Well, I am currently using a pretty old book by H.E White "Atomic Spectra", and he defined spin orbit interaction energy as the product of the resultant frequency and the projection of spin angular momentum on the orbital angular momentum. My question is why? On what basis did he defined the spin orbit interaction energy as such.
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asked Aug 24, 2015 at 15:49
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$\begingroup$ I don't know the answer to that, Roshan. But there's a bit about this on hyperphysics, which I think is good for delivering understanding. $\endgroup$– John DuffieldCommented Aug 24, 2015 at 15:58
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There's a semiclassical way of deriving the Hamiltonian term corresponding to the spin orbit interaction in an atom, but I don't know if this is what you're looking for (the correct way would be using the relativistic correction included in Dirac's equation), anyhow:
Consider the classical picture of an atom orbiting the nucleus, now in the electron's frame the nucleus of course appears to be rotating the electron, this orbiting leads to a magnetic field equal to \begin{align} \mathbf{B} = \frac{E\times \mathbf{v}}{c^2} \tag{1} \end{align} which you can obtain by doing the Lorentz transformations of the fields in SR (I showed you this derivation recently here). Now the $E$ field felt by the electron can be written as the gradient of its potential energy, $\mathbf{E} = -\nabla V(\mathbf{r})$ or in polar coordinates: $-\frac{\mathbf{r}}{r}\frac{dV(\mathbf{r})}{dr}: (*)$ The spin orbit term results from the interaction of this $\mathbf{B}$ field with the electron's spin: \begin{align} H = -(1/2) \mathbf{m}\cdot \mathbf{B} \tag{2} \end{align} Now by substituting $(*)$ in $(1)$ you end up with a $\mathbf{r}\times \mathbf{v}$ term which you can express as the orbital anglular momentum $\mathbf{L},$ and the magnetic moment $\mathbf{m}$ in $(2)$ is equal to: $$ \mathbf{m} = \frac{ge\hbar}{2m_e}\mathbf{S} $$ with $g$ the Landé factor and the factor $1/2$ the Thomas factor. Inserting everything back into $(2)$ we obtain: \begin{equation} H = \frac{e\hbar^2}{2m_e c^2 r}\frac{dV(\mathbf{r})}{dr}\mathbf{S}\cdot \mathbf{L} \tag{3} \end{equation} In case your model is hydrogen-like then you can substitute the $1/r\frac{dV}{dr}$ term in $(3)$ by its corresponding Coulomb potential. If you plan to look at a more recent book covering such topics, Stephen Blundell's book comes recommended.
