Labeling the eigenstates of the many electron Hamiltonian

Question

I'm taking a course on atomic and molecular physics and there is a chapter about finding suitable quantum numbers that label the eigenstates of a given Hamiltonian. The lecture notes say that it basically comes down to finding a complete set of operators that commute with the Hamiltonian and with each other. The eigenstates can then be labeled using the eigenvalues of these operators. This makes sense to me. However it brings up a few questions:

• For the hydrogen atom (and ignoring spin-orbit coupling etc), why are the eigenstates labeled by $n$, $l$ and $m_l$ instead of $n$, $l$, $m_l$ and $m_s$
• The course notes contain the following sentence: "The eigenstates of the many-electron system can thus be labeled by $L$ and $S$". With $L$ the total azimuthal quantum number and $S$ the total spin quantum number. But $\hat{L}_z$ and $\hat{S}_z$ also commute with the Hamiltonian in addition to $\hat{L}^2$ and $\hat{S}^2$. So why aren't $M_L$ and $M_S$ necessary as well in order to uniquely label the eigenstates? (Again, this part of the lecture notes ignores everything related to spin-orbit coupling)

Answer 1

I think they are skipping over a few things. Usually spin in hydrogen atom orbitals is semi-ignored and you just say that there are 2 electrons in each state, and each state is labeled $n$, $l$, and $m_l$.

The next question is a little more complicated. Indeed $H$, $L^2$, $S^2$, $L_z$, and $S_z$ all commute. However a state that has a definite value for these quantities does not necessarily have a well defined total angular momentum defined $J = L + S$. This has the operator, \begin{align*} J^2 &= (L + S)^2, \\ J^2 &= L^2 + S^2 + 2L\cdot S. \end{align*} Note that $J^2$ does not commute with $L_z$ and $S_z$. So instead we choose the set of quantum numbers, $J^2$, $L^2$, $S^2$, and $J_z$. It is usually prefered to consider a state with well defined angular momentum than one with well defined $L_z$ and $S_z$.