jjshan said:
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?
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.
a model ans said that combination of spin and isospin part should be symmetric, is that correct?
please explain me elaborately
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.