# Singlet and Triplet Terms misunderstanding

1. Nov 6, 2014

### Eats Dirt

I am a little unclear on why in atomic physics that the total orbital momentum terms have to be specifically singlet or triplet states, for example

G -Singlet
F -Triplet
D -Singlet
P -Triplet
S -Singlet

I think it has to do with the pauli exclusion principle but don't really understand the process.

If someone could walk me through why that would be great! (edit: 2 electron case)

2. Nov 6, 2014

### Einj

It is simply due to the fact that you are dealing with 2 electrons, i.e. two spin-1/2 particles. When you consider the two electrons system as a whole you are building a new Hilbert space out of the tensor product of the two Hilbert spaces of the single electrons. Let's focus on the spin only. What you are doing is building:
$$|j,_z\rangle=|1/2,\pm1/2\rangle\otimes|1/2,\pm1/2\rangle,$$
where $J=S_1+S_2$ is the total spin. Now, it turns out that, in Quantum Mechanics, when you compose two spin-1/2 states, the total angular momentum can be either J=0 (single) or J=1 (triplet). Most of the times the atomic interaction depends on the total spin of the system and so it turns out that it is determined by the fact that the system is in a single/triplet.

For example, if you have an interaction of the kind:
$$H=\lambda \vec{S}_1\cdot\vec{S}_2,$$
where $\lambda$ is coupling constant, then you can write:
$$J^2=(S_1+S_2)^2=S_1^2+S_2^2+2S_1\cdot S_2 \Rightarrow S_1\cdot S_2=\frac{J^2-S_1^2-S_2^2}{2}=\frac{j(j+1)-3/2}{2}.$$
Then if you are in a single case $j=0$ and the energy shift is given by $\Delta E_{sing}=-(3/4)\lambda$, while if you are in a triplet $j=1$ and $\Delta E_{tripl}=+(1/4)\lambda$.

3. Nov 6, 2014

### Eats Dirt

$$^{2S+1}L_{J}$$