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Spin orbit coupling

  1. Feb 27, 2013 #1
    I don't know if you are familiar with it, but in pertubationt theory for hydrogen it is handy to look for eigenstates of J = L + S since this commutes with the hamiltonian (L and S are not separately conserved).
    My book then says that the good quantum numbers are: n,l, j, mj
    I must admit I'm not used to this idea of good quantum numbers - my book hasn't introduced the term properly (I'm guessing it is just the quantum numbers belongning to the set of eigenvectors that diagonalizes the pertubation) and the thing that bothers me the most is: Why is l a good quantum number but not s? Surely these should be treated on eqaul footing since they are just the length of the total orbital and spin angular momentum respectively. And why would l be a good quantum number when L is not conserved separately.
     
  2. jcsd
  3. Feb 27, 2013 #2

    DrDu

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    Well, j and mj are certainly good quantum numbers as they refer to the total angular momentum which is conserved in an isotropic system.
    l and s are only approximately good in so far as the perturbation is weak and will lift in lowest order the splitting of the degenerate states with different j but same s and l.
    s is probably not mentioned as it is always 1/2 and this doesn't have to be written down all the time.
     
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