I've recently studied the addition of angular momenta in quantum mechanics, and faced a massive confusion during the addition of spin, in a two-electron system.
When adding the two spin-1/2 electrons, the two possible values of total spin is either $0$ or $1$. For the former, we have a singlet state, as there is only one possible wavefunction corresponding to that, and in case of the latter, we have a triplet state, as there are three possible wavefunctions i.e. configurations, all of which are symmetric.
Later however, as I moved on to many electron atoms, I came across singlet, doublet, triplet, quartet etc, which were defined on the basis of their spectra and spin 'multiplicity'. For singlet, I saw that $s=0$, and for triplet $s=1$. However, I couldn't reconcile the idea of doublet and the other states with my initial understanding. Plainly speaking, is the 'spectroscopic L-S coupling' singlet, doublet, etc the same as the 'addition of spin' singlet, triplet, etc? If that is the case, there shouldn't be any possible way two-electron systems can have a doublet state, as the total spin is always going to be either 0 or 1. Am I correct in saying this? For a doublet state, there should be an odd no. of electrons in total, right?
Best Answer
If you combine $l=1$ and $s=\frac 1 2$, the $|J, M\rangle$ states, in terms of $|L_z; S_z\rangle$ are:
$$\left |\frac 12,\frac 12 \right\rangle = \sqrt{\frac 1 3} \left|0; \frac 12\right\rangle -\sqrt{\frac 23}\left|1; -\frac 12\right\rangle$$
$$\left|\frac 12,-\frac 12\right\rangle = \sqrt{\frac 2 3} \left|-1; \frac 12\right\rangle -\sqrt{\frac 13}\left|0; -\frac 12\right\rangle$$
which forms a doublet, and a quartet:
$$\left|\frac 32,\frac 12\right\rangle = \sqrt{\frac 2 3} \left|0; \frac 12\right \rangle +\sqrt{\frac 13}\left|1; -\frac 12\right\rangle$$
$$\left|\frac 32,-\frac 12\right\rangle = \sqrt{\frac 1 3} \left|-1; \frac 12\right\rangle +\sqrt{\frac 23}\left|0; -\frac 12\right\rangle$$
$$\left|\frac 32,\pm\frac 32\right\rangle=\left|\pm 1; \pm\frac 12\right\rangle$$