Is this something due to change in Hamiltonian due to relativistic correction. i.e. $H$ being $$(\frac{e^2}{4\pi {\epsilon _0}}) \mathbf{S \cdot L}$$ instead of $$(\frac{e^2}{8\pi {\epsilon _0}}) \mathbf{S \cdot L}$$ like in Thomas Precession.

Ref :-Page 278 DJ Griffiths Introduction To Quantum Mechanics.

  • $\begingroup$ Which Hamiltonian? $\endgroup$ – ZeroTheHero Apr 6 at 4:55
  • $\begingroup$ Hamiltonian for electron orbiting around nucleus in magnetic field of proton. $\endgroup$ – user256265 Apr 6 at 5:07
  • $\begingroup$ Why would you want (or expect) the perturbation to commute? After all, it’s a perturbation and if it did commute you would already have the eigenstates so there wouldn’t be any fun in finding the eigenvalues... $\endgroup$ – ZeroTheHero Apr 6 at 5:16
  • $\begingroup$ oh !! electron was in non-perturbed state if there was no influence of proton but by mag field the perturbation happens and that's why hamiltonian is no longer same and no longer commuting with L ? I am sorry but it seems to me I am having hard time correlating everything here in the book. $\endgroup$ – user256265 Apr 6 at 5:25

Why don't they commute?

You can work out the maths. It does not have anything to do with the factor but the form.

$$ \begin{align} [\mathbf L \cdot \mathbf S, \mathbf L_x] & = [L_xS_x+L_yS_y +L_zS_z, L_x] \\ & = S_x[L_x,L_x] + S_y[L_y,L_x] + S_z[L_z,L_x] \\ & = 0 - S_y (i \hbar L_z) + S_z (i \hbar) L_y \\ & = i\hbar (\mathbf L \times \mathbf S)_x\end{align}$$

Similarly, you can work out other two components. Clearly, Hamiltonian does not commute with $\mathbf L$. The same kind of calculation goes with $\mathbf S$ with $\mathbf L \rightarrow \mathbf S$.

The spin-orbit interaction implies the presence of a torque between the orbital and the spin moments. The $z$ components cannot be constants of motion since they are precessing. What we hence need is the total angular momentum $\mathbf J =\mathbf L+\mathbf S$.

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