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Understanding the baryon wavefunction

  1. May 8, 2013 #1
    If a baryon wavefunction is [itex]\Psi = \psi_{spatial} \psi_{colour} \psi_{flavour} \psi_{spin}[/itex],
    and we consider the ground state (L=0) only.
    We know that the whole thing has to be antisymmetric under the interchange of two quarks. We know that colour is antisymmetric (always colourless) and spatial is symmetric. Therefore, the combination of flavour and spin has to be symmetric.
    That's fine, I understand that. However, I'm very uncertain about what the 'interchange of two quarks' actually means. Interchange how?

    For the 'easy' example of |uuu>. How is the flavour symmetric?

    Basically, what does it mean to be symmetric in the quark model?
  2. jcsd
  3. May 9, 2013 #2
    In the notation, interchanging two quarks just means changing which quark you write first and which one you write second. Nature does not label them "quark 1" and "quark 2"--all she knows is "there are two quarks"--so any expression you write down had better be agnostic about which quark is "first" and which is "second" (up to an overall minus sign in the case of fermions).

    Anyway, the best way to get started is probably with examples. Suppose there were particles composed of two quarks. Here are some flavor-symmetric states:

    |uu>, |dd>, (|uu> + |dd>)

    Here is a flavor-antisymmetric state:

    (|ud> - |du>)

    [If you swap the first and second quark, you get (|du> - |ud>), which is the negative of the original state.]

    Here is a state without definite flavor symmetry (it is a linear combination of a flavor-symmetric and a flavor-antisymmetric state):


    [If you swap the first and second quark, you get |du>, which is not a multiple of the original state.]

    Here is a three-particle example. What is the only totally symmetric flavor state with one u, one d, and one s quark? It is

    (|uds> + |usd> + |dus> + |dsu> + |sud> + |sdu>)

    For example, on exchanging the first and second quark you get the state

    (|dus> + |sud> + |uds> + |sdu> + |usd> + |dsu>)

    which you can verify is the same state as we started with.
  4. May 12, 2013 #3
    Thank you very much. I've been searching everywhere for a clear concise explanation! The answer is actually quite simple when you know about it!
    Thanks again
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