Spherical Coordinates

Since angular momentum can be represented as a generator of rotations you can use the equation for an infinitesmial rotation to construct the coordinates of angular momentum in spherical coordinates.

The cartesian coordinates x,y,z can be written in spherical as follows:

$x=r\sin \theta \cos \phi \!$ $y=r\sin \theta \sin \phi \!$ $z=r\cos \theta \!$

Denote the state $\langle \mathbf {r} \!\,|=\langle \mathbf {r} \!\,\theta \phi |$

If you choose $\alpha \!$ along the z-axis then the only coordinate that will change is phi such that $\phi \rightarrow \phi +\alpha$. Now the state is written as:

$\langle \mathbf {r} \!\,\theta \phi |\left(1+{\frac {i}{\hbar }}\alpha L_{z}\right)=\langle \mathbf {r} \!\,\theta \phi +\alpha |$

Working to first order in alpha the right hand side becomes:

$\langle \mathbf {r} \!\,\theta \phi |+\alpha {\frac {\partial }{\partial \phi }}\langle \mathbf {r} \!\,\theta \phi |$

Therefore $L_{z}={\frac {\hbar }{i}}{\frac {\partial }{\partial \phi }}$

Now choose $\alpha \!$ along the x-axis then the cartesian coordinates are changed such that: $x\rightarrow x\!$ , $y\rightarrow y-\alpha z\!$, and $z\rightarrow z+\alpha y\!$,

from these transformations it can be determined that $\delta \theta =-\alpha \sin \phi \!$ since $\delta z=\alpha y\!$ and since x does not change it can be determined that $\delta \phi =-\alpha \cot \theta \cos \phi \!$.

This means that the original state is now written as: $\langle \mathbf {r} \!\,\theta \phi |\left(1+{\frac {i}{\hbar }}\alpha L_{x}\right)=\langle \mathbf {r} \!\,\,\theta -\alpha \sin \phi \,\phi -\alpha \cot \theta \cos \phi |$

Expanding the right hand side of the above equation as before to the first order of alpha the whole equation becomes: $\langle \mathbf {r} \!\,\theta \phi |L_{x}={\frac {\hbar }{i}}\left(-\sin \phi {\frac {\partial }{\partial \theta }}\,-\cot \theta \cos \phi {\frac {\partial }{\partial \phi }}\!\right)\langle \mathbf {r} \!\,\theta \phi |$

Therfore $L_{x}={\frac {\hbar }{i}}\left(-\sin \phi {\frac {\partial }{\partial \theta }}\,-\cot \theta \cos \phi {\frac {\partial }{\partial \phi }}\!\right)$

Using the same techinque, choose $\alpha \!$ along the y-axis and the coordinates will change in a similar fashion such that it can be shown that $L_{y}={\frac {\hbar }{i}}\left(\cos \phi {\frac {\partial }{\partial \theta }}-\cot \theta \sin \phi {\frac {\partial }{\partial \phi }}\!\right)$