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 $K\neq0$?

1 Answer 1


Some of the statements you make above are not quite correct, rather that correct individual points I think that it is simpler to start from scratch, as it were, and try to explain what happens.

Two quantum numbers are needed to define the rotational energy of a symmetric top molecule, such as CH$_3$I or benzene. The total angular momentum is characterised by $J$ and this generally does not lie along the 'figure' axis of the molecule which is the C-I bond axis in CH$_3$I for example. This quantum number has integer values starting at zero, i.e. $0,\, 1,\,2\cdots$ The total angular momentum is $ \hbar\sqrt{J(J+1)}$. The second quantum number $K$ is the projection of $J$ along the figure axis and so has $2J+1$ values ranging from $-J$ to $+J$ and therefore has values $0,\,\pm 1,\, \pm 2, \cdots \pm J$. The size of the projection quantum number is $K\hbar$ and clearly as it is the projection along the figure axis $K$ cannot exceed $J$.

The energy levels of any vibrational level form stacks of $J$ levels rather like those of a rigid rotor but where each $K$ value has its own stack beginning with $J=K$. Each type of vibration has sets of $K$ and associated $J$ levels for each vibrational quantum number.

We are considering rotation-vibrational transitions. The selection rule for parallel transitions, i.e. those with a dipole moment along the figure axis always have $\Delta K=0$ and then $\Delta J = 0, \pm 1$ unless $K=0$ then transitions with $\Delta J=0 $ are forbidden. The reason that $\Delta K = 0$ is that radiation cannot induce transitions between different $K$ values because rotation about the figure axis induces no rotating dipole moment. (The dipole lies along the axis in this type of transition). When $K=0$ there is no angular momentum component along the figure axis and so if the total angular momentum is also zero there is no rotating dipole to couple to the radiation.

In the case of perpendicular transitions the dipole has a component perpendicular to the figure axis. Bending vibrations can induce perpendicular transitions. The selection rules are $\Delta K=\pm 1$ and $\Delta J = 0,\pm1$. In this case changing $K$ does induce a change in the dipole and so transitions can occur.


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