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Fermions in bound states and their wavefunctions

  1. Aug 10, 2009 #1
    Hello all,

    This may be my very first post on Physics Forums. I am a 1st year physics grad student and need some help on something that's been bugging me. Suppose we have two spin half particles in a bound state. The total spin will either be 0 or 1. The spin 0 state, for example, would be symmetric (even parity right?) so it would need an antisymmetric spatial wavefunction to make the overall wavefunction antisymmetric since we have fermions? But then I thought the overall wavefunction may be symmetric because the total spin is that of a boson?

    Rephrased, my question is this: would the total wavefunction have to be antisymmetric since we are dealing with fermions, or would it be symmetric since the total spin is that of a boson?
    Which is it and why?

    If we came along and didn't know that there were two fermions in there would we think it was a boson?

    Does the fermions being in a bound state matter? What about the shape of the potential?
  2. jcsd
  3. Aug 10, 2009 #2


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    Welcome to PF!

    Hello ZombieCat! Welcome to PF! :smile:
    uhhh? :confused: did you use to be Schrodinger's cat? :biggrin:
    It's symmetric because it is a boson …

    a bound state of an even number of fermions is a boson.

    That's why mesons are bosons, but protons and neutrons are fermions … they're two quarks and three quarks respectively! :wink:
  4. Aug 10, 2009 #3


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    Sorry, wrong. The spin-0 state is antisymmetric, and the spin-1 state is symmetric.
  5. Aug 11, 2009 #4
    Re: Welcome to PF!

    Haha! I guess this zombie cat USED to be Schrodinger's cat, but is now the quantum mechanically undead. Thanks Tiny Tim!

    As for the symmetry of the spin-0 state, (ud-du)/sqrt(2), I understand that the spins are opposed to make this happen and this state should be antisymmetric... I guess I'm getting confused about the difference between symmetry and parity, (-1)^L. Does this not apply here? Why not?
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