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Degeneracy of the 3d harmonic oscillator

  1. Apr 18, 2007 #1
    Hi!

    I'm trying to calculate the degeneracy of each state for 3D harmonic oscillator.
    The eigenvalues are

    En = (N + 3/2) hw

    Unfortunately I didn't find this topic in my textbook.
    Can somebody help me?
     
  2. jcsd
  3. Apr 18, 2007 #2
    What does the "N" operator look like? Is it just [tex]a^\dagger a[/tex] or something else?
     
  4. Apr 18, 2007 #3
    Hi,
    sorry I made a mistake in the formula. It should be

    En = (n + 3/2) hw
     
  5. Apr 19, 2007 #4
    [itex]n[/itex] is the sum of 3 quantum numbers, [itex]n_x, n_y, n_z[/itex]; for the lowest rung, which is [itex]n=0[/itex], there's no degeneracy. For [itex]n=1[/itex], there's 3 fold degeneracy (one of the quantum numbers is 1, and rest is 0). For [itex]n=2[/itex], there's 6 fold degenertacy (one of them may be 2 and rest 0, or two of them may be 1) and so on. The result is, for nth energy level, there's 3n degeneracy.
     
  6. Apr 20, 2007 #5
    Thanks for the answer!!
     
  7. May 7, 2007 #6
    Watch out!!

    No that's not right.
    You can't work it out for the first couple of cases and then presume the trend continues like that.
    In fact, the degeneracy g(n) is:
    g(n) = (1/2)(n+1)(n+2)
     
  8. May 7, 2007 #7

    nrqed

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    And even this, I think, is nt completely right. There are some levels with additional accidental degeneracy which don't fit the pattern.
     
  9. May 7, 2007 #8
    Really? Which ones?
    The formula can be derived like this:
    n = n1 + n2 + n3
    where 1,2,3 are three orthogonal directions.
    Choose n1 then
    n2 + n3 = n - n1
    Can always pick n - n1 + 1 different pairs of n2, n3.
    Sum over n1 from 0 to n:

    Sum(n - n1 +1) = Sum(n - 1) - Sum(n1)
    =(n - 1)*(n - 1) - (1/2)n(n + 1)
    =(1/2)(n+1)(n+2)

    Maybe there's a mistake somewhere?
     
  10. May 7, 2007 #9

    nrqed

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    I was thinking about the cubic infinite square well which is more tricky since it involves the sum of squares, so never mind my comment about accidental degeneracy.

    Your formula may be right but I don't understand how you got from (n - 1)*(n - 1) - (1/2)n(n + 1) to (1/2)(n+1)(n+2). Those two expressions are not equal. Moreover, the sum [itex] \sum_{n_1 = 0}^n [/itex] gives (n+1), not (n-1), right?

    EDIT: Actually, the formula does not seem to work. I may have missed some states, but I can't get it to work for n=10, say.
     
    Last edited: May 7, 2007
  11. May 7, 2007 #10
    sorry - i was being silly
    Sum(n - n1 +1) = Sum(n + 1) - Sum(n1)
    =(n + 1)*(n + 1) - (1/2)n(n + 1)
    =(1/2)(n+1)(n+2)
     
  12. May 7, 2007 #11
    Does work for n = 10, look:
    State
    3 x 10, 0, 0
    3 x 8, 1, 1
    3 x 6, 2, 2
    3 x 4, 3, 3
    3 x 2, 4, 4
    3 x 0, 5, 5
    6 x 9, 1, 0
    6 x 8, 2, 0
    6 x 7, 3, 0
    6 x 7, 2, 1
    6 x 6, 4, 0
    6 x 6, 3, 1
    6 x 5, 4, 1
    6 x 5, 3, 2
    Degeneracy = 6 x 8 + 3 x 6 = 6 x 11 = 66

    The formula for g(n) gives (1/2)*(11)*(12) = 66

    So it works.
     
    Last edited: May 7, 2007
  13. May 7, 2007 #12

    nrqed

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    Yes, you are completely right. I had missed one entry.
    Good job. Thanks for the correction.

    If you know a formula for the infinite cubic well, let me know...It's in that case that accidental degeneracy are a killer.

    Regards

    Patrick
     
  14. Mar 7, 2010 #13
    Thank you everyone.. i too was stuck in it and it helped me a lot
     
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