The selection rules for symmetric top molecules are
$\Delta J = 0, \pm1$ and $\Delta K = 0$ for $K \neq 0$ (parallel transition)
$\Delta J = \pm1$ and $\Delta K = 0$ for $K = 0$ (parallel transition)
$\Delta J = 0, \pm1$ and $\Delta K = \pm1$ for $K = 0$ (perpendical transition)
My question concerns the possibility of a $\Delta J=0$ transition. For example, the only way how $\Delta J =0$ is possible for linear molecules is with degenerate normal modes (e.g. $\ce{CO2}$ bend), as these can "evolve" into one another and give rise to vibrational angular momentum, which ensures that conservation of angular momentum can still hold regardless of $\Delta J$.
The fact that $\Delta J =0$ is allowed for all transitions of symmetric tops means that there is vibrational angular momentum associated with all of them. My questions are:
- What is the origin of vibrational angular momentum if the normal modes involved are not degenerate? Can you help me visualise it, for example for the $\ce{C-Cl}$ stretch of $\ce{CH3Cl}$ (a parallel non-degenerate transition)?
- Why is the selection rule different for $K=0$ and $K\neq0$?