Confusion on quantum numbers

Question

So, I've known for a long time the famous quantum numbers $n, l, m, s$ and I thought these were all of the quantum numbers, and then when applying the Schrödinger equation to orbital angular momentum and getting the spherical harmonics, with their numbers $l$ and $m$, I thought, okay here they are that's all. But recently, I've been taught that angular momentum is not only composed of the orbital angular momentum, but also the intrinsic angular momentum, the spin, $\vec J = \vec L \otimes 1\!\!1 + 1\!\!1 \otimes \vec S$. And with this, I'm introduced too to the quantum numbers $j$ and another $m$, which I think can be specified by $m_j$, and also $s$ and $m_s$.

I'm confused by so many $m$'s. Are the quantum nubers that I initially wrote the only ones and the other ones can be derived from these ones? Are all $m, m_j, m_l, m_s$ different between them or there's one that englobes them all? Is the $s$ that you get from $\vec S$ the same $s$ as my initial one? What's the physical meaning of all these quantum numbers? Are there any other quantum numbers that I haven't encountered yet?

Answer 1

Hydrogen atom with and without Coulomb potential

For H-atom with Coulomb potential and no perturbations such as spin-orbit interaction, relativistic correction etc, $n,l,m_l,m_s$ are good (conserved) quantum numbers because the operators $\textbf{L}^2,L_z,\textbf{S}^2,S_z$ commute with the Hamiltonian, and hence can be used to label the states. $n,l,m_l,s,m_s$ labels the eigenvalues of $H,\textbf{L}^2,L_z,\textbf{S}^2, S_z$ respectively. Since $s=1/2$ for a spin-1/2 particle, $s$ is omitted from the set $n,l,m_l,s,m_s$.

With spin-orbit interaction, it is the operators $H,\textbf{L}^2,\textbf{S}^2, \textbf{J}^2,J_z$ commute with the hamiltonian where $\textbf{J}=\textbf{L}+\textbf{S}$. Hence, $n,l,s,j,m_j$ are the complete set of good quantum numbers ($s$ is redundant because $s=1/2$).

So in a nutshell, $m_l$ is the eigenvalue of $L_z/\hbar$, $m_s$ is that of $S_z/\hbar$ and $m_j$ is that of $J_z/\hbar$.

General theory of quantum angular momenta

Any angular momentum $\textbf{J}$ is defined as a set of three hermitian operators $\textbf{J}\equiv (J_1,J_2,J_3)$ which satisfy the following commutation relations $$[J_i,J_j]=i\hslash\epsilon_{ijk}J_k\hspace{0.5cm}\text{i,j=1,2,3}.$$ The symbol $\textbf{J}$ is introduced here for a general angular momentum vector: it can stand for $\textbf{L}$, $\textbf{S}$, $\sum \textbf{L}$, $\sum \textbf{S}$ or $\textbf{L}+\textbf{S}$.

Answer 2

Brief answer.

for each electron we assign $n$, $l$, $s$

e.g. 2p electron $n=2$, $l=1$, $s={1 \over 2}$

In the presence of a magnetic or electric field we need to think about $m_l$ and $m_s$ the projections of $l$ and $s$ in the direction of the magnetic or electric field

e.g. 2p electron $n=2$, $l=1$, $m_l = +1,0,-1$, $s={1 \over 2}$, $m_s=+{1 \over 2}, -{1 \over 2}$ - there are several possible $m_l$ and $m_s$ values.

If we consider all the electrons of an atom then we need to combine the individiual $l$ contributions of each electron to give $L$ and combine all the $s$ to give $S$... The total orbital ang. momentum and total spin ang. momentum combine to give the total ang. momentum $J$. - and there are, of course, $m_L$, $m_S$ and $m_J$ values... For an example see this calculation of the term states for carbon in the ground state

Note this is not as bad as it appears as each closed subshell contributes overall zero to $L$, $S$ and $J$. eg for carbon 1s$^2$, 2s$^2$, 2p$^2$ we only need to consider the two 2p electrons.