# Angular Momentum Operators

1. Mar 9, 2009

### latentcorpse

How does one obtain the formulae for the angular momentum operators in spherical polar coordinates i.e.

$\hat{L_x}=i \hbar (\sin{\phi} \frac{\partial}{\partial{\theta}} + \cot{\theta} \cos{\phi} \frac{\partial}{\partial{\phi}}$
$\hat{L_y}=i \hbar (-\cos{\phi}{\phi} \frac{\partial}{\partial{\theta}} + \cot{\theta} \sin{\phi} \frac{\partial}{\partial{\phi}}$
$\hat{L_z}=-i \hbar \frac{\partial}{\partial{\phi}}$
$\hat{L}^2=\hbar^2 \left[\frac{1}{\sin{\theta}} \frac{\partial}{\partial{\theta}} \left(\sin{\theta} \frac{\partial}{\partial{\theta}} \right) +\frac{1}{\sin^2{\theta}} \frac{\partial^2}{\partial{\phi}} \right]$

???

2. Mar 10, 2009

### malawi_glenn

Use the del-operator in spherical coordinates.

3. Mar 10, 2009

### latentcorpse

so $\hat{L}=\hat{r} \times \hat{p}$

where $\hat{p}=-i \hbar \nabla$

do i use $\nabla$ as the gradient operator in spherical polars?

also do i write r in terms of x,y,z or in terms of r,theta,phi?

4. Mar 10, 2009

### malawi_glenn

have you even tried?

5. Mar 10, 2009

### latentcorpse

yes. my notes run through it for the Cartesian case and subsitute r=(x,y,z) and $\nabla=(\partial_x,\partial_y,\partial_z)$.
regardelss of which of the combinations i of r and del iuse above i can't get the right answer.

surely if i sub in for del in spherical polars i'm actually computing $\hat{L_r},\hat{L_\theta},\hat{L_\phi}$ rather than $\hat{L_x},\hat{L_y},\hat{L_z}$, no???

6. Mar 10, 2009