Let me start by answering
Big-Daddy said:
ISpin-orbit coupling is a purely relativistic effect?
No, I was a bit clumsy in my phrasing. In constructing a relativistic theory for the electron, which Dirac did, you end up needing spin to construct an equation that is Lorentz-invarient. But spin is not in itself a relativistic phenomenon, but an instrinsic property of the electron.
If you start from the Dirac equation for one electron and assume that relativistic effects are small, you can obtain a series expansion in terms of ##v/c## that you can use as corrective terms (or perturbation) in the Hamiltonian for the hydrogen atom. Spin-orbit coupling is one of those terms. Actually, you get (assuming a fixed nucleus)
$$
\hat{H} = m_e c^2 + \frac{\hat{P}^2}{2m_e} + V(R) - \frac{\hat{P}^4}{8 m_e^3 c^2} + \frac{1}{2 m_e^2 c^2} \frac{1}{R} \frac{d V(R)}{dR} \hat{L} \cdot \hat{S} + \frac{\hbar^2}{8 m_e^2 c^2} \Delta V(R) + \ldots
$$
The terms are in order: (1) mass energy of the electron; (2) kinetic energy of the electron; (3) Coulomb potential; (4) relativistic correction to the momentum; (5) spin-orbit coupling; (6) Darwin term (due to the Compton wavelength of the electron).
Note that the interaction between the spin of the electron and the spin of the nucleus is not included here, but can be added as an additional term.
Big-Daddy said:
If you see the Hamiltonian specified on page 3 here:
http://www.phys.ubbcluj.ro/~vchis/cursuri/cspm/course2.pdf
What would I add to that to make it as exact as we know how to make it?
For molecules (and for atoms with more than one electron), things are more complicated. You have additional terms due to spin-spin interactions between electrons.