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Pure Vibrational Dipole Transitions

Pure vibrational transitions are between states where only vibrational quantum numbers are changed,
$\displaystyle \vert Km_K, v, \alpha\rangle \to \vert Km_K, v', \alpha\rangle.$     (2.13)

Such transitions then depend on matrix elements of the dipole operator,
$\displaystyle \langle v\vert {\bf d}_\alpha \vert v'\rangle,$     (2.14)

where \bgroup\color{col1}$ \vert v\rangle$\egroup is an harmonic oscillator eigenstate (we write \bgroup\color{col1}$ v$\egroup instead of \bgroup\color{col1}$ n$\egroup now), and
$\displaystyle {\bf d}_\alpha = \langle \alpha \vert {\bf d} \vert\alpha\rangle$     (2.15)

is the diagonal matrix element of the dipole operator between the adiabatic electronic eigenstates \bgroup\color{col1}$ \vert\alpha\rangle$\egroup.

Remember that the harmonic potential came from the Taylor expansion of the Born-Oppenheimer energy,

$\displaystyle U_\alpha(r)$ $\displaystyle \approx$ $\displaystyle U_\alpha(r_\alpha) + \frac{1}{2}\frac{d^2}{dr^2} U_\alpha(r=r_\alpha)(r-r_\alpha)^2$  
$\displaystyle \leadsto \hat{H}_{\rm osc}$ $\displaystyle =$ $\displaystyle \frac{\hat{p}^2}{2\mu}+\frac{1}{2}m\omega_\alpha^2\hat{x}^2=
\hbar\omega_\alpha\left(a^{\dagger}a+\frac{1}{2}\right)$ (2.16)
$\displaystyle \omega_\alpha^2$ $\displaystyle =$ $\displaystyle \frac{1}{\mu} \frac{d^2}{dr^2} U_\alpha(r=r_\alpha)$ (2.17)

where the harmonic oscillator coordinate \bgroup\color{col1}$ x=r-r_\alpha$\egroup.

The dipole moment operator \bgroup\color{col1}$ {\bf d}_\alpha$\egroup depends on the electronic wave functions \bgroup\color{col1}$ \alpha$\egroup and thus parametrically on the coordinate \bgroup\color{col1}$ x$\egroup that describes the internuclear separation. We Taylor-expand

$\displaystyle {\bf d}_\alpha(x) = {\bf d}_\alpha(0) + {\bf d}_\alpha'(0) x + O(x^2).$     (2.18)

For transitions between \bgroup\color{col1}$ v$\egroup and \bgroup\color{col1}$ v'\ne v$\egroup, one therefore has to linear approximation
$\displaystyle \langle v\vert {\bf d}_\alpha \vert v' \rangle$ $\displaystyle =$ $\displaystyle {\bf d}_\alpha'(0)
\langle v\vert x \vert v' \rangle
= {\bf d}_\a...
...0)\sqrt{\frac{\hbar}{2\mu\omega}}
\langle v\vert a+a^{\dagger} \vert v' \rangle$  
  $\displaystyle =$ $\displaystyle {\bf d}_\alpha'(0)\sqrt{\frac{\hbar}{2\mu\omega}}
\left(\delta_{v+1,v'} \sqrt{v+1} + \delta_{v-1,v'} \sqrt{v} \right).$ (2.19)

The vibrational selection rule thus is
$\displaystyle \Delta v = \pm 1.$     (2.20)

The corresponding energy differences determine the transition frequency,
$\displaystyle \Delta \varepsilon_{\rm vib}(v) = \hbar\omega_\alpha,$     (2.21)

which means that a purely vibrational, harmonic spectrum just consists of a single spectral line!


next up previous contents index
Next: Vibration-Rotation Spectra Up: Pure Vibration Previous: Recap of the Harmonic   Contents   Index
Tobias Brandes 2005-04-26