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I Is baryon antisymmetry always true?

  1. Mar 13, 2017 #1

    CAF123

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    The baryon wavefunction is comprised of the direct product of contributions forming different Hilbert spaces such that $$|\Psi \rangle = |\text{spin} \rangle \otimes |\text{flavour}\rangle \otimes | \text{colour}\rangle \otimes |\text{space}\rangle. $$ The necessity for a colour degree of freedom is usually motivated in the literature from the delta(++) spin 3/2 containing quark content ##uuu##. It is flavour symmetric by inspection, is symmetric in the spin quantum numbers and for lowest lying states, has symmetric space state. The state thus contains identical fermions but is overall symmetric under interchange of any of the quarks. This is in violation of the PEP – the resolution was of course the addition of the colour degree of freedom which is necessarily antisymmetric so as to conform to the principle.

    It’s then said that the generic wavefunction for a baryon is overall antisymmetric. Is this only the case where the baryon wavefunction is of the form ##\epsilon_{ijk} \psi^{(1)}_i \psi^{(1)}_j \psi^{(1)}_k## or ##\epsilon_{ijk} \psi^{(1)}_i \psi^{(1)}_j \psi^{(2)}_k## where in the former case we have all three qqq the same and in the latter only two are identical?

    I say this because if we consider one of the baryons with no flavour symmetry e.g ##|uds \rangle## this is a state with no identical fermions so does the the PEP (i.e total wavefunction antisymmetry) have to hold for this case? (I guess analogously to the fact that the mesons have no requirement of antisymmetry because the content of the valence quarks is ##q \bar q## and this is never two identical quarks)
     
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  3. Mar 13, 2017 #2

    mfb

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    You have the color degrees of freedom for every baryon, but if the quark flavors are different, you don't have to worry about exchanges of quarks. The "exchange of color" is still relevant (there is nothing special about "green", for example).
     
  4. Mar 13, 2017 #3

    CAF123

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    Yes, but does baryonic states with no flavour symmetry of the quarks have to obey Fermi-Dirac statistics? Just by looking at the flavour content of such states we see that it is not possible to have any two identical quarks and so the requirement of PEP/Fermi-Dirac statistics/antisymmetry of wave function perhaps need not be obeyed by such configurations?
     
  5. Mar 13, 2017 #4

    mfb

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    All the up quarks in an uds baryon follow the Fermi-Dirac statistics. For a single quark that is not a really interesting property.
     
  6. Mar 13, 2017 #5

    CAF123

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    The observable 'uds' states are in a flavour symmetric combination (the 10), flavour antisymmetric (the 1), and two mixed flavour symmetries (the two 8's) - so the observable states are either all antisymmetric or symmetric in flavour. But for the case of uds quark content we can never have any two of of u,d or s in the same quantum state. So why do we still say that the baryon wavefunction for these states still has to satisfy Fermi-Dirac stats? The states with u,d and s quark content do not have identical particles so why do we still say that they have to have an overall wavefunction satisfying antisymmetry? (I'm just trying to understand why the PEP/Fermi dirac stats should be enforced on such states)
     
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