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I Why is spin-orbit coupling neglected in Helium atom?

  1. Mar 9, 2017 #1
    I was looking at grotrian diagrams for helium and I see that there is no splitting of energy levels due to spin-orbit coupling. In my book it is said that spin-orbit coupling in helium is small and can be neglected but no further explanation is given. At the same time we do spin-spin coupling between electrons in He which obviously is not negligible.

    If there is no spin-orbit coupling then there is no fine structure in helium spectral lines like in alkali metals, right? So why is that we can neglect spin-orbit coupling in He?
     
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  3. Mar 9, 2017 #2

    blue_leaf77

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    Spin-orbit coupling is the first-order portion of relativistic effect in atoms or molecules. For light atoms such as He, the nuclear attraction is not so strong to cause the electrons to move in relativistic speed, hence the relativistic effect is small and in turn the spin-orbit coupling is also small.
    In contrast, spin-spin coupling is independent of relativistic motion because it's solely governed by Pauli exclusion principle - any system with more than one identical particles will show up to some extent the effect of Pauli principle.
     
  4. Mar 9, 2017 #3
    But we have spin-orbit interaction taken into account for hydrogen atom fine structure and hydrogen is even lighter than helium.

    Here http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html they say:

     
  5. Mar 9, 2017 #4

    blue_leaf77

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    It's not like the spin-orbit effect He completely vanishes, it's there but very small (still larger than hydrogen I think). The good thing with hydrogen is that the spin-orbit coupling can be analytically incorporated perturbatively as well as exactly through the Dirac equation. The same thing in any other atom with more than one electron is almost imposible to do. That's why in most introductory books on atoms and molecules the author never bothers to account for spin-orbit effect analytically.
     
  6. Mar 10, 2017 #5
    So to sum it up, if there is no spin-orbit coupling (or we neglect it) then there is no fine structure in He?
    By fine structure I mean spectral lines split in more than one component (with no external magnetic field present) if we look at it with high enough resolution.

    For example on this Grotrian diagram for He https://www.physics.byu.edu/faculty/christensen/Physics 428/FTI/Helium Grotrian Diagram'.htm
    there is no vertical separation between terms 3D0,1,2 on the right (triplet) side of diagram. Does it mean that for example any transition from 33D0,1,2 to say 23P0,1,2 gives just one line? Like all D terms have same energy no matter value of J.

    Please correct me if my understanding is wrong, here is my logic: if there is spin-orbit coupling then there is split of energy levels which gives line splitting (small but observable). So if there is no spin-orbit coupling there is no fine structure?

    However on this diagram they included J value in term symbol (low right corner) which means that they did perform LS spin-orbit coupling but still they didn't draw separation of energy levels. Confused...
     
  7. Mar 10, 2017 #6

    blue_leaf77

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    Yes.
    Yes.
    Yes.
    If they shall include spin-orbit, then there is no reason to exclude nuclear spin. In turn, if they do include nuclear spin, they should also take the radio wave level Lamb's shift. And if they include Lamb's shift, the gravitational effect should not be left behind.
    It will be an endless of an addition of correction terms. The reason why that diagram does not include spin-orbit is just because it's too small to be of significant importance for the topic being discussed.
    Take a look at this database from NIST http://physics.nist.gov/PhysRefData/Handbook/Tables/heliumtable5.htm and see how small the splitting for 3P0,1,2 levels is compared to the difference between this level with 1S.
     
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