Question

In Appendix G. the operator for the square of the angular momentum is shown to be

${\hat{\mathbf{L}}}^{\mathbf{2}}\mathbf{=}\mathbf{-}{\sout{\mathbf{h}}}^{\mathbf{2}}\left[csc\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+cs{c}^2\theta \frac{{\partial }^2}{\partial {\varphi }^2}\right]$

Use this to rewrite equation (7-19) as${\hat{\mathbf{L}}}^{\mathbf{2}}\mathbf{⚇}\mathit{\Phi }\mathbf{=}\mathbf{-}\mathit{C}{\sout{\mathbf{h}}}^{\mathbf{2}}\mathbf{⚇}\mathit{\Phi }$

Short answer

${\hat{L}}^2\theta \Phi =-C{\sout{h}}^2\theta \Phi$

Step-by-step solution

 Given data

The angular momentum operator is given as:

${\hat{L}}^2=-{\sout{h}}^2[csc\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+cs{c}^2\theta \frac{{\partial }^2}{\partial {\varphi }^2}]$

Where, $\sout{h}$ is the reduced Planck’s constant.

 Calculation

The separation of the variable of the above equation can be written as:

$$\begin{gathered} \left(csc\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial }{\partial \theta }\right)+cs{c}^2\theta \frac{{\partial }^2}{\partial {\varphi }^2}\right)\Theta \Phi =C\Theta \Phi \\ \left(csc\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial \Theta \Phi }{\partial \theta }\right)+cs{c}^2\theta \frac{{\partial }^2\Theta \Phi }{\partial {\varphi }^2}\right)=C\Theta \Phi \\ \frac{1}{\Theta }csc\theta \frac{\partial }{\partial \theta }\left(sin\theta \frac{\partial \Theta }{\partial \theta }\right)+cs{c}^2\theta \frac{1}{\Phi }\frac{{\partial }^2\Phi }{\partial {\varphi }^2}=C \end{gathered}$$

Now we can write the operator as:

$$\begin{gathered} {\hat{L}}^2=-{\sout{h}}^2\left[csc\theta \frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)cs{c}^2\theta \frac{{\partial }^2}{\partial {\varphi }^2}\right] \\ {\hat{L}}^2⊝\Phi =-{\sout{h}}^2\left[csc\theta \frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)cs{c}^2\theta \frac{{\partial }^2}{\partial {\varphi }^2}\right]⊝\Phi \\ =-{\sout{h}}^2\Phi csc\theta \frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)-{\sout{h}}^2\Phi cs{c}^2\theta \frac{{\partial }^2\Phi }{\partial {\varphi }^2} \\ =-{\sout{h}}^2⊝\Phi \frac{1}{⊝}csc\theta \frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial \Phi }{\partial \theta }\right)cs{c}^2\theta \frac{1}{\Phi }\frac{{\partial }^2\Phi }{\partial {\varphi }^2} \\ {\hat{L}}^2⊝\Phi =-C{\sout{h}}^2⊝\Phi \end{gathered}$$

 Conclusion

Thus, it can be written as ${\hat{L}}^2⊝\Phi =-C{\sout{h}}^2⊝\Phi$.