# Rovibrational Transitions

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## Ro-Vibrational Transitions

### 1 Order of Magnitude Energies

The Born-Oppenheimer approximation allows us to treat the electrons in a molecule as a cloud– they are much less massive and therefore have much higher velocities than the nuclei. If ${\displaystyle L}$ is the molecular size, typical electrons have momentum ${\displaystyle \hbar /a}$ and the electronic energy spacings can be expressed as:

${\displaystyle E_{elec}\sim {\frac {\hbar ^{2}}{mL^{2}}}\,\!}$

The nuclei feel an equivalent potential that only depends on the internuclear distance and the electronic state. The internuclear potential has a minimum, and vibrations about the minimum can be roughly modeled as a harmonic oscillator. This potential is about ${\displaystyle {\frac {1}{2}}M\omega ^{2}\xi ^{2}}$, where ${\displaystyle \omega }$ is the vibration frequency and ${\displaystyle \xi }$ is the displacement from equilibrium. If the displacement is the same order as the size of the molecule, the electronic energies should be about ${\displaystyle \hbar ^{2}/2mL^{2}}$:

${\displaystyle {\frac {1}{2}}M\omega ^{2}L^{2}\sim {\frac {\hbar ^{2}}{2mL^{2}}}\,\!}$
${\displaystyle E_{vib}\sim \hbar \omega \sim \left({\frac {m}{M}}\right)^{1/2}{\frac {\hbar ^{2}}{mL^{2}}}\sim \left({\frac {m}{M}}\right)^{1/2}E_{elec}\,\!}$

The nuclei can also rotate, and have rotational energies that depend on their angular momentum. If ${\displaystyle J}$ is the quantum angular momentum number,

${\displaystyle E_{rot}\sim {\frac {\hbar ^{2}J(J+1)}{2I}}\,\!}$
${\displaystyle I\sim ML^{2}\,\!}$

For small ${\displaystyle J}$,

${\displaystyle E_{rot}\sim {\frac {m}{M}}E_{elec}\,\!}$

So the electronic, vibrational, and rotational energy states have contributions that scale with the electron-to-nucleus mass ration:

${\displaystyle E_{elec}:E_{vib}:E_{rot}\sim 1:\left({\frac {m}{M}}\right)^{1/2}:\left({\frac {m}{M}}\right)\,\!}$

### 2 Rotation-Vibration Spectra

While it is possible to have a pure rotational spectrum, a pure vibrational spectrum is very unlikely: energies required to excite vibrations are much larger than those required to excite rotation. However, a combination of rotation and vibrational modes can be excited.

Rotational energies can be described using the angular momentum number ${\displaystyle J}$:

${\displaystyle E_{rot}={\frac {\hbar ^{2}}{2I}}J(J+1)\,\!}$

where ${\displaystyle J=0,\ 1,\ 2,\ ...}$, ${\displaystyle I}$ is the moment of inertia, and ${\displaystyle \mu }$ is the reduced mass:

${\displaystyle I=\mu L^{2}\,\!}$
${\displaystyle \mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}\,\!}$

The vibrations can again be modeled as a harmonic oscillator:

${\displaystyle E_{vib}=\hbar \omega (n+1/2)\,\!}$

where ${\displaystyle n=0,\ 1,\ 2,\ ...}$ and ${\displaystyle \omega ={\sqrt {k/m}}}$ where ${\displaystyle k}$ is the effective spring constant.

The total energy of rovibrational transitions, then, is:

${\displaystyle E_{rovibe}=\hbar \omega (n+1/2)+{\frac {\hbar ^{2}}{2I}}J(J+1)\,\!}$

The selection rules for rovibrational transitions tell us that ${\displaystyle \Delta n=+1}$ and ${\displaystyle \Delta J=\pm 1}$. ${\displaystyle \Delta J=+1}$ is called the R branch, and ${\displaystyle \Delta J=-1}$ is called the P branch. This notation matches the videos above, but is opposite the notation in Rybicki & Lightman. We can use our rovibe energy expression to find the frequency of emitted photons for the R branch:

${\displaystyle \Delta J=+1\,\!}$
${\displaystyle E_{2}-E_{1}=\hbar \omega (n+1+1/2)-\hbar \omega (n+1/2)+{\frac {\hbar ^{2}}{2I}}(J+1)(J+2)-{\frac {\hbar ^{2}}{2I}}J(J+1)\,\!}$
${\displaystyle E_{2}-E_{1}=\hbar \omega +{\frac {\hbar ^{2}}{I}}(J+1)=\hbar \omega _{obs,R}\,\!}$
${\displaystyle \omega _{obs,R}=\omega +{\frac {\hbar }{I}}(J+1)\,\!}$

and the P branch:

${\displaystyle \Delta J=-1\,\!}$
${\displaystyle E_{2}-E_{1}=\hbar \omega (n+1+1/2)-\hbar \omega (n+1/2)+{\frac {\hbar ^{2}}{2I}}(J-1)(J)-{\frac {\hbar ^{2}}{2I}}J(J+1)\,\!}$
${\displaystyle E_{2}-E_{1}=\hbar \omega -{\frac {\hbar ^{2}}{I}}J=\hbar \omega _{obs,P}\,\!}$
${\displaystyle \omega _{obs,P}=\omega -{\frac {\hbar }{I}}J\,\!}$

However, we should note that the potential is not perfectly harmonic; the slight asymmetries we can see in the potential cause the bond length to increase as ${\displaystyle n}$ increases. So if ${\displaystyle n}$ increases, ${\displaystyle L}$ increases, causing ${\displaystyle I}$ to increase. Our rovibrational energy expression depends on ${\displaystyle I^{-1}}$, so the energy decreases as ${\displaystyle n}$ increases. This effectively pulls all of our spectral lines to the left, which decreases the separation between lines on the R branch and increases the separation between lines on the P branch. Similar effects can also be found when comparing the expressions for the observed frequencies. Putting everything together, we can plot our rovibrational spectrum.