# Dominant contribution of v = 0 → v = 1 transition in vibrational spectroscopy

Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.

After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($$\Delta v = \pm 1$$), my professor's notes state that:

Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $$v = 0 \to v = 1$$ transition and is called fundamental transition.

I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $$v = 0 \to v = 1$$ will take place?

• A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $\nu$ (nu). The Greek letter $\nu$ is used to refer to a frequency; so we can have a transition for which $\Delta E = h\nu$, for example. Feb 3 '19 at 21:23

The reference should really be to the Boltzmann distribution where the population is proportional to $$\exp(-E_n/(k_\mathrm BT))$$ where $$E_n$$ is the energy of the $$n^{th}$$ energy level. This shows that the population is greatest when $$n$$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $$k_\mathrm BT$$ at room temperature $$\approx 210$$ wavenumbers you can see that the population of $$n= 0$$ levels can be very large compared to that when $$n\gt 0$$.