I'm trying to understand the relationship between symmetry correction and rotational quantum states, particularly in the case of dipoles with identical atoms.
For an angular momentum quantum number $J=1$ there are 3 quantum states according to $g(J)=2J+1$. Each quantum state determines the z-vector of the angular momentum $M$. The spacial direction of the angular momentum $M$ for each quantum state is then:
The orbital plane of a dipole is perpendicular to $M$. I can see that I can re-orentiate scenario a) in space to make it look like scenario c).
Thus for $J=1$ there are actually 2 quantum states instead of 3.
I noticed that, at any value of $J$, this can be done for each pair of quantum states that have the same z-vector momentum $M_z$ but in opposite directions. The only quantum state at any $J$ that isn't part of a pair is when $M_z=0$
This would make me deduce that the number of quantum states for a dipole after symmetrical correction $\sigma$ is equal to $\frac{2J}{\sigma=2}+1$
However, here's a snippet of a source
Apart from this discussing the density of states, why does it show that after symmetrical correction the number of quantum states is $\frac{2J+1}{\sigma}$ instead of $\frac{2J}{\sigma}+1$?
