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Units

Absorption or emission of infrared light can cause transitions between energy levels in the harmonic oscillator. Usually, only transitions between adjacent energy levels ($\Delta n = 1$) occurs with a large intensity. The frequency of the photon $\nu_{photon}$ thus needs to match
$\displaystyle h \nu_{photon}$ $\textstyle =$ $\displaystyle \Delta E = h \nu.$ (20)

In this particular case, the frequency of the photon $\nu_{photon}$ must be the same as the frequency $\nu$ of the oscillator. Often, $\nu_{photon}$ is measured in wavenumbers (cm$^{-1}$) instead of Hertz. Since the photon obeys $c = \lambda \nu$, the reciprocal wavelength of the photon $1/\lambda$ can be written as
$\displaystyle \frac{1}{\lambda}$ $\textstyle =$ $\displaystyle \frac{\nu}{c}
= \frac{\omega}{2 \pi c}
= \frac{1}{2 \pi c} \sqrt{ \frac{k}{\mu} }.$ (21)

Note that units of cm$^{-1}$ will result for $1/\lambda$ if, for example, $c$ is in cm s$^{-1}$, $k$ is in J m$^{-2}$, and $\mu$ is in kg. Because of the numerical equivalence between $\nu_{photon}$ and $\nu$, one also frequently reports $\nu$ in units of cm$^{-1}$ using the above conversion. Unfortunately, spectroscopists frequently denote this $\nu$ as $\omega$, which as we have seen is defined differently above ($\nu$ and $\omega$ differ by a factor of $2 \pi$). When $\omega$ is reported in cm$^{-1}$, this is actually $\nu$ in cm$^{-1}$, not the $\omega$ defined above. One reason for this unfortunate discrepancy in notation is that spectroscopists prefer to use $\nu$ to refer to fundamental (actually observed) frequencies, which differ from the harmonic (model) frequencies because the potential wells in diatomic molecules are not strictly harmonic but contain an anharmonic contribution.

What are the units of $k$? The SI units would be J m$^{-2}$, but these are not really natural units for the very small energies and distances involved in a quantum oscillator. Frequently, these are reported in mdyn Å$^{-1}$. The conversion is

$\displaystyle \frac{\rm mdyn}{\rm\AA}$ $\textstyle =$ $\displaystyle \frac{ 10^{-8} {\rm N} }{ 10^{-10} {\rm m} }
\times \frac{\rm m}{\rm m}
= 10^2 \frac{\rm J}{{\rm m}^2}.$ (22)


next up previous
Next: About this document ... Up: Quantum Harmonic Oscillator Previous: Quantum Harmonic Oscillator
David Sherrill 2002-10-16