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(*restored equation*)

$\begin{displaymath}-\hbar e^{-\imath \phi} \left ( \frac{\partial}{\partial \theta} - \imath \cot \theta \frac{\partial}{\partial \phi} \right ) \end{displaymath}$

Therefore, using the same example as before, we have

$\begin{displaymath}L^- = -\hbar e^{-\imath \phi} \left ( \frac{\partial}{\parti... ...phi} \right ) \cos \theta = -sin \theta \hbar e^{-\imath \phi} \end{displaymath}$

To clinch the example, let's up ladder on $3-cos^2\theta$ . For the up laddering we have

$\begin{displaymath}-\hbar e^{+\imath \phi} \left ( \frac{\partial}{\partial \the... ...\theta \frac{\partial}{\partial \phi} \right )(3-\cos^2\theta) \end{displaymath}$

(49)

which is

$\begin{displaymath}-\hbar e^{+\imath \phi} \left ( \frac{\partial}{\partial \the... ...) \vert 2,0> = -2 \cos \theta \sin \theta\hbar e^{\imath \phi} \end{displaymath}$ = -2 \cos \theta \sin \theta\hbar e^{\imath \phi} \end{displaymath}">

(50)

i.e., |2,-1>. We do this again, i.e.,

$\begin{displaymath}L^-\vert 2,+1> = -\hbar e^{+\imath \phi} \left ( \frac{\parti... ...left ( -K \cos \theta \sin \theta\hbar e^{\imath \phi}\right ) \end{displaymath}$ = -\hbar e^{+\imath \phi} \left ( \frac{\parti... ...left ( -K \cos \theta \sin \theta\hbar e^{\imath \phi}\right ) \end{displaymath}">

2001-12-26