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Degeneracy and Symmetrization Requirement

  1. Feb 15, 2010 #1
    I am encountering two-particle systems for the first time, and I am a little confused about something.

    Consider a infinite square well with two noninteracting particles, both fermions (though, disregard spin degrees of freedom, this is a strictly 1-d scenario).

    So I go to write down the states of the system corresponding to different energies. Consider the case:

    [tex] \psi_{23} = \frac{1}{\sqrt{2}} \left[ \psi_2(x_1) \psi_3(x_2) - \psi_3(x_1) \psi_2(x_2) \right] [/tex]


    [tex] \psi_n (x) = \sqrt{\frac{2}{a}} \sin n \pi x / a [/tex]

    This corresponds to energy

    [tex] E = \frac{13 \hbar^2 \pi^2}{2 m a^2} [/tex]

    This is all good. What I don't get is this: why isn't this energy degenerate? In other words, isn't the state:

    [tex] \psi_{32} = \frac{1}{\sqrt{2}} \left[ \psi_3(x_1) \psi_2(x_2) - \psi_2(x_1) \psi_3(x_2) \right] [/tex]

    Also an energy eigenfunction of the Hamiltonian, with the same eigenvalue. Aren't the two states distinct?

    Now, the symmetrization requirement says:

    [tex] \psi(x1,x2) = -\psi(x2,x1)[/tex]

    for fermions, which is the case above. I feel like this is relevant. But I still see there being two separate states with the same energy. Isn't this the definition of degeneracy?
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 15, 2010 #2
    It does get very confusing with the sums and differences and symmetrizations etc, but I wonder if you haven't made the very simple mistake of writing out two states which are the same as each other except for a negative sign? So they're actually the same state.
  4. Feb 15, 2010 #3
    Ah, its as simple as that. If two eigenvectors are the same up to a scale factor, they are the same eigenvector. I was forgetting that.

  5. Feb 15, 2010 #4
    The two states you wrote down are linearly dependent. They only differ up to an overall minus sign.

    [tex]\psi_{23} = -\psi_{32}[/tex]

    Two states are considered degenerate when they are not linearly dependent.

    You can in fact construct an infinite number of states which are all eigenstates of the Hamiltonian. Namely,

    [tex] \psi = e^{i\theta}\psi_{32}[/tex]

    where [itex]0\leq \theta< 2\pi[/itex]. All these states are eigenstates, with the values [itex]\theta=0[/itex] and [itex]\theta=\pi[/itex] corresponding to your two functions. But you do not consider these states as "new states", since the overall phase doesn't matter.

    EDIT: too slow..
  6. Feb 15, 2010 #5
    That was helpful too, I hadn't thought about the phase factor. Thanks!
  7. Feb 15, 2010 #6


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    I don't understand what you have written. How can you have fermions without considering spin? Are you just saying that you are only going to accept spatial wavefunctions that satisfy the anti-symmetry property of fermionic wavefunctions? That is fine I guess, but the issue with that is there is another class of solutions where the spatial part of the wavefunction is symmetric under exchange, but the antisymmetry is provided by the spin portion of the wavefunction.

    All of this is usually worked out in some detail in a standard textbook ... it can often be found under the heading of "Pauli Exclusion Principle" in the context of excited states of the helium atom.
  8. Feb 15, 2010 #7
    I understand, this is just an example I am pulling from Griffiths.
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