If a many-electron Hamiltonian $H$ commutes with ${\vec L}^2, {\vec S}^2, {\vec J}^2$, and $J_z$ but not $L_z$ and $S_z$, the energy eigenstates are designated by $|LSJM_J\rangle$. Since there is no external field (no breakdown of rotational symmetry) that distinguishes different values of $M_J$, the energy cannot depend on $M_J$ and therefore, such a state, for a given value of $J$ will have a $(2J+1)$-fold degeneracy.

It is clear that the energy eigenstates $|LSJM_J\rangle$ are not labeled by $M_L, M_S$ because $H$ does not commute with $L_z, S_z$. But is there any chance that the energy depends on $M_L$ and/or $M_S$? Does the absence of an external field (rotational symmetry) also exclude the possibility of energy depending on $M_L$ and/or $M_S$?


1 Answer 1


$M_L$ and $M_S$ are actually still good quantum numbers for stretched states even for non-zero external fields.

You have already established that the energy does not depend on $M_J$. The number of states is "conserved" as you break symmetries. Each state in the broken-LS coupling régime is "adiabatically" connected to the zero-field LS coupling régime. Meaning you can connect an $|M_L, M_S\rangle$ state at $|\mathbf{B}|\neq 0$ to an $|M_J\rangle$ state as $|\mathbf{B}|\rightarrow 0$.
So why would the energy depend on $M_S$ or $M_L$, if it does not depend on $M_J$?

Physically, in order for the energy to depend on the quantum number $M_L$ or $M_S$, you need an interaction that couples to the orbital ($L$) or spin ($S$) angular momentum of the electron(s) respectively. This could, for instance, be an electric field (Stark effect) or a magnetic field (Zeeman effect).
So if you have no external field of any kind, and therefore no interaction that would change the energy of the electron(s), why would there be such a dependence in the energy?

From the "symmetry" perspective, notice that it is slightly less trivial that simple rotational symmetry.
An external electric field along $z$ breaks rotational symmetry, but the Stark effect causes the energy splitting to go as $|m_J|$. To break the symmetry between $\pm m_J$, you need to also break time-reversal symmetry. With, for instance, an external magnetic field: the Zeeman effect, in fact, results in the energy splitting to go as $\propto m_J$.


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