In general, the diatomic vibrational potential is not exactly equal to the harmonic oscillator. Let's write the potential as a sum of the harmonic oscillator potential plus a correction:
The correction is called anharmonic because it is not harmonic. It leads to corrections in the harmonic oscillator eigenenergies, which can be written (in the spectroscopic notation) as the expansion,
where and are called anharmonic constants, and the dots represent higher order terms. The infrared active transitions will be observed in the spectrum at the energy differences corresponding to , i.e.,
This expression should always be positive since it corresponds to the energy difference from a lower energy state to a higher energy state. However, near the n quantum number which satisfies,
the difference will go from positive to negative. The physical explanation for this is that the true potential is only bound for a finite region in energy, and so admits only a finite number of bound states. Thus the last n for which you get a positive energy difference corresponds to the highest allowed bound state. As you include the higher order energy corrections, this argument will be more complicated, but the main points are:
Another interesting feature of the anharmonic oscillator is that since its eigenstates are not exactly those of the harmonic oscillator, the calculation of the dipole moment is no longer exact. This leads to the possibility of having infrared active transitions with , etc., at , etc.. These transitions are called overtones, and though possible, are much, much weaker than the harmonic infrared active transition.
An often-quoted example of an anharmonic potential is the , given by,
where is the dissociation energy. A sketch of this potential is drawn in Figure 22.
Figure 22: Morse Potential Energy Diagram
It turns out that the Morse Potential can be solved analytically! The calculation is somewhat tedious, but the rather interesting result is that the energies are to within experimental accuracy equal to:
which is to say that the expansion does not involve any terms higher in order than . As a consequence, the Morse Oscillator exhibits anharmonic effects, but is fairly tame because the leading order correction leads to the correct result. This has made it a fairly often-cited model system.