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Isospin and quark model, why isospin is not considered in quark model

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    why the isospin wave function was not considered in quark model? only the others (flavour, color, space, spin) are considered?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 15, 2013 #2

    fzero

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    Isospin is an SU(2) subgroup of the approximate SU(3) flavor symmetry. If you have already accounted for flavor, it is not necessary to add something new to take isospin into account.
     
  4. Mar 15, 2013 #3
    please explain some thing more than that.
    and the total wave function of baryons should be anti symmetric under the exchange of particles(fermions), then the combination of spin part and flavour part should be symmetric because color part is anti symmetric and spatial part is symmetric (note! here spatial wave function is product of spin, flavour, space and color. isospin is not consider.) what about the combination of spin part and isospin part interms of particle exchane?
    a model ans said that combination of spin and isospin part should be symmetric, is that correct?
    please explain me elaborately
     
  5. Mar 15, 2013 #4

    fzero

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    The total wave function must be completely antisymmetric. The color part is also completely antisymmetric because baryons have zero color. Therefore, the combination of spin, flavor (which as I said includes isospin) and space must be totally symmetric.

    Details on building up the baryonic wavefunctions can be found, for example, in http://www.physics.umd.edu/courses/Phys741/xji/chapter3.pdf, but I can summarize a bit. In those notes, only 2 flavors of quark are considered, so they're dealing precisely with isospin. Since the isospin and spin symmetries are both SU(2), the component wavefunctions can be worked out in parallel. In combining 3 (iso)spin 1/2 particles, we get 8 states, of varying symmetry properties, which are labeled by the total (iso)spin quantum numbers. We can then work out all possible states formed by products ##\psi_\mathrm{spin}\psi_\mathrm{isospin}##, as well as their total symmetry under particle exchange. Finally, we have the spatial part of the wavefunction, for which exchange properties are determined by the orbital angular momentum. For instance, the ##L=0## state is symmetric under particle exchange, while the ##L=1## state is antisymmetric.

    For any given identified particle, we need to match its quantum numbers and relative mass up with our table of states.

    It sounds like whatever you were reading was considering an ##L=0## state. Since the orbital wavefunction is symmetric, then the spin-isospin part must also be symmetric in this case.
     
  6. Mar 15, 2013 #5
    thanks a lot.
    could you explain me how the isospin is related with flavour in L=0 state?
    if isospin part is symmetric then what about flavour?
     
  7. Mar 16, 2013 #6

    fzero

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    Isospin symmetry is precisely the flavor symmetry if we restrict to the states composed of the up and down quarks. With 2 quarks we build the states from the doublet

    $$\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} = \begin{pmatrix} u \\ d \end{pmatrix}.$$

    If we include the strange quark, we have SU(3) flavor symmetry and the triplet

    $$\begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \end{pmatrix} = \begin{pmatrix} u \\ d \\ s \end{pmatrix}.$$

    If we were to construct wavefunctions from products of these triplets, a subset of states would be the same as the ones that we constructed from the isospin doublets, but we would get additional states that have at least one strange quark.
     
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