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Spin-Orbit coupling.

  1. Dec 3, 2011 #1
    Hey everyone,

    I've been reading about the Spin-Orbit coupling effect in the Hydrogen atom.
    However, there's something I don't quite understand.
    The effect is being explained and calculated like this: we move to the rest frame of the electron, in which the electron has no orbital angular momentum (spin stays, however). We then consider the magnetic field appearing caused by the proton, circulating around the electron in his rest system.
    At a certain point, a relation is derived, connecting the magnetic field of the proton with the orbital angular momentum of the electron in the rest frame. ([itex]\vec{B}=\frac{1}{4\pi\epsilon_{0}}\frac{e}{mc^{2}r^{3}}\vec{L}[/itex])
    From this point onward a new Hamiltonian element is being developed (including some relativistic corrections) as a perturbation of the original Hamiltonian, which was derived in the lab (/proton) rest frame.
    My issue is this:
    What is the source of this effect in the lab system? Since there's no magnetic field in this system, there should be no splitting (based on the SO effect)- yet there has to be one (it cannot depend on system of reference). And more then that - how can we analyze the effect in a certain system, and then just add it to the Hamiltonian of the original system, which was the rest frame of the proton? In the electron's rest frame there isn't any kinetic energy of the electron, for example, so the energy levels are completely different....

    I'd really appreciate an explanation :-)

  2. jcsd
  3. Dec 3, 2011 #2
    In classical mechanics, the Lagrangian of a system of particles transforms as:
    L' = L - \vec{V} \cdot \vec{P} + \frac{1}{2} M V^{2}
    where [itex]\vec{V}[/itex] is the relative velocity of the two frames, [itex]\vec{P} = \sum{m_a \, \vec{v}_a}[/itex], and [itex]M = \sum_{a}{m_a}[/itex] is the total mass of the system.

    On the other hand, the quantity:
    \sum_{a}{\vec{p}'_a \cdot \vec{v}'_a} = \sum_{a}{\vec{p}_a \cdot \vec{v}_a} - 2 \vec{V} \cdot \vec{P} + M V^2
    follows the specified transformation rule. Thus, the Hamiltonian:
    H' \equiv \sum_{a}{\vec{p}'_a \cdot \vec{v}'_a} - L' = H - \vec{V} \cdot \vec{P} + \frac{1}{2} M V^2

    In the CM frame, [itex]\vec{P} = 0[/itex], so the Hamiltonian differs only by the total kinetic energy of the system. Therefore, if we calculate any addition to the Hamiltonian in one frame, it will have the same contribution in any other as well.
  4. Dec 4, 2011 #3
    Ok, I think I got it - thanks!
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