The Symmetric Top

Examples: CH3I, C6H6, NH3

If we designate with I|| the molecular axis and with I^ the inertia moment perpendicular to the molecular axis then for I|| > I^(for instance, benzene) the top will rotate like a plate (oblate) and for I||< I^ it will move like cigar (prolate CH3I). The energy is given by (classical physics):

E = (JA² + JB²)/2I^ + JC²/2I||

The transition to quantum mechanics will be much easier if we take into account the fact that energy values corresponding to J2 and chosen axis are obtained simultaneously. Since

J² = JA² + JB² + JC²

we obtain energy values:

E = J²/2I^ + JC² · [(2I||)-1− (2I^)-1]

If now we substitute squared kinetic moment J² with eigenvalues of quantum mechanical kinetic moment operator, J² → J(J+1)h² and designate the eigenvalues of corresponding components on molecular axis with K (JC y® Kh y) we obtain the quantum mechanical energy levels:

EJK = B · J(J+1) + (A - B)K2 |  | K = 0, ±1, ±2, ..., ±J
B = h / (4πcI^) | A = h / (4πcI||) | J = 0, 1, 2, ...

Extreme cases of the symmetric top rotation
a) around molecular axis | b) perpendicular to molecular axis

The K sign obviously doesn't have influence on energy since it depends on K2. Since kinetic moment may take 2J + 1 values (MJ = 0, ± 1, ± 2, ± 3, ... ± J), the symmetric top level degeneracy is only 2(2J + 1)-fold. The exception is K = 0 because this level has (2J + 1)-fold degeneracy. K-sign shows direction of rotation, but nevertheless the rotational energy doesn't depend on direction of rotation.

For one extreme case when |K| = J, J is parallel to molecular axis as much as possible and molecule rotates mainly around its axis and energy levels essentially depend on the rotational constant A (resp. I||):

EJ,K = J = BJ + AJ2 ≈ AJ2

For another extreme case, K = 0, the molecule rotates perpendicular to its axis. Therefore it doesn't have kinetic moment in the molecular axis direction. The rotational energy is then given by:

EJ, K=0 = BJ(J+1)