.

(*restored equation*)

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

Downladdering proceeds in the same way. For L^- we have

$$L_x -\imath L_y = \imath \hbar \left ( \sin \phi \frac{\par... ...\sin \phi}{\sin\theta} \frac{\partial}{\partial \phi} \right )$$ (48)

Again, grouping by partial derivatives we obtain

$$L_x -\imath L_y = \left ( \imath \hbar \sin \phi - \hbar \co... ...+ \hbar \cot \sin \phi \right ) \frac{\partial}{\partial \phi}$$

which is,

$$L_x -\imath L_y = \hbar\left ( \imath \sin \phi - \cos \phi ... ...+ \sin \phi \right )\cot \theta \frac{\partial}{\partial \phi}$$