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

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?

• This is a very confusing question. There are 6 electrons in carbon. Also: exchange symmetry is under particle interchange, so if they are orbiting each other, the $(-1)^L$ parity is important since particle exchange flips their relative coordinate; however, if they're orbiting another particle, you can't use that fact. But I can't tell if that is what your are asking. – JEB Oct 23 '18 at 3:44
• @jeb he is talking about $p$ electrons. L of complete subshells is zero... – user153036 Oct 23 '18 at 4:34

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.