In rotational-vibrational spectroscopy, the "fundamental" transition is the one in the lowest electronic state between the first vibrational level ($\nu'=1$) and the ground level ($\nu''=0$). Since the rotational quantum number J also changes (the selection rule here is usually $\Delta J = \pm1$), we do not get a single transition, but a molecular band of transition lines.

For a diatomic molecule (in this case CO), this spectrum looks something like this: Simulated vibration-rotation line spectrum of carbon monoxide (source: Wikipedia)

This band of CO can be seen in absorption. My question is: Why can we see it at all, if it involves a vibrational transition between $\nu''=0$ and $\nu'=1$ in the same electronic state, which has a transition probability of zero?

Here is a Morse potential to show what I mean:

Morse potential and eigenfunctions

As can be seen, the wave functions of the lowest two levels are pretty much orthogonal to each other (if that is the right word, my QM is a bit rusty) - the integral of their product will be close to zero. However, according to the usual literature (e.g. Herzberg's "Spectra of Diatomic Molecules"), the transition probability is proportional to the square of this integral, so the transition probability would be zero and we would never see this transition.

Let me elaborate a bit to show you where this relation comes from.

In a Morse potential, the wave functions are given by $$\Psi = \Psi_e \Psi_\nu(r-r_e) \frac{1}{r} \Psi_r(\theta,\phi)$$ As far as I know, this is already the Born-Oppenheimer approximation at work. If we want to know the intensity of a transition, we need to look at the transition moment $$R=\int \Psi'^{*}M\Psi''d\tau$$ However, since our wave function is a product of the electronic, vibrational and rotational wave functions, we can split up this integral. In "Spectra of Diatomic Molecules", Herzberg gives this for the $R_z$ component as: $$R_z=\int {\Psi_e}'^{*}{\Psi_\nu}'M{\Psi_e}''{\Psi_\nu}''d\tau_e dr \int \sin \theta \cos \theta {\Psi_r}'^{*}{\Psi_r}'' d\theta d\phi$$ The second integral is a constant for a given J' and J'' combination and, apparently, can therefore be neglected in the calculation of the band intensity. What we are left with is the first integral. With the electric moment $M=M_e+M_n$ (one part depending on the electrons and the other on the nuclei), it can be shown that the part depending on the nuclei becomes zero, and R becomes $$R=\int {\Psi_\nu}'{\Psi_\nu}'' \int M_e {\Psi_e}'^{*}{\Psi_e}''d\tau_e$$

So this derivation shows that R is proportional to the integral over the product of the two vibrational wave functions. And like I said earlier, in the Morse potential of diatomic molecules, this should be very close to zero for the two lowest vibrational states!

So what am I missing, why can we see this fundamental transition? Is one of the approximations in this derivation wrong? One possibility I can see is that the vibrational wave functions are not exactly the same as J changes, so that the integral gets a value higher than zero. But if this effect was so strong that it explains why we can see these bands, then it would be something we cannot ignore. We do ignore it all of the time, however, and the intensities of the rotational structure are usually only calculated with a Boltzmann factor for the initial state.

Any help would be appreciated!

  • $\begingroup$ When the light interacts with the molecule, the eigenstates of total Hamiltonian are not orthogonal any more so the transitions happens. This is my idea. $\endgroup$ – Bettertomo May 14 '18 at 8:20

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