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Rotation Energy Levels and Degeneracy.

  1. Aug 23, 2006 #1

    Clausius2

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    Solving the Schrödinger equation in spherical coordinates for a diatomic gas, one finds that the rotational energy leves are given by:

    [tex]\epsilon_l=K\cdot l(l+1)[/tex] where [tex]l=0,1,2...[/tex] is the rotational quantum number and K is a constant.

    It is said that each energy level shows a degeneracy of [tex]g_l=2l+1[/tex].

    I understand Degeneracy occurs if for different energy levels one has the same value of energy. Is that right?. Is every quantum number representing an energy level? If that, [tex]\epsilon[/tex] is a single valued function of [tex]l[/tex], so I cannot have the same energy for different quantum numbers. How is the thing of [tex]g_l[/tex] obtained, and how is it physically interpretable for let's say [tex]l=1[/tex]?.

    Thanks in advance.
     
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  3. Aug 23, 2006 #2

    Galileo

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    Degeneracy occurs when you have different eigenstates having the same energy. In this case the quantum number 'l' is apparantly not enough to completely specify the state, you also need a quantum number 'm' which ranges from -l to +l in integer steps, so that gives a degeneracy level of 2l+1.
     
  4. Aug 23, 2006 #3

    Clausius2

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    So that means that for different [tex]l[/tex] which is a multiple of the eigenvalue I have to have the same energy.

    I don't see that. Can you elaborate that a little bit more?. Also I don't see the logic of the the "so that" you are using.

    Thanks.
     
  5. Aug 25, 2006 #4

    Galileo

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    No, you can't have different l's and the same energy, because in this case E=l(l+1)K, so states with different l's have different energies. There's another quantum number needed to specify a state, which does not affect the energy, that's m.

    Consider the hydrogen atom. An eigenstate requires three quantum numbers to characterize: n, l and m, where n is a positive integer, l can range from 0 to n-1 and m from -l to l (all in integer steps). In this case the energy is only dependent on n. So all states with the same n, but different l and m have the same energy. The degeneracy of the n'th energy level is n^2 (it's just counting, see below).

    Well, I haven't looked at or solved the Schrodinger equation for this case. I assumed it was similar to a rigid rotor system. The angular part of the wave function are described by spherical harmonics which have two quantum numbers l and m. m ranges from -l to l in integer steps, so given l, m can take the values l, l-1, l-2, ..., 0, -1, -2, ..., -l. If you count, that are 2l+1 possible values for m (1,2,..,l gives l values, -1,-2,..,-l another l, plus the 0 gives 2l+1 total).
     
    Last edited: Aug 25, 2006
  6. Aug 25, 2006 #5

    Gokul43201

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    It is the rigid rotor, isn't it?

    Clausius, you could think of the energy of the molecule as depending only on the magnitude and not the orientation of the angular momentum vector. For any given [tex]l[/tex], you can picture each [itex]m_l[/itex] as corresponding to a different orientation of the angular momentum about some fixed axis (with the constraint that there are only a limited number of such allowed orientations).
     
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