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Angular Momentum Operators

  1. Mar 9, 2009 #1
    How does one obtain the formulae for the angular momentum operators in spherical polar coordinates i.e.

    [itex]\hat{L_x}=i \hbar (\sin{\phi} \frac{\partial}{\partial{\theta}} + \cot{\theta} \cos{\phi} \frac{\partial}{\partial{\phi}}[/itex]
    [itex]\hat{L_y}=i \hbar (-\cos{\phi}{\phi} \frac{\partial}{\partial{\theta}} + \cot{\theta} \sin{\phi} \frac{\partial}{\partial{\phi}}[/itex]
    [itex]\hat{L_z}=-i \hbar \frac{\partial}{\partial{\phi}}[/itex]
    [itex]\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][/itex]

    ???
     
  2. jcsd
  3. Mar 10, 2009 #2

    malawi_glenn

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    Use the del-operator in spherical coordinates.
     
  4. Mar 10, 2009 #3
    so [itex]\hat{L}=\hat{r} \times \hat{p}[/itex]

    where [itex]\hat{p}=-i \hbar \nabla[/itex]

    do i use [itex]\nabla[/itex] 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?
     
  5. Mar 10, 2009 #4

    malawi_glenn

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    have you even tried?
     
  6. Mar 10, 2009 #5
    yes. my notes run through it for the Cartesian case and subsitute r=(x,y,z) and [itex]\nabla=(\partial_x,\partial_y,\partial_z)[/itex].
    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 [itex]\hat{L_r},\hat{L_\theta},\hat{L_\phi}[/itex] rather than [itex]\hat{L_x},\hat{L_y},\hat{L_z}[/itex], no???
     
  7. Mar 10, 2009 #6

    malawi_glenn

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