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Degeneracy in 3-d box

  1. Mar 30, 2007 #1
    This isn't a homework question, rather a question about something stated in my book and an online source.

    When is there degeracy in a 3-D rectangular box when none of the sides are of equal length?

    I understand that when there are two or more state functions that have same energy level there is degeneracy... but more general, I read something online about the ratio of quantum numbers n must be equal to and integer (yet not equal to eachother), and/or there must be a linear combination of the lengths of the sides (i.e a=.5*b=.25*c, a being length in x, etc.)

    Can someone better explain this to me?

    As always, thanks in advance!
     
  2. jcsd
  3. Mar 30, 2007 #2

    Galileo

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    All you need to know are the eigenvalues. If the lengths of the box are Lx, Ly and Lz, then the eigenvalues are:

    [tex]E_{nlm}=\frac{\hbar^2}{2m}(k_x^2+k_y^2+k_z^2)[/tex]
    where [tex]k_x = \pi n/L_x[/tex]
    [tex]k_y = \pi l/L_y[/tex]
    [tex]k_z = \pi m/L_z[/tex] .
    with n,l and m integers.

    You have degeneracy when you can find different values of n,l and m with the same value for E.
     
  4. Mar 30, 2007 #3
    There may be no degeneracy if you can't express the lengths of the box in terms of the other lengths in relation to integer values. Degeneracy only arises when there is some sort of symmetry in the system, and clearly that's not the case for some random box lengths.
     
  5. Mar 31, 2007 #4

    I understand the symmetry.

    Expressing the lengths in terms of other lengths as integers, you mean possibly some kind of multiple length of two sides in relation to the first would be suitable for degeneracy?
     
  6. Mar 31, 2007 #5

    Dr Transport

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    This is called accidental degeneracy.......
     
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