19.4: The Rigid Rotor

The rigid rotor is a simple model of a rotating stick in three dimensions (or, if you prefer, of a molecule). We consider the stick to consist of two point-masses at a fixed distance. We then reduce the model to a one-dimensional system by considering the rigid rotor to have one mass fixed at the origin, which is orbited by the reduced mass \(\mu\), at a distance \(r\). The cartesian coordinates, \(x,y,z\), are then replaced by three spherical polar coordinates: the co-latitude (zenith) angle \(\theta\), the longitudinal (azimuth) angle \(\phi\), and the distance \(r\). The TISEq of the system in spherical coordinates is:

\[ - \dfrac{\hbar^2}{2I} \left[ \dfrac{1}{\sin \theta} \dfrac{\partial}{\partial \theta} \left(\sin\theta\dfrac{\partial}{\partial \theta} \right) + \dfrac{1}{\sin^2 \theta} \dfrac{\partial^2}{\partial \phi^2} \right] \psi(r) = E_{\ell} \psi(r), \label{20.4.1} \]

where \(I=\mu r^2\) is the moment of inertia. After a little effort, the eigenfunctions can be shown to be the spherical harmonics \(Y_{\ell}^{m_{\ell}}(\theta, \phi)\).\(^1\) The eigenvalues are simply:

\[ E_{\ell} = \dfrac{\hbar^2}{2I} \ell(\ell+1), \label{20.4.2} \]

where \(\ell=0,1,2,\ldots\) is the azimuthal quantum number, and \(m_{\ell}=-\ell, -\ell+1, \ldots, \ell-1, \ell\) is the magnetic quantum number. Each energy level \(E_{\ell}\) is \((2\ell+1)\)-fold degenerate in \(m_{\ell}\). Notice that the energy does not depend on the second index \(m_{\ell}\), and the functions with fixed \(m_{\ell}=-\ell,-\ell+1,\dots,\ell-1,\ell\) have the same energy. Since this problem was, in fact, a one-dimensional problem, it results in just one quantum number \(\ell\), similarly to the previous two cases. The index \(m_{\ell}\) that appears in the spherical harmonics will assume some importance in future chapters.

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1. For a description of the spherical harmonics see here