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Quantum Physics, when is spin-orbit coupling necessary?

  1. Jun 7, 2008 #1
    When considering a simple hydrogen atom, which essentially is an electron moving in a spherical electric field, you don't need to take spin-orbit coupling into account. For larger atoms you do. I don't understand my books' very brief explanation of this. Im thinking that the electron has as much spin in the hydrogen atom as it has in any other. Or is it just that the effect is negligible for hydrogen?
     
  2. jcsd
  3. Jun 7, 2008 #2
    I may try an answer.

    The effect is not negligible for hydrogen. The fact is that the classical treatment of hydrogen atom (without spin-orbit coupling etc..) allows one to retrieve the historical spectroscopy results of the hydrogen atom which were not very accurate compared to what we can do nowadays. Actually this is not the only reason. Indeed, since the hydrogen atom has spherical symmetry, spin-orbit coupling first does not affect the ground state and second allow one to see different energy state for different value of the quantum number [tex]l[/tex]. In atoms with more than one electron, there is no spherical symmetry and these particular effects don't appear which implies, if I remember well, that the difference in energy levels for different [tex]l[/tex] is of the order of the spin-orbit coupling.

    Practically, the model of an atom structure and its refinement are made using perturbation theory and in order to be coherent with the accuracy you chose to study you have to take into account all the terms of the same order and that is what is done in atomic physics (this is just that there is no equivalent "order zero" structure for polyelectronic atoms that corresponds to the simplest one for the hydrogen atom).
     
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