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Angular Momentum

Neglecting the geometric phase terms, Eq. (E.5.2), we have in three spatial dimensions
$\displaystyle \mathcal{H}_{{\rm n},\alpha}^{\rm rel}$ $\displaystyle =$ $\displaystyle \frac{{\bf p}^2}{2\mu} + E_\alpha(r)
= - \frac{\hbar^2}{2\mu}\Delta_{\bf r} + E_\alpha(r)$  
  $\displaystyle =$ $\displaystyle - \frac{\hbar^2}{2\mu} \left(\frac{\partial^2}{\partial r^2}+\fra...
...}\frac{\partial}{\partial r}\right)
+ \frac{{\bf J}^2}{2\mu r^2} + E_\alpha(r),$ (1.13)

where \bgroup\color{col1}$ {\bf J}$\egroup is the relative angular momentum operator of the nuclei. We have a three-dimensional problem which however due to the radial symmetry of \bgroup\color{col1}$ E_\alpha(r)$\egroup is reduced to a one-dimensional radial eqaution, very much as for the case of the hydrogen atom! We could write the eigenfunctions of \bgroup\color{col1}$ \mathcal{H}_{{\rm n},\alpha}^{\rm rel} $\egroup as
$\displaystyle \Psi({\bf r}) = R(r) Y_{JM}(\theta,\phi)$     (1.14)

with the corresponding angular quantum numbers \bgroup\color{col1}$ J$\egroup and \bgroup\color{col1}$ M$\egroup of the nuclear relative motion separated off in the spherical harmonics.

Instead of dealing with the angular momentum operator of the nuclei, one would rather descrive rotations of the whole molecule by the total angular momentum \bgroup\color{col1}$ {\bf K}$\egroup of the molecule

$\displaystyle {\bf K} = {\bf J} + {\bf L} + {\bf S},$     (1.15)

where \bgroup\color{col1}$ {\bf L}$\egroup is the total angular momentum of all electrons and \bgroup\color{col1}$ {\bf S}$\egroup is the total spin.



Subsections
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Next: Spin Up: Vibrations and Rotations in Previous: Born-Oppenheimer Approximation   Contents   Index
Tobias Brandes 2005-04-26