# Determining symmetry correction by looking at rotational quantum states

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

• 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. May 14, 2020 at 8:52
• @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?
– Phy
Jun 12, 2020 at 15:46