Ground state of Phosphorus Problem

[r_tea]

This problem is 1.1b out of "Atomic Physics" by Budker, Kimball, and Demille. There are solutions in the book, but I am confused:

I'm asked to find the ground state configuration of Phosphorus, which is has 3 P-state valence electrons. Following Hund's rule, we want to find a state with largest total spin (S) and largest total angular momentum (L) (I use little l and s to refer to single particle states). So for 3 electrons, picking the largest S state is easy: we get S=3/2 (i.e. |m_s=1/2 \rangle|m_s=1/2 \rangle|m_s=1/2 \rangle).

Since the spin part is chosen to be symmetric, we must construct an antisymmetric spatial wavefunction. We know all electrons are in the P manifold, so our choices of states for each particle are
l_i=1, m_{l_i}=1,0,-1

The authors go on to use a Slater determinant to find a totally anti-symmetric combination of these states for 3 particles, which coincides with the total angular momentum state |L=0, m_L=0\rangle. Great! So then the ground state will be
|S=3/2, m_S= \text{4 possible values}\rangle|L=0, m_L=0\rangle.

But what the authors don't address is the total angular momentum L=2 state, which should also be anti-symmetric (since symmetry alternates between L=3,2,1,0). And also, since L=2>L=0, it should have a lower energy according to Hund's rules, no? That is my confusion.

Thanks.

 

[r_tea]

Actually, I think I solved my own problem. It is true that for two identical spins the total angular momentum states follow the symmetry pattern:
L_\text{total,max} \rightarrow \text {Symmetric}
L_\text{total,max}-1 \rightarrow \text {Antisymmetric}
L_\text{total,max}-2 \rightarrow \text {Symmetric}
\text{etc.}

However, for three particles, I'm pretty sure this statement is not true. For example, in my problem I had the addition of three l=1 particles. I think the states actually look like...
L_\text{tot}=3 \rightarrow \text {Symmetric}
L_\text{tot}=2 \rightarrow \text {No def symmetry}
L_\text{tot}=1 \rightarrow \text {No def symmetry}
L_\text{tot}=0 \rightarrow \text {Antisymmetric}

Hence, only the Ltot=0 case would work for my problem.

Does anyone know any more about this many-particle addition of angular momentum? E.g., maybe the alternating symmetry works for addition of even numbers of identical particles?