Symmetric Top

Examples: $CH_3I$, $C_6H_6$, $NH_3$.

Denoting $I_{\Vert}$ as the moment of inertia which is parallel and $I_{\bot}$ as the moment of inertia which is perpendicular to the molecular axis, we can consider two cases: $I_{\Vert} > I_{\bot}$ and $I_{\Vert} < I_{\bot}$. In the first case we have an oblate top (which is like a pancake) and in the second case a prolate top (which is like a cigar). Within the world of atoms and molecules, benzene is an example for the first, while methyliodide an example for the second. For both types, the rotational energy can be presented as sum of two terms:

$$E = \frac{J_A^2+J_B^2}{2J_{\bot}} + \frac{J_C^2}{2I_{\Vert}}$$ (39)

Transition to quantum mechanics is eased by the fact that the energy values respective ${\bf J}^2$ and one selected axis obtained simultaneously. For ${\bf J}^2$, which is

$${\bf J}^2 = J_A^2 + J_B^2 + J_C^2$$ (40)

the energy values are defined by

$$E = \frac{J^2}{2I_{\bot}} + J_C^2\cdot\left(\frac{1}{2I_{\Vert}}-\frac{1}{2I_{\bot}}\right)$$ (41)

We now replace the square of the angular momentum by the eigenvalues of the quantum mechanical angular momentum operator ${\bf J}^2 -> J(J+1)\hbar^2$ and use the quantum number $K$ as projection of the momentum ${\bf J}$ onto the molecules axis (i.e. $J_C \Psi = K \hbar\Psi$). For the quantum mechanical energy levels, we finally obtain:

$$F_{JK} = B\cdot J(J+1) + (A-B)K^2 \mbox{, with } J=0,1,2,\ldots \: K=0,\pm 1,\ldots , \pm J,$$ (42)

where the rotational constants $A$ and $B$ are

$$A=\frac{\hbar}{4\pi c I_{\Vert}} \:\:\:\:\:\: B=\frac{\hbar}{4\pi c I_{\bot}}$$ (43)

Obviously, the sign of $K$ does not affect the energy as it dependents on the square of $K$. Therefore, all levels are doubly degenerated. There are $2J+1$ projections of the angular momentum onto the external axis ( $M_J = 0, \pm 1, \pm 2, \pm 3, ... \pm J$) and therefore, each energy level of the symmetric top is $2(2J+1)$-fold degenerated. For $K = 0$ we have $(2J+1)$-fold degeneracy. Though there is no definition for clockwise/anti-clockwise and +/-, the sign indicates two possibilities of rotation with the same energy.

We now consider two extreme cases of a rotating symmetric top:$\vert K\vert\simeq J$ and $K = 0$.

For the first case $\vert K\vert\simeq J$, when $J$ has maximum projection onto the molecular axis, the rotation is mainly proceeds around this axis and the rotational constant A is decisive for the energy levels.

$$E_{J,K=J} = B\cdot J + A J^2 \approx A\cdot J^2$$ (44)

For the other case $K = 0$ the molecule rotates perpendicular to its axis. Therefore, the projection of the angular momentum on the axis is zero and the rotational energy becomes

$$E_{J,K=0} = B\cdot J(J+1)$$ (45)

Auf diesem Webangebot gilt die Datenschutzerklärung der TU Braunschweig mit Ausnahme der Abschnitte VI, VII und VIII.