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Rotational Energy and Degeneracy

  1. Nov 13, 2015 #1
    I have a question regarding the rotational energy equation, derived based on the rigid rotor assumption:

    ##\epsilon = k\theta_r J (J+1)##
    where k = boltzmann constant, and J is the rotational quantum number.

    The degeneracy is 2J+1.

    Let's assume the constant quantity ##k\theta_r## = 1 and that the energy level is 2Joules.
    Thus 2 = J(J+1) => J = +/- 1. The only physical solution is J = 1.
    Thus degeneracy is 3 for J = 1.

    Here is where I am confused. I thought degeneracy is when you have an energy level that DOES NOT consist of a unique set of quantum numbers. So if the degeneracy is 3, doesn't that mean I should have 3 different J values that can give me an energy level of 2 Joules?
    I think I am understanding this incorrectly, because for 2 joules, the only solution is J = 1. Doesn't that mean the degenearcy is 1?
     
  2. jcsd
  3. Nov 13, 2015 #2

    Nugatory

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    Staff: Mentor

    There are three different and linearly independent solutions to the Schrodinger equation with ##J=1##. That's where the three-fold degeneracy comes from.

    Google for "spherical harmonics" to see how this can be; the three solutions are multiples of the three functions ##Y^1_m## with ##m## equal to -1, 0, or 1.
     
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