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Degenerate states of 2 particles in a 1D harmonic oscillator potential

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    "Two non-interacting particles are placed in a one-dimensional harmonic oscillator potential. What are the degeneracies of the two lowest energy states of the system if the particles are
    a)identical spinless bosons
    b)identical spin-1/2 fermions?

    2. Relevant equations

    3. The attempt at a solution
    I think degeneracy is equal to n^2
    for b) two fermions can be in the same spatial state if they are in different spin states. one would be in spin up and spin down so this would affect the degeracy. Apart from that..I'm not sure
  2. jcsd
  3. Apr 29, 2009 #2
    Re: Degeneracies

    exactly so there is no degeneracy for fermions (at least for the wavefunction of the entire state)

    and for bosons there is no such pauli exclusion principle.

    i would like someone to confirm however, if we are asked for the degeneracy are we meant to write a number i.e. 2 for the boson case or are we expected to write a spiel about pauli exclusion etc.

    also isn't degeneracy given by something like [itex]g_n =\sum_{l=0}^n l(l+1)[/itex] - can someone tell me whether this is true or not?
  4. Apr 29, 2009 #3
    Re: Degeneracies

    I am not sure about your equation

    Why is there no degeneracy for fermions - is that just because there are only two particles? What would the degeneracy be if there were three particles? What do you mean at least for the wavefunction of the entire state?
  5. Apr 30, 2009 #4
    Re: Degeneracies

    didn't read your question properly. my above argument is tru for the ground state only. but in the first excited state it's possible to have two fermions with different wavefunctions and still have the same degenaracy. i.e. if you use the symmetric spatial wavefrunction then you need to put it in a spin singlet state - can you figure out the other possibility?
  6. Apr 30, 2009 #5
    Re: Degeneracies

    actually it looks like the degeneracy relation is [itex]g_n=\sum_{l=0}^{n-1} (2l+1)=n^2[/itex]
  7. May 1, 2009 #6


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    Science Advisor
    Gold Member

    Re: Degeneracies

    The number of degeneracies is just the number of configurations you can have to get a specific energy. How many different ways can you arrange 2 fermions/bosons to get them to have the lowest total energy?

    @Latentcorpse: It looks like your formula is for the degeneracies of the hydrogen atom (or a hydrogenic atom), I don't think that applies to this problem...? (Correct me if I'm wrong)
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