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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:

enter image description here

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).

enter image description here

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

enter image description here

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 $\frac{2J}{\sigma}+1$?

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  • $\begingroup$ The density of states calculation is correct, the multiplicity is $2J+1$ and these levels are degenerate. I don't follow why you want to ignore the $m_z=0$ case. It should be treated in the same way as the others. $\endgroup$
    – porphyrin
    May 14, 2020 at 8:52
  • $\begingroup$ @porphyrin I am not ignoring $m_z=0$, I am considering $m_z=1$ and $m_z=-1$ to be the same quantum state in space. So any non-zero value of $m_z$ and its negative counterpart should be treated as 1 quantum state. The number of quantum states that have a negative counterpart is $2J$ so this should be divided by $\sigma$. The $+1$ part is reserved for $m_z=0$ which does not have a negative counterpart and thus is a quantum state by itself. Furthermore, how can a dipole at $J=2$ have $\frac{2J+1}{\sigma} = 2,5$ quantum states according to the source while it should be integers? $\endgroup$
    – Phy
    Jun 12, 2020 at 15:46

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