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We need to form L2, i.e.,

\begin{displaymath}L^2 = L_x \cdot L_x
+ L_y \cdot L_y
+ L_z \cdot L_z
\end{displaymath} (23)


\begin{displaymath}L_x^2 = - \hbar^2
\left (
\sin \phi \frac{\partial}{\partial...
...\cos \phi}{\sin\theta} \frac{\partial}{\partial \phi}
\right )
\end{displaymath} (24)


\begin{displaymath}L_y^2 = - \hbar^2
\left (
\cos \phi \frac{\partial}{\partial...
...\sin \phi}{\sin\theta} \frac{\partial}{\partial \phi}
\right )
\end{displaymath} (25)


\begin{displaymath}L_z^2 = - \hbar^2
\frac{\partial
\frac{\partial}{\partial \phi}
}{\partial \phi}
\end{displaymath} (26)

For Lx2 we have, upon expansion

 
$\displaystyle \frac{L_x^2 }{ - \hbar^2}=
\sin \phi \frac{\partial \sin \phi \fr...
...}{\partial \theta} \Rightarrow \sin^2 \phi \frac{\partial^2}{\partial \theta^2}$      
$\displaystyle + \frac{\cos \theta \cos \phi}{\sin\theta} \frac{\partial\sin \ph...
... \cot \theta \cos \phi \sin \phi\frac{\partial^2}{\partial
\theta\partial \phi}$      
$\displaystyle \sin \phi \frac{\partial \frac{\cos \theta \cos \phi}{\sin\theta}...
...\cos \phi \sin \phi}{\sin\theta}\frac{\partial^2}{\partial \phi\partial \theta}$      
$\displaystyle + \frac{\cos \theta \cos \phi}{\sin\theta} \frac{\partial \frac{\...
...rac{\cos^2 \theta \cos^2 \phi}{\sin^2\theta}
\frac{\partial^2}{\partial \phi^2}$     (27)

For Ly we have

 
$\displaystyle \frac{L_y^2}{ - \hbar^2 }=
\cos \phi \frac{\partial \cos \phi \fr...
...}}{\partial \theta}
\Rightarrow \cos^2\phi \frac{\partial^2}{\partial \theta^2}$      
$\displaystyle - \frac{\cos \theta \sin \phi}{\sin\theta} \frac{\partial \cos \p...
...in \phi}{\sin\theta} \cos \phi
\frac{\partial^2}{\partial \theta \partial \phi}$      
$\displaystyle -\cos \phi \frac{\partial\frac{\cos \theta \sin \phi}{\sin\theta}...
...s \theta \sin \phi}{\sin\theta}
\frac{\partial^2}{\partial \phi\partial \theta}$      
$\displaystyle + \frac{\cos \theta \sin \phi}{\sin\theta} \frac{\partial \frac{\...
...rac{\cos^2 \theta \sin^2 \phi}{\sin^2\theta}
\frac{\partial^2}{\partial \phi^2}$     (28)

and, of course

   \begin{displaymath}
\frac{L_y^2}{ - \hbar^2 }=
\frac{\partial^2}{\partial \phi^2}
\end{displaymath}

It follows, adding Equations 27, 28 and 7, that

\begin{eqnarray*}\sin^2 \phi \frac{\partial^2}{\partial \theta^2}
\nonumber \\
...
...tial \phi^2}
\nonumber \\
+ \frac{\partial^2}{\partial \phi^2}
\end{eqnarray*}



next up previous
Next: . Up: Alternative Formulations for Angular Operators Previous: .

2001-12-26