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?

  • $\begingroup$ Related : Total spin of two spin- 1/2 particles. $\endgroup$
    – Frobenius
    Commented Aug 1, 2021 at 20:15
  • $\begingroup$ You can have a doublet when $j=1/2$, and this can occur in coupling $\ell=1$ or $\ell=0$ with $s=1/2$. For two spin-$1/2$s see physics.stackexchange.com/q/632973/36194 $\endgroup$ Commented Aug 1, 2021 at 20:56
  • $\begingroup$ @ZeroTheHero shouldn't spin multiplicity be equal to 2 for a doublet ? In that case total spin must be 1/2. That is only possible if there is exactly one unpaired electron, and so the system must have total odd no. of electrons, only then you have a chance that one of them would be unpaired, provided all others are paired up, right ? I don't see, how j is entering the picture, as I thought doublet only depends on s. $\endgroup$ Commented Aug 1, 2021 at 21:02
  • 1
    $\begingroup$ Yes. $2j+1=2$ when $j=1/2$. This can happen for an odd number of electrons. $\endgroup$ Commented Aug 1, 2021 at 21:59

1 Answer 1


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:

$$\left|\frac 32,\frac 12\right\rangle = \sqrt{\frac 2 3} \left|0; \frac 12\right \rangle +\sqrt{\frac 13}\left|1; -\frac 12\right\rangle$$

$$\left|\frac 32,-\frac 12\right\rangle = \sqrt{\frac 1 3} \left|-1; \frac 12\right\rangle +\sqrt{\frac 23}\left|0; -\frac 12\right\rangle$$

$$\left|\frac 32,\pm\frac 32\right\rangle=\left|\pm 1; \pm\frac 12\right\rangle$$


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