Question 5.P.3: Find the rotational energy levels of a diatomic molecule.

Consider two molecules of masses m_{1} and m_{2} separated by a constant distance \vec{r}. Let r_{1} and r_{2} be their distances from the center of mass, i.e.,m_{1}r_{1}=m_{2}r_{2}. The moment of inertia of the diatomic molecule is

I=m_{1}r^{2}_{1} +m_{2}r^{2}_{2}\equiv \mu r_{2},             (5.216)

where r=|\vec{r}_{1}-\vec{r}_{2}| and where μ is their reduced mass, \mu =m_{1}m_{2}/(m_{1}+m_{2}). The total angular momentum is given by

|\hat{\vec{L}}|=m_{1}r_{1}r_{1}\omega +m_{2}r_{2} r_{2} \omega =I\omega =\mu r^{2} \omega                   (5.217)

and the Hamiltonian by

\hat{H}=\frac{\hat{\vec{L}}^{2} }{2I} =\frac{\hat {\vec{L}}^{2}}{2\mu r^{2}} .                    (5.218)

The corresponding eigenvalue equation

\hat{H}|l,m〉=\frac{\hat{\vec{L}}^{2}}{2\mu r^{2}}|l,m〉= \frac{l(l+1)\hbar ^{2} }{2\mu r^{2}} |l,m〉,                  (5.219)

shows that the eigenenergies are (2l + 1)-fold degenerate and given by

E_{l} =\frac{l(l+1)\hbar ^{2} }{2\mu r^{2}}.                (5.220)