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Hydrogen spin-orbit interaction

  1. Feb 21, 2010 #1
    For a particular energy level in hydrogen, with quantum numbers n and l, one will find when considering the spin-orbit interaction, the level is split into two fine structure levels with energy separation:

    [tex]\Delta E_{s.o.}=\beta_{nl}(l+1/2)[/tex]

    I was trying to prove this result. The spin of an electron is 1/2. Therefore there are two possible values for the total angular momentum, as the spin can be either +-1/2. Therefore using (and the relevant energy spin-orbit equation):

    [tex]\Delta E_{s.o.}=E_{j=l+1/2}-E_{j=l-1/2}[/tex]

    gives the first result. However, when you follow the proof through I am confused because in the energy spin-orbit equation (not stated here) you use s=1/2 for both possible energy states. However you use l=+-1/2 for the angular momentum. Why is this? Surely you'd use s=+-1/2 for the electron spin also?
  2. jcsd
  3. Feb 21, 2010 #2


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    Angular momentum is always positive. However, the total angular momentum j is a vector sum of l and s, so it may assume values spanning from |l-s| to l+s, when both l and s are non-zero. So, it is not that you use positive and negative values for angular momentum in the equation, it is that you need to consider all the ways that l and s can couple to yield j. Clear?
  4. Feb 21, 2010 #3
    Yeah, i understand why l is positive. However, my question was concerning the spin. When you prove the result you use only the positive value for spin (s=1/2), for both the j=1+1/2 and j=l-1/2 states.

    [tex]E_{s.o.}=\frac{\beta}{2} [j(j+1)-l(l+1)-s(s+1)][/tex]

    You use positive values for l and s in the above equation. I don't understand why the positive values for s, since as you said, you have to consider all the ways l and s can couple to yield j.
  5. Feb 21, 2010 #4


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    right ... and for different couplings you have different values of j. If you assume we are talking about the 2p orbital, then l=1, s=1/2, and we have two possibilities:

    2P3/2 state: l=1, s=1/2, j=l+s=3/2 --> E=beta/2(15/4 - 8/4 - 3/4) = beta/2

    2P1/2 state: l=1, s=1/2, j=|l-s|=1/2 --> E=beta/2(3/4 - 8/4 - 3/4) = -beta
  6. Feb 21, 2010 #5
    Okay, so strictly l and s are always positive. It is just for different coupling, j will take different values: |l-s|< or = j < or = l+s ?
  7. Feb 21, 2010 #6


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    You got it. :wink:
  8. Feb 21, 2010 #7
    The spin-orbital interaction (L S coupling) is gotten using the Law of cosines as follows, (though the sign is a little different),

    [tex]2 \hat{L} \cdot \hat{S} = (\hat{L}+\hat{S})^2(=\hat{J}^2) - \hat{L}^2 - \hat{S}^2 = (J(J+1) -l(l+1)-s(s+1)) \hbar^2[/tex]

    Here we have to consider the precession of the orbital and spin movement, so we use [tex]J(J+1),l(l+1),s(s+1)[/tex] instead of [tex]J^2, l^2, s^2[/tex].

    The s is positive, but the figures of the triangle are different between J=L-1/2 and J=L+1/2.
    The three sides of both triangles are [tex]\sqrt{J(J+1)}, \sqrt{l(l+1)}, \sqrt{s(s+1)}[/tex]. (Here, only J is different.)
    This means the angles between [tex]\sqrt{l(l+1)}[/tex] and [tex]\sqrt{s(s+1)}[/tex]are different in both cases.

    Strange to say, the equation of the spin-orbital interaction is gotten by the "classical" methods.
    Thomas used the classical rotation and classical relativity to get this equation. (This value coincided with the Sommerfeld's equation.)

    So Pauli said in the Novel lecture, as follows,
    Although at first I strongly doubted the correctness of this idea because of its classical-mechanical character, I was finally converted to it by Thomas' calculations on the magnitude of doublet splitting.
    On the other hand, my earlier doubts as well as the cautious expression <<classically non-describable two-valuedness>> experienced a certain verification during later development....
    Last edited: Feb 21, 2010
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