$L$-$S$ coupling with more than two electrons in the last shell

Question

Is there a easy way to calculate all the terms for complicate configurations with more than two electrons in the last shell?

If I try to find the terms in $L$-$S$ coupling $[Ar]3d^3$ or $[Ar]3d^5$ looks very hard. When I have just two electrons in the last level there are rules that make easier to compute it and respect Pauli, for example, I can just do $S=|s_1-s_2|, ..., s_1+s_2$ and $L=|l_1-l_2|,...,l_1+l_2$ and the only terms that satisfy Pauli are the ones with $L+S=$even number

In the first example $[Ar]3d^3$ I would try to do it in this way: First calculate the terms for the first two electrons and find

$$ 3d^2 \to S=0,1 \quad L=0,1,2,3,4 $$

Applying Pauli

$$ S=0, L=0 \to \quad ^1S \quad \quad \quad \quad S=1, L=0 \to \quad Impossible $$

$$ S=0, L=1 \to \quad Impossible \quad \quad \quad \quad S=1, L=1 \to \quad ^3P $$

$$ S=0, L=2 \to \quad ^1D \quad \quad \quad \quad S=1, L=2 \to \quad Impossible $$

$$ S=0, L=3 \to \quad Impossible \quad \quad \quad \quad S=1, L=3 \to \quad ^3F $$

$$ S=0, L=4 \to \quad ^1G \quad \quad \quad \quad S=1, L=4 \to \quad Impossible $$

So I have this terms $^1S, ^3P, ^1D, ^3F, ^1G$. Now I should do the same for the third electron with each of this term $^1S + 3d \quad \to \quad L_1=0, S_1=0, l_2=2, s_2=1/2 \quad \to \quad L=2, S=1/2$ $^3P + 3d \quad \to \quad L_1=1, S_1=1, l_2=2, s_2=1/2 \quad \to \quad L=1,2,3; S=1/2,3/2$ $^1D + 3d \quad \to \quad L_1=2, S_1=0, l_2=2, s_2=1/2 \quad \to \quad L=0,1,2,3,4; S=1/2$ $^3F + 3d \quad \to \quad L_1=3, S_1=1, l_2=2, s_2=1/2 \quad \to \quad L=1,2,3,4,5; S=1/2,3/2$ $^1G + 3d \quad \to \quad L_1=4, S_1=0, l_2=2, s_2=1/2 \quad \to \quad L=2,3,4,5,6; S=1/2$

But now, how can I identify the terms that I have to delete to respect Pauli?

Answer 1

This will not be very satisfactory answer but yes there is way, which requires the use of Young diagrams or partitions to label representations of the permutation group.

When you have 2 particles, the $S$ or $L$ states can be only symmetric or antisymmetric. It's then a matter of combining the proper states.

When you have 3 or more particles, you can have states with mixed symmetry. For instance, with three spin-1/2 particles, you have $S=3/2$ states with are fully symmetric: the permutation symmetry is labelled by the partition $\{3\}$ In addition, there are two sets of $S=1/2$ states with mixed symmetry the permutation symmetry is labelled by the partition $\{2,1\}$. It's no so easy to draw Young diagrams with this editor so I will dispense with that part.

In an obvious notation, the two different $S=1/2, M_s=1/2$ states are \begin{align} \vert\textstyle\frac{1}{2}\frac{1}{2}\rangle_1&= \frac{1}{\sqrt{2}}\left(\vert+\rangle_1\vert-\rangle_2\vert+\rangle_3- \vert-\rangle_1\vert+\rangle_2\vert+\rangle_3\right), \tag{1} \\ \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_2&= -\frac{2}{\sqrt{6}}\vert +\rangle_1\vert+\rangle_2\vert-\rangle_3 +\frac{1}{\sqrt{6}}\vert +\rangle_1\vert-\rangle_2\vert+\rangle_3 +\frac{1}{\sqrt{6}}\vert -\rangle_1\vert+\rangle_2\vert+\rangle_3 \, .\tag{2} \end{align} You will notice that the state in (1) is antisymmetric w/r to permutation of spins 1 and 2, but if you permute spins 1 and 3, or spins 2 and 3, it does not come back to itself: rather it goes to a linear combination of the state (1) and (2). The state of (2) does not have any obvious permutation symmetry but again if you permute any two spins, you will get a linear combination of (1) and (2).

The same occurs when you combine multiple states with the same $\ell$: some of the possible total $L$ values will be repeated, and for these values of $L$ the states will have mixed symmetry. To best understand the permutation symmetries of your complete set of $3$ $\ell=2$ states requires plethysms and Schur functions (wiki is not helpful for either), or a really painful hand calculation. After some work I did find the following:

• The set of fully symmetric states (partition $\{3,0\}$) contains states with $L=6,4,3,2,0$,
• The set of partially symmetric states (partition $\{2,1\}$) contains states with $L=5,4,3,2,2,1$ ($L=2$ is repeated twice),
• The set of fully antisymmetric states (partition $\{1,1,1\}$) contains states with $L=3$ and $L=1$.

Now there's a rule which comes up when studying the permutation group, and it is that to construct a fully symmetric state you must combine states with the same type of permutation symmetry, while to construct fully antisymmetric states you must combine states with conjugate permutation symmetries. I don't want to get into the definition of conjugate symmetries but suffice it to say that, to construct fully antisymmetric states you would need to combine:

1. the possible values of $L$ for partition $\{1,1,1\}$ with the possible values of $S$ for partition $\{3\}$: thus here you get values of $J$ in $L\otimes S=3\otimes 3/2$ and $1\otimes 3/2$.
2. the possible values of $L$ for partition $\{2,1\}$ with the possible values of $S$ for partition $\{2,1\}$: thus here you get two values of $J$ (one for each set of spin states with $S=1/2$) in $5\otimes\frac{1}{2}, 4\otimes\frac{1}{2}, 3\otimes\frac{1}{2}$, a total of 4 sets $2\otimes\frac{1}{2}$ since $L=2$ appear twice, and $S=1/2$ appear twice, and finally a pair for $1\otimes\frac{1}{2}$.
3. Because there are not $S$ states labelled by the partition $\{1,1,1\}$, it is not possible to construct antisymmetric states with value of $L$ for the partition $\{3\}$. Thus in particular no states with $J=13/2$ can be fully antisymmetric; the largest $J$ value will be 11/2, and will occur in combining $L=5$ with $S=1/2$ states.