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Lande g-factor and total angular momentum conservation

  1. Sep 16, 2012 #1
    I'm reading about the derivation of the lande' g-factor which comes about when one considers an electron moving about a nucleus which is put in an external magnetic field. This gives rise to a perturbative hamiltonian

    [tex] H = - (\vec \mu_s + \vec \mu_s) \cdot \vec B_{ext} = \frac{e}{2m} \vec{B}_{ext} \cdot (\vec J+ g \vec S)[/tex][/tex]

    and to find the associated energy (expectation value of H) one encounters the problem that the spin S is not conserved in this situation but and one then states that the toatal spin J is conserved and that S will be precessing about J. Therefore the average value of S which is interessting for the expectation value can be expressed as

    [tex] \vec S_{av} = \frac{(\vec S \cdot \vec J)}{J^2} \vec J[/tex]

    and we can express

    [tex]2 \vec S \cdot \vec J = J^2 + S^2 - L^2[/tex]

    and the problem is essentially solved from there. Now what I wonder about is how one really figures out that the total angular momentum _is_ conserved. I would like a classical (and QM) proof of this statement. Could anyone lead me in the right direction?
     
  2. jcsd
  3. Sep 16, 2012 #2

    Bill_K

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    "The effect can be classified as weak or strong according to whether the magnetic energy is small or large in comparison with the spin-orbit energy. The Zeeman Effect refers to the weak-field case, while Paschen-Back Effect refers to the strong-field case. In the weak-field case, the magnetic energy has matrix elements between states of different j for but not between states of the same j and different m. We neglect the former, because of the relatively large energy separation between states of different j. Thus the magnetic energy is diagonal with respect to m for each j and shifts the energy of each of the states by its expectation value for that state."
     
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