It is a system of two particles separated by a distance d with no possibility of vibration $r=d$. The energy of the rotor is kinetic, therefore $V=0$. We are only interested in rotational energy.

Since r is constant, we can omit the factor $R(r)$ in the wave function, which will be given by a spherical harmonic. \begin{equation} \Psi =Y_{l,m}(\theta,\varphi) \end{equation} We set up the Schrödinger equation: \begin{equation} \left[-\frac{\hbar^2}{ 2\mu}\left(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right)+\frac{1 }{2\mu d^2}\hat{l}^2+V(r)\right]Y_{l,m}(\theta,\varphi)=EY_{l,m}(\theta,\varphi ) \end{equation} The derivatives with respect to $r$ are null, since the function does not depend on this variable.

\begin{equation} \frac{1}{2\mu d^2}\hat{l}^2 Y_{l,m}(\theta,\varphi)=EY_{l,m}(\theta,\ varphi) \end{equation} Since $\hat{l}^2\Psi=l(l+1)\hbar^2\Psi$ and changing $l$ by the rotational quantum number $J$, we are left with: \ begin{equation} \frac{1}{2\mu d^2}J(J+1)\hbar^2\cancel{Y_{l,m}(\theta,\varphi)}=E\cancel{Y_ {l,m}(\theta,\varphi)} \end{equation} \begin{equation} E=\frac{J(J+1)\hbar^2}{2\mu d^2}\;\ ;\;\; J=0,1,2.... \end{equation}

  • The lowest level is E=0, so we have no rotational energy at the zero point. \item The energy levels of the rigid rotor are degenerate, the energy of the rotor depends on J, but the wave function depends on J and m.
  • For each value of J, m takes values $-J,......,0,.....,+J$. Each value of J has $2J+1$ values of m. This implies that each energy level is $2J+1$ times degenerate.
  • The rotational levels of a diatomic molecule can be approximated by the energies of the rigid rotor.