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Degeneracy of rotationally invariant potentials

  1. May 7, 2008 #1
    I can appreciate the degeneracy of an infinite cubical well, in which there are three different directions, and hence three different separation constants from Schrodinger's equation which determine three separate n's (for lack of a better word.. principal quantum numbers, i suppose. it really bothers me that you can excite only one direction of movement.. wouldn't this imply that energy is not a scalar?) But what about for spherically symmetrical potentials, in which the only direction of movement which has an effect on the energy is radial movement? I mean, I know that three dimensions means there will definitely be degeneracy, but where does this come from? How can we only excite one 'dimension' of movement? Can I draw any parallels between the cubical well and the spherical well?
  2. jcsd
  3. May 9, 2008 #2


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    What do you mean by, "excite only one direction of movement". I don't understand.

    Also, note that in general, the energy in a spherically symmetric potential V(r) can depend on the angular quantum numbers l and m. Only for the special cases when the potential is of the form 1/r or r^2, does the energy depend exclusively on the radial quantum number n.
    Last edited: May 9, 2008
  4. May 10, 2008 #3


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    Try an example -- say a spherical well.
    Reilly Atkinson
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