Spherical Coordinates and the Angular Momentum Operators

The transformation from spherical coordinates to Cartesian coordinate is.

$$\begin{eqnarray*} x&=&r\sin\theta\cos\phi \\ y&=&r\sin\theta\sin\phi \\ z&=&r\cos\theta \\ \end{eqnarray*}$$

The transformation from Cartesian coordinates to spherical coordinates is.

$$\begin{eqnarray*} r&=&\sqrt{x^2+y^2+z^2} \\ \cos\theta&=&{z\over\sqrt{x^2+y^2+z^2}} \\ \tan\phi&=&{y\over x} \\ \end{eqnarray*}$$

We now proceed to calculate the angular momentum operators in spherical coordinates. The first step is to write the ${\partial\over\partial x_i}$ in spherical coordinates. We use the chain rule and the above transformation from Cartesian to spherical. We have used $d\cos\theta=-\sin\theta d\theta$ and $d\tan\phi={1\over\cos^2\phi}d\phi$. Ultimately all of these should be written in the sperical cooridnates but its convenient to use $x$ for example in intermediate steps of the calculation.

$$\begin{eqnarray*} {\partial\over\partial x}&=&{\partial r\over\partial x}{\part... ...tial r} -{1\over r}\sin\theta{\partial\over\partial\theta} \\ \end{eqnarray*}$$

Bringing together the above results, we have.

$$\begin{eqnarray*} {\partial\over\partial x}&=&\sin\theta\cos\phi{\partial \over... ...tial r} -{1\over r}\sin\theta{\partial\over\partial\theta} \\ \end{eqnarray*}$$

Now simply plug these into the angular momentum formulae.

$$\begin{eqnarray*} L_z&=&{\hbar\over i}\left(x{\partial\over\partial y}-y{\parti... ...al \theta} +i\cot\theta{\partial\over\partial \phi}\right) \\ \end{eqnarray*}$$

We will use these results to find the actual eigenfunctions of angular momentum.

$$\begin{eqnarray*} L_z&=&{\hbar\over i}{\partial\over\partial\phi} \\ L_\pm&=&... ...al \theta} +i\cot\theta{\partial\over\partial \phi}\right) \\ \end{eqnarray*}$$