I am quite confused about the origin and justification of the rotational selection rules of methanol, and I would appreciate any help or references. As far as I understand, there are three irreducible representations of the symmetry group of methanol ($A_1$, $A_2$,and $E$) and the two $A$-type are linked by dipole-allowed radiative transitions. Now, I think the literature also calls these ($A_1$ and $A_2$) $A^+$ and $A^-$ representations, I suppose because the parity of the eigenfunctions which generate the $A_1$ and $A_2$ are $+1$ and $-1$, respectively. Is this correct?

From what I have studied, because the electric dipole generates the $A_2$ representation, usually the allowed transitions follow the selection rule $A_1\leftrightarrow A_2$ (because $A_2\times A_2=A_1$), and I have seen this in many references. However, a lot of strong $A$-type methanol rotational lines are $++$ or $--$. I have noticed that all of these are associated with transitions with $|\Delta J|=1$ and all of the $+-$ and $-+$ transitions are associated with $\Delta J=0$. In fact, I have seen this statement as a selection rule in some papers (in this reference, they say that the $+-$ refers to K-type doubling ... ?), but it is not at all clear to me where does these rules come from.

I assume that the symmetry group of methanol is $C_{3v}$ (in books sometimes they call it $G_6$), and the representations above are usually the three which appear in the character tables. I think the group $C_{3v}$ represent permutations and permutation-inversions involving the protons of only the methyl group, because the other proton is too far and cannot be easily swapped with the others. The $J$ above represent the total rotational angular momentum of the molecule. Apparently, the usual approach is to express the wavefunctions as linear combinations of the product of symmetric top rotor wavefunctions and the wavefunctions associated with the (hindered or not) rotation of the methyl group around the CO axis. I am interested in transitions at the torsional ground state ($\nu_t=0$) so I guess we can assume that the rotational potential is large. The $K$ I mention briefly above is the projection of the angular momentum in the main principal axis of the molecule, which is roughly aligned with the CO bond. Since methanol is asymmetric, there are two energy levels with the same $J$ and $K$ in the rotational spectrum, which is what I think they call the $K$ doubling. I am in no way an expert on this, so if you want to point out inaccuracies in what I have just wrote I would be thankful.

  • $\begingroup$ crosspost of physics.stackexchange.com/questions/291290/… $\endgroup$
    – Mithoron
    Nov 7 '16 at 0:07
  • $\begingroup$ It would help is you state what you consider the point group to be, C1, Cs or something else. Are you worried about the restricted rotation of the OH group or do you ignore this, in which case the molecule is C3v. Have a look at chemistry.stackexchange.com/questions/58609/… which may help in identifying transitions. $\endgroup$
    – porphyrin
    Nov 7 '16 at 9:00
  • $\begingroup$ Thank you for your comment. I include a paragraph with more details. $\endgroup$
    – user37085
    Nov 7 '16 at 12:55

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