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What does a general PE have to to with a S-O perturbation? Stumped!

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that for a general potential energy [itex]V(r)[/itex] that the form of the spin-orbit Hamiltonian becomes

    [itex]\hat{H}_{S-O}=\frac{1}{2m_{e}^{2}c^{2}|\hat{\textbf{r}}|}\frac{d\hat{V}}{dr} \hat{\textbf{L}}\cdot\hat{\textbf{S}}[/itex]

    Suggestion: Start with [itex]\textbf{B} = -( \textbf{v} /c) \times \textbf{E}[/itex]

    2. Relevant equations
    The normal form of the spin-orbit Hamiltonian is

    [itex]\hat{H}_{S-O}=\frac{Ze^{2}}{2m_{e}^{2}c^{2}|\hat{\textbf{r}}|^{3}}\hat{\textbf{L}}\cdot\hat{\textbf{S}}[/itex]


    3. The attempt at a solution
    I'm sorry but I don't have much. What does the hint have to do with anything?
    I can see that if you plug in [itex]V(r) = \frac{-Ze^{2}}{r}[/itex] you get back the expression for the regular spin-orbit Hamiltonian. The book derives the regular Hamiltonian by setting

    [itex]\textbf{B} = \frac{-Ze \textbf{v} \times \textbf{r}}{cr^{3}}[/itex]

    [itex]\textbf{μ} = \frac{ge}{2m_{e}c} \textbf{S}[/itex]

    So,

    [itex]\hat{H}_{S-O} = -\textbf{μ} \cdot \textbf{B}[/itex]

    And then the author handwaves the factor of two that comes from the Thomas precession by saying we made a relativistic error "somewhere".

    I don't understand from the potential energy will arise from the hint. If I plug in the relativistic B field, I'll get a triple product involving magnetic moment, velocity and the electric field. What does this have to do with a general potential? I don't see where the insertion of the coulomb potential occurred during the derivation. Something to do with the B field? I "see" it in there.
     
  2. jcsd
  3. Jun 22, 2013 #2

    TSny

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    You need to relate the potential energy V(r) to the electric field E. The potential energy is V(r) = -e##\phi(r)## where ##\phi(r)## is the electric potential. How is E related to ##\phi##?
     
    Last edited: Jun 22, 2013
  4. Jun 22, 2013 #3
    Oh wow, that was "a could've had a V8" moment I was having here.

    [itex]\vec{\textbf{E}} = -∇\Phi = \frac{1}{e}\frac{dV}{dr}[/itex]

    I thought of this but I threw it out because "but this would only hold if all the potential energy is in the electric field!"

    I just realized all the potential energy is in the electric field in a simple picture of hydrogen. For some reason I was thinking I need to put in some magnetic effects or something.

    ...Is it? I'm sort of confused now. Why is all the potential energy due to the electric field? Does it have something to do with the fact that magnetic fields and electric field are "the same thing" in gaussian units? Isn't spin-orbit coupling due to the magnetic field(s) an electron sets up? Is there an effective B field set up by intrinsic spin?

    I think I got the problem down, but now I'm confused about the physics.
     
  5. Jun 23, 2013 #4

    TSny

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    There is a general result of electromagnetism (and relativity) that if there exists only an electric field E in one frame of reference, then there will generally exist both an electric field and a magnetic field B in another frame moving relatively to the first frame. B is given approximately by B = -v/c x E where v is the velocity of the "moving" frame relative to the first frame.

    In the frame of reference at rest relative to the nucleus, the only significant field is assumed to be a central electric field associated with V(r). But, in a reference frame moving with one of the orbiting electrons, there is both an electric field and a magnetic field. It's this magnetic field that exists in the frame of the electron that couples with the spin to form the spin-orbit interaction.
     
  6. Jun 23, 2013 #5
    Thanks so much, it's all coming together now!
     
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