# Are the $+\leftrightarrow+$ and $-\leftrightarrow-$ methanol rotational transitions dipole-allowed?

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.

• Would this rather be a question for Chemistry SE? – Steeven Nov 6 '16 at 23:05
• Thanks for your comment. Maybe you are right, I will left it here a few days before closing it and I will ask in the other community also. Please feel free to downvote the question if you think it is outside the scope of the Physics SE. – Enredanrestos Nov 6 '16 at 23:15
• @Steeven It may or may not be on topic on Chemistry SE, but it's certainly on topic here. (In fact, I'm quite happy to sponsor a bounty on this one if it doesn't get an answer.) – Emilio Pisanty Nov 7 '16 at 13:41
• @EmilioPisanty Sure, I also have not voted for closing. But I have a slight feeling that the OP could find more people within this field in the chemistry aisle. – Steeven Nov 7 '16 at 16:56
• The $\pm$ in the A levels of methanol is a splitting induced in the $\pm K$ levels by the slight asymmetry of the methanol molecule. The overall parity of the levels is given by $\pm(-1)^{J+\nu_t}$ for $K$ even and odd, respectively, with $\nu_t$ the quantum number associated with the torsional vibration. See also E. Herbst, J. K. Messer, F. C. D. Lucia, and P. Helminger, J. Mol. Spectrosc. 108, 42 (1984). – Paul Nov 7 '16 at 21:02