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    The pi electrons in benzene can be modelled as a particle on a ring, and from looking in some books I can find the answer to one question about the degeneracy: that the n=0 level is singly degenerate but that each of the subsequent n=1,2 are doubly degenerate. But I can't understand how you can have an n=0 level, or how to explain that the rest of the levels are doubly degenerate and that the n=0 level isn't???

    thanks
    Dan

    Edit:

    there is a slideshow here of some information on what I'm on about:

    http://docs.google.com/viewer?a=v&q=...nXY4apt8BJvg9g

    it refers to j=0,1,2,3 etc. is this the angular momentum and how come its changed to dealing with this now?
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    (Original post by danhirons)
    The pi electrons in benzene can be modelled as a particle on a ring, and from looking in some books I can find the answer to one question about the degeneracy: that the n=0 level is singly degenerate but that each of the subsequent n=1,2 are doubly degenerate. But I can't understand how you can have an n=0 level, or how to explain that the rest of the levels are doubly degenerate and that the n=0 level isn't???

    thanks
    Dan

    Edit:

    there is a slideshow here of some information on what I'm on about:

    http://docs.google.com/viewer?a=v&q=...nXY4apt8BJvg9g

    it refers to j=0,1,2,3 etc. is this the angular momentum and how come its changed to dealing with this now?
    Hmm...the easiest way I can think of to explain this is to use a different wavefunction for the particle-on-a-ring. It has the same periodicity as the other one, so it satisfies the boundary conditions (\Psi(0) = \Psi(2n\pi)).

    That wavefunction is:

    \Psi(\theta) = \exp(i n \theta)

    It satisfies our boundary conditions, since \exp(0) = 1, \exp(i 2 n \pi) = 1 and its energy using their Hamiltonian is:

    \mathcal{H}\Psi(\theta) = \frac{\hbar^2 n^2}{2 m r^2}\exp(i n \theta)

    So E = \frac{\hbar^2 n^2}{2 m r^2}

    For n = 0 the energy is 0 and there is no other state with energy 0.

    However, E(n = 1) = \frac{\hbar^2 1^2}{2 m r^2} = \frac{\hbar^2 (-1)^2}{2 m r^2} = E(n = -1). This is true for all n, so all of them are doubly degenerate, since you can have plus or minus n giving you the same answer. n = 0 is singly degenerate since plus/minus zero has no real meaning.

    Not sure about the benzene issue, since actually the third level is singly degenerate. I've never thought about the particle on the ring approach to benzene.
 
 
 
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