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?


2 Answers 2


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.



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.

Intuitively, there is no reason that when considering the spin and angular momentum of the system to expect that they will stay the same. But we can be sure that the total is conserved throughout the system.

*Sorry, I'm lazy and I don't want to type up all the tensor products


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