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Zeeman Effect - average shift is zero?

  1. May 23, 2014 #1
    1. The problem statement, all variables and given/known data

    2zgd7hw.png

    Part (a):Find the first order shift in energy
    Part(b): What is the degeneracy after perturbation? Find Show average shift in energy is zero.

    2. Relevant equations



    3. The attempt at a solution

    I've shown part (a), the troubling part is part (b).

    Part (b)
    With the perturbation, the degeneracy is lifted. Hence degeneracy = 0.

    For ##j=l+\frac{1}{2}##, shift is ##\Delta E_+ = \frac{1}{4}mc^2 \alpha^4\frac{(l+\frac{1}{2})(l+\frac{3}{2}) - l(l+1) - \frac{3}{4}}{n^3 l(l+\frac{1}{2})(l+1)}##

    For ##j=l-\frac{1}{2}##, shift is ##\Delta E_- = \frac{1}{4}mc^2 \alpha^4\frac{(l-\frac{1}{2})(l+\frac{1}{2}) - l(l+1) - \frac{3}{4}}{n^3 l(l+\frac{1}{2})(l+1)}##.

    The total perturbation is given by the sum of positive perturbation and negative perturbation:

    [tex]\delta E= \Delta E_+ + \Delta E_-[/tex]
    [tex]\delta E = \frac{mc^2 \alpha^4}{2n^3 l}[/tex]

    Clearly, the average perturbation for a given n and l is not zero. How does it even become zero then?
     
  2. jcsd
  3. May 23, 2014 #2
    j = l + 1/2 and j = l - 1/2 have different degeneracies. They contribute different weights to the average.
     
    Last edited: May 23, 2014
  4. May 23, 2014 #3
    Each has a weight of ##2l-1##?
     
  5. May 23, 2014 #4
    No, they have weights 2j+1 for two different values of j
     
  6. May 23, 2014 #5
    Oh yeah, because when two angular momenta ##j_1## and ##j_2## are combined, each state in ##J = j_1 + j_2## ranges from -J to +J in steps of 1.

    I keep forgetting that this is a two-angular momentum problem, not a simple one with only spin.

    [tex]\delta E = (2j_+ +1)\Delta E_+ + (2j_- +1)\Delta E_-[/tex]

    I worked it out, and it appears to be zero.
     
    Last edited: May 23, 2014
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