Total Orbital Angular Momentum and its relation to $M_L$ & Symmetry

Question

I am currently working through the physics governing how electrons arrange themselves within an atom into orbitals, and I have two related questions. Note that all the atoms I talk about are concerning its ground states.

The first is: how is total orbital angular momentum, $L$ (capital $L$) related to $M_L$? I know that for two electrons (say Carbon), $L=l_1+l_2$, ..., $|l_1-l_2|$; so $L$ can be $2$, $1$, $0$. Is this because if $l=1$ (say for valence of Carbon), then $m_l$ can be $1$, $0$, $-1$? Then why can $L$ not be $-2$ or $-1$? And, what happens for Nitrogen where there are three valence electrons? Is $L=l_1+l_2+l_3$, ..., $|l_1-l_2-l_3|$, giving $3$, $2$, or $1$? I know that $L=0$ is a possibility, so I am confused there as well.

My next question is concerning the symmetry arguments of identical particles. I know that the total wavefunction must be anti-symmetric because electrons are fermions. Going back to Carbon, when determining $L$ values, we first look at $L=2$ (because of Hund's second rule). However, the spin wavefunction is symmetric according to Hund's first rule, so we have to find an anti-symmetric spatial wavefunction. I read that we can rule out $L=2$ because $L=2$ is symmetric, and $L=1$ is anti-symmetric so that is the correct $L$. I do not understand how we can prove that $L=2$ is symmetric. I've read somewhere that $L=2$ is symmetric because the "top of the ladder" i.e. the $|2,2\rangle$ state is symmetric so the rest must be symmetric. Please explain why that is the case. Following that, why must we have 6 symmetric states and 3 antisymmetric states?

Answer 1

One way to think of the $L$ is that is it related to the eigenvalue of the square of the total angular momentum operator $\hat L\cdot \hat L$. Then $$ \hat L\cdot \hat L \, \psi_{n\ell m}(r,\theta,\phi)=\hbar^2 L(L+1)\psi_{n\ell m}(r,\theta,\phi) $$ Since the length of $\vec L\cdot \vec L$ is necessarily non-negative, the eigenvalue of $\hat L\cdot \hat L$ should be non-negative, which implies for integer $L$ that $L\ge 0$ and thus eliminates the possibility of $L=-1$.

Now $\hat L\cdot \hat L $ commutes with the projection $\hat L_z$ and the states $\psi_{n\ell m}(r,\theta,\phi)$ solutions to the time-independent Schrodinger equation are chosen to have fixed value of $L$. Since $\sqrt{L(L+1)}$ is the "length" of the angular momentum vector, the projection $M_L$ cannot be greater than the length and indeed $L$ by itself is also the largest possible $M_L$ value in a set of eigenstates of $\hat L\cdot \hat L $ with eigenvalue $L(L+1)$.

Finally, if you have three electrons, you need to first combine the first two of them, and then combine the result of this with the last one. Thus, combining three particles with $\ell=1$ will give, in the first step, $L=0,1,2$, and combining these with the last $\ell=1$ will give $$ L_{tot}=(L=0)\times (\ell=1)+ (L=1)\times (\ell=1)+(L=1)\times (\ell=1)= 1+0+1+2+1+2+3\, . $$ Note that the final list of $L_{tot}$ contains $1$ three times and $2$ twice so some values of $L_{tot}$ can be repeated (this never happens with only two angular momenta).

Now the symmetry question. To see that the $L=2$ states are necessarily symmetric start with the $L=2,M_L=2$ state which is necessarily $$ \vert L=2,M_L=2\rangle = \vert \ell_1=1,m_1=1\rangle_1 \vert \ell_2=1,m_2=1\rangle_2\, , \tag{1} $$ and is obviously symmetric w/r to interchange of particle index $1$ and $2$. $\vert L=2,M_L=2\rangle$ must be as (1) (up to an overall phase) because, in the set of states $\vert \ell_1=1,m_1\rangle_1\vert \ell_2=1,m_2\rangle_2 $ there is only one way of constructing an eigenstate of $L_z=L_z^{(1)}+L_z^{(2)}$ with eigenvalue $M=2$, and it is by combining the $m_1=1$ and $m_2=1$ states.

To construct the other states with $L=2$ one should act on $\vert L=2,M_L=2\rangle$ with the other angular momentum operators, which are of the form $L_k= L_k^{(1)}+L_k^{(2)}$. Because $L_k$ is clearly symmetric under permutation of particle index $1$ and $2$, the action of $L_k$ does not change the symmetry properties of states under permutation, so all states with $L=2$ will be symmetric under permutation since the one state $\vert L=2,M=2\rangle$ of (1) is clearly symmetric under permutation.

One can show that the $L=1, M=1$ state is of the form $$ \vert L=1,M=1\rangle=\frac{1}{\sqrt{2}} \left(\vert \ell=1,m_1=1\rangle_1\vert \ell=2,m_2=0\rangle_2 -\vert \ell=1,m_1=0\rangle_1\vert \ell=2,m_2=1\rangle_2\right)\, . \tag{2} $$ One way to see this is by using Clebsch-Gordan tables. Clearly, (2) is antisymmetric. Likewise, one easily shows that the $L=0$ state is fully symmetric. See also this post.