# Why does Hamiltonian no longer commutes with $\mathbf L$ and $\mathbf S$ in presence of spin orbit coupling?

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.

• Which Hamiltonian? – ZeroTheHero Apr 6 at 4:55
• Hamiltonian for electron orbiting around nucleus in magnetic field of proton. – user256265 Apr 6 at 5:07
• 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... – ZeroTheHero Apr 6 at 5:16
• 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. – user256265 Apr 6 at 5:25

\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 $$\mathbf J =\mathbf L+\mathbf S$$.