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Spin-Orbit Coupling

  1. Aug 2, 2014 #1
    Can anyone explain why the energy levels separate as a result of spin-orbit coupling?

    Also, what determines the *new* number of "sub-levels" which are obtain after the original energy levels are separated as a result of spin-orbit coupling?

    Please give examples, if you want, using the hydrogen, helium and lithium atoms (all of which are assumed to be in the ground state).
     
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  3. Aug 2, 2014 #2

    WannabeNewton

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    The interaction term ##H_I \sim \vec{S}\cdot \vec{L}## in the Hamiltonian corresponding to the spin-orbit coupling, where ##\vec{L} = -i\hbar \vec{x}\times \vec{\nabla}## is the orbital angular momentum and ##\vec{S}## is the spin angular momentum, breaks the degeneracy in the energy levels because the energy eigenvalues computed from first order time-independent perturbation theory are dependent on the spin eigenvalues. Intuitively this makes perfect sense.

    Consider first a classical analogue of an electron in circular orbit around a stationary proton. In this frame the proton just has a static Coulomb field. In the rest frame of the electron on the other hand the proton is now in orbit around it and generates a magnetic field. The magnetic dipole moment of the electron couples to this magnetic field and as a result the energy of the electron will change depending on the orientation of its magnetic dipole moment relative to the local axis of the magnetic field. In the original frame, where the electron is orbiting around the proton, this splitting of energy is therefore due to the orbital motion of the electron coupling to its spin##^{\dagger}##. The quantum mechanical situation is essentially the same with the caveat that for stationary states of the electron we cannot define its rest frame because these are not momentum eigenstates. Regardless the basic idea is still the same because using the classical argument above we can write down a classical Hamiltonian which includes spin-orbit coupling and then promote it to an operator and add it as a perturbation to the background Hamiltonian of the hydrogen atom for Schrodinger's equation.

    The first order perturbation theory of Schrodinger's equation.

    You will find the aforementioned calculation in quite literally every QM textbook in the section on time-independent perturbation theory. It will usually be listed under the calculation of the fine-structure of Hydrogen. So take your pick of your favorite QM textbook and look there, there's no point in reproducing the calculation here.

    ##^{\dagger}## There is a relativistic correction to the spin-orbit coupling due to Thomas precession which occurs even if the particle in circular orbit is uncharged. It is due to the fact that the rest frame of a particle in circular orbit is really a continuous sequence of momentarily comoving inertial frames of changing orientation relative to the global inertial frame of the proton. It can be shown that this changing orientation of the family of momentarily comoving inertial frames due to the circular orbit leads to a precession of a gyroscope (classically) and spin operator (quantum mechanically) relative to the global inertial frame.
     
  4. Aug 2, 2014 #3

    Dr Transport

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    Adding spin-orbit coupling reduces the symmetry of the system. For example, in a semiconductor, without spin you have a 3 fold degenerate system at the top of the valence band. If you add spin, you get 6 states which are degenerate. If you add in the spin-orbit coupling, you split off two of the states, leaving a 4-fold degenerate (heavy and light holes) and a double degenerate spin-split off set of states separated from the original by the spin-orbit coupling energy. Look in any solid state text book and you'll see a more extensive coverage.
     
  5. Aug 3, 2014 #4
    WannabeNewton, thank you but I have read quite a few "famous" books on quantum mechanics about the topics (Abers, Melissinos, Griffiths etc.) and I did not find them conclusive. Usually I ask a question on physics forums only *after* searching a lot. Some detailed numerical examples in regard to how the degeneracy is lifted (regarding the electron-orbiting-around-proton example) would be nice. In fact this is where the problem seems to always occur in the books I've read - steps are skipped under the assumption that they are "too easy".
     
  6. Aug 3, 2014 #5
    Could you please write an example of calculations of how this happens? (assuming that only first-order perturbation matters). Here, in the detailed steps of the calculation, is where the problem always seems to occur.
     
  7. Aug 3, 2014 #6
    In fact, no offense but this is what I feel like saying when I read all these books of quantum mechanics "dear authors, please throw your intuition out the window and give me the detailed calculations. Please don't skip steps and feel free to assume that I am the dumbest person on Earth and thus you should not skip any step in the calculations, no matter how "obvious" it would seem to you. Thus, for example, please don't write "3+4+5 = 12". Instead, write "3+4+5 =7 + 5 = 12". If this style of writing seems exaggerated to you, do it nevertheless. It will keep you away from skipping steps (or from not explaining things in detail) when it comes to more complicated calculations. By the way, it applies not only to calculations but also to the way in which authors describe or explain concepts, theories etc. From this point of view (i.e. writing in detail), 50 pages are *much* more easy to read than 5 pages."
     
    Last edited: Aug 3, 2014
  8. Aug 3, 2014 #7

    dextercioby

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    The spin orbit coupling appears in the non-relativistic limit of the Dirac equation: it's a term extra to the standard Hamiltonian of the Schroedinger equation whose energy levels, say for the H-atom, are known. With the new Hamiltonian, you can treat this extra term as a stationary perurbation. We can compute with the known theory corrections to the H-atom energy levels, or we say the energy level gets a splitting into 2 new ones.
     
  9. Aug 3, 2014 #8
    Thank you dextercioby, could you please write the calculations of this 2 new levels? (I know, dumb question, I apologize, please feel free to consider that I am dumb but please write the detailed calculations).
     
  10. Aug 3, 2014 #9

    dextercioby

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    We can't spill out the details here, just guide you to the literature: check out Greiner's QM - Special Chapters, Page 312 or better Bransden and Joachain - Physics of Atoms and Molecules.
     
    Last edited: Aug 3, 2014
  11. Aug 3, 2014 #10

    WannabeNewton

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    The calculation in Griffiths looks extremely detailed to me. The calculation is also very straightforward so filling in the missing steps should be easy. You cannot expect authors to hold your hand through derivations at this level anymore. If you're having trouble without hand-holding now then you're going to have a very, very, very, very, very, very, very hard time when it comes to QFT.

    Read the following, it's about as detailed a discourse as you'll find with spin-orbit coupling in time-independent perturbation theory: http://wps.aw.com/wps/media/objects/2529/2590609/sample_chapters/Chapter09.pdf
     
  12. Aug 3, 2014 #11
    WannabeNewton, I am aware that the attitude you describe ("you can't expect authors to hold your hand") is predominant in most (all?) institutions of research/education and I think it is a huge mistake. It is one of the major reasons why websites like physics forums exist. I do expect authors not only to hold my hand but to *be* my hand in such situations, if you will. But thank you very much for the link. It is really of great help to me.
     
    Last edited: Aug 3, 2014
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