- #1
wotanub
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Homework Statement
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]
Homework 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]
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.