We can approximate a diatomic molecule as a rigid bodies, ignoring their vibrational motion for now.

Conditions:

• The two masses $m_1$ and $m_2$ separated by a distance $r_0$

• they are free to rotate around their common CoM

• Its moment of inertia is given by

  $$ \begin{aligned} I &= m_1\left(\frac{m_2}{m_1+m_2}\,r_0\right)^2+m_2\left(\frac{m_1}{m_1+m_2}\,r_0\right)^2 \\ &= \frac{m_1\,m_2}{m_1+m_2}\,r_0^2 \equiv \mu\,r_0^2

  \end{aligned} $$

  where $\mu$ is the reduced mass

• Classically the kinetic energy and angular momentum are given by

  $$ \begin{aligned} E&={\textstyle{\frac12}}I\,\omega^2\\ L&=I\,\omega

  \end{aligned} $$

• Since $L$ is conserved we can write kinetic energy as

  $$ E = \frac{L^2}{2I} $$

• The potential energy function is zero therefore we expect in quantum mechanics

  $$ \widehat{H}=\frac1{2I}\,\widehat{L}^2 $$

• We know the eigenvalues and eigenfunctions of $\widehat{L}^2$ we can immediately write down the rotational energy levels of a diatomic molecule

  $$ E_\ell = \frac{\hbar^2}{2I}\,\ell(\ell+1) $$

  💃 Example: for an $H_2$ molecule $\mu=m/2=0.84 \times 10^{-27}$ , $r_0=0.74 \r{ \text A}$

  $$ \Rightarrow\frac{\hbar^2}{2I}=7.5\times10^{-3}\,\mathrm{eV} $$

  🗒️ Notes:

  • This energy, via the Boltzmann constant $E\sim kT$ corresponds to a temperature $T \sim 90K$ so rotational excited states are populated at room temperature, unlike the vibrational states
  • This energy corresponds to a photon wavelength $\lambda \approx 0.16 \, \text{mm}$ in the far infrared towards microwaves. Most diatomic molecules have similar bond length, but higher masses and therefore smaller energy, so they give longer wavelengths, more towards microwaves.

Microwave spectroscopy

Experimenters in microwave spectroscopy typically quote their results in wave number, measured in $\text{cm}^{-1},\widetilde{\nu}$

$$ \widetilde{\nu}=\frac1\lambda=\frac{f}{c}=\frac{E}{hc}=\frac{h}{8\pi^2cI}\ell(\ell+1)\equiv B\ell(\ell+1) $$

🗒️ Note: to give $B$ in $\mathrm{cm}^{-1}$, $c$ should be used in $\mathrm{cm\,s}^{-1}$)

The energy diagram is:

• Only transitions $\Delta l=\pm 1$ are allowed by emission/absorption of one photon

  Therefore we get absorption lines at $\widetilde{v}=2B,4B,6B,8B,\ldots$

• $B$ can be measured from the microwave absorption spectrum and hence $I$ and $r_0$

• In practice diatomic molecules vibrate and rotate simultaneously. The vibration spectrum we calculated previously are a more complicated because the vibrational transition can only occur by a single photon if it is accompanies by a rotational transition with $\Delta l=\pm 1$.

  Thus each vibrational infrared absorption line is actually a series of very close but equally spaces lines with the centre one missing.