


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    Elwin, Not at all. Good questions. A hadron is colorless, which means it's a color singlet, i.e. the color part is totally antisymmetric.
    The space part is (almost always) L=0, so totally symmetric.
    That leaves the spin part and flavor part, which must both have the same symmetry "S". So that (antisymmetric x symmetric x S x S) works out to be antisymmetric overall, for a fermion state.

    As I said, the decuplet has a symmetric flavor state, so the spin state must be the same - symmetric also.

    Ryder focused on the decuplet because the octet which contains the proton is a bit more complicated to describe. It has "mixed" symmetry (neither totally symmetric nor totally antisymmetric) As likewise the coupling of the three spins to total spin 1/2 is mixed.
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