# Good /not good quantum numbers in spin-orbit coupling

Given that the Hamiltonian associated with the spin-orbit interaction can be expressed in terms of the total orbital angular momentum and total spin operators as: $$H_{SO} = -\frac{e}{2m_e ^2 c^2} \frac{1}{r} \frac{d \phi}{dr} \mathbf{\hat{L} \cdot \hat{S}}$$ the form of this operator indicates that the spin-orbit interaction couples $$\mathbf{\hat{L}}$$ and $$\mathbf{\hat{S}}$$ so that they no longer have fixed $$z$$ components. Why are $$\mathbf{\hat{L_z}}$$ and $$\mathbf{\hat{S_z}}$$ no longer conserved quantities, and why are $$M_L$$ and $$M_s$$ not good quantum numbers in the presence of spin-orbit coupling?

Simply show that the Hamiltonian does not commute with $$\mathbf L$$ or $$\mathbf S$$, i.e.,* $$[\mathbf L \cdot \mathbf S, \mathbf L] \neq 0 \,\,\,\text{and} \,\,\, [\mathbf L \cdot \mathbf S, \mathbf S] \neq 0.$$
Hence $$m_L$$ and $$m_S$$ are not good quantum numbers, because the Hamiltonian isn't diagonalizable in the $$\mathbf L$$ or $$\mathbf S$$ basis.
However, $$\mathbf J = \mathbf L + \mathbf S$$ does commute, and $$m_J = m_L + m_S$$ is a good qunatum number.
The interaction means that there is a torque between the orbital and the spin moments. The $$z$$ components are not longer constants of motion but are precessing.