QM Angular momentum addition with known projection ($m$ quantum number)

Question

Certainly a lot of questions have been asked about this topic, but nowhere I found a concrete answer to my question. I guess there is something very basic about this I don't quite understand yet.

When we add two momenta, say $L$ and $S$, the resulting total angular momentum can range anywhere from $|L-S|$ to $L+S$ in integer steps. I wonder if this is only the case when $m_L$ and $m_S$ are unknown. I think about the vector model where L and S "precess" on a cone around the z direction. When I add them, say $L=S=1$ it makes (in my understanding) an important difference whether eg $m_L=m_S=+1$ or $m_L=-m_S=1$. In the latter case I don't think $J=L+S$ can take on the value $+2$ because L and S point in "different directions", though by the rules of angular momentum coupling $J=2$ is possible as long as $m_J=0=m_L+m_S$. This question is also inspired by one of Hund's rules and the statement that closed subshells of atoms don't contribute any angular momentum. If we have a closed subshell, for example 2D, and add all the projections $m_L$ of electrons in this subshells together, we get zero - good. But why is the total angular momentum zero as well? I mean it could be anything as long as it's projection is zero right? (by general rules for coupling). Basically what I want to know is whether we can determine the value of $J=L+S$ completely if we know $m_L$ and $m_S$ beforhand, for example: What is the total angular momentum of 2 P-electrons (each having l=1) with one of them having $m_l=1$ and the other one having $m_l=-1$?

Answer 1

Your specific question of the total angular momentum of two such $2p$ states can be answered by looking at the Clebsch-Gordon coefficients. That is with the Condon and Shortley phase convention, the Clebsch-Gordon coefficients are $C^{20}_{111-1} = \frac{1}{\sqrt{6}}$, $C^{10}_{111-1} = \frac{1}{\sqrt{2}}$, $C^{00}_{111-1} = \frac{1}{\sqrt{3}}$, so that $$ |\ell_1=1~ m_1=1 ~\ell_2=1~ m_2=1\rangle = \frac{1}{\sqrt{6}}|J=2~ M=0 ~\ell_1=1~\ell_2=1\rangle + \frac{1}{\sqrt{2}}|J=1 ~M=0 ~\ell_1=1~\ell_2=1\rangle+\frac{1}{\sqrt{3}}|J=0~ M=0 ~\ell_1=1~\ell_2=1\rangle$$ and there is an amplitude to be in each of those $J$ states.

Your question about Hund's rules giving a zero angular momentum is related but somewhat different. These require the single particle orbital radial states to be the same for the closed shell so that rotating an orbital gives a linear combination of the others. This along with the required antisymmety makes the angular momentum zero.

A trivial example of this would be a spin half up and a spin half down. Without antisymmetry, you would have $$|\uparrow \downarrow\rangle = \frac{1}{2}\left [ |\uparrow \downarrow\rangle + |\downarrow \uparrow\rangle \right ] +\frac{1}{2}\left [|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle \right ].$$ The first term is the $S=1$, the second the $S=0$. So you have both values. If it is required to be antisymmectric, only the second $S=0$ term survives.

A way to see that the closed shell gives zero angular momentum, is to calculate L_x, L_y, and L_z. As you say, L_z is zero since the $m$ values add to zero. If you write $L_x = \sum_i L_{ix}$, that is the sum of the single particle angular momenta in the $x$ direction, and then remember that $L_x = (L^++L^-)/2$, then, for a closed shell, every raising operator either gives zero if it acts on an $m=\ell$ state, or it makes an orbital have the same angular momentum as another. In the Slater determinant wave function, this gives zero since to particles are in the same orbital. The lowering operator acts similarly, it gives zero if $m=-\ell$ or it doubly occupies an orbital. So $L_x=0$. $L_y=-i(L^+-L^-)$, so again it is zero. Since $L_x=L_y=L_z=0$, $L^2=0$.