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Angular Momentum Question

  1. Jun 21, 2008 #1
    In http://web.utk.edu/~tbarnes/website/qm1/qm1hw_2006/hw1_2006/hw1_2006.pdf , on page 82, where it converts the momentum operators from Cartesian to spherical coordinates, I am a bit confused on how they get from 5.74-5.76 to 5.80-5.82 using the given relations. Could someone help me? Thanks a bunch :smile:
     
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  3. Jun 21, 2008 #2

    Fredrik

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    That pdf has 3 pages.
     
  4. Jun 21, 2008 #3
  5. Jun 21, 2008 #4

    Fredrik

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    That's a long and tedious calculation that starts by noting that if f depends on x,y,z and eacho of those depend on the spherical coordinates, we can write down the derivatives with respect to the spherical coordinates...

    [tex]\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} +\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta} +\frac{\partial f}{\partial z}\frac{\partial z}{\partial \theta}[/tex]

    ...and so on. This shows that the derivative operators in spherical coordinates are linear combinations of the derivative operators in cartesian coordinates, so you get a system of equations which is possible to solve by the method of of adding a (well-chosen) multiple of one equation to another (lots of times).
     
  6. Jun 22, 2008 #5
    Hmm... could an easier way be

    [tex]\frac{\partial f}{\partial z}=\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial z} +\frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial z} +\frac{\partial f}{\partial r}\frac{\partial r}{\partial z}[/tex] ?

    You can compute [tex]\partial \theta / \partial z,[/tex] for example, by differentiating implicitly with [tex]z = r \cos \theta[/tex]... would this work?
     
  7. Jun 22, 2008 #6

    Fredrik

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    Unfortunately no. There is a formula for the derivative of the inverse of the map [itex](r,\phi,\theta)\mapsto (x,y,z)[/itex], but figuring out what to put into it involves a procedure that's equivalent to the one I described. However, I forgot that it's actually quite easy to express [itex](r,\phi,\theta)[/itex] as a function of (x,y,z) instead of the other way round. Maybe it's easier to start out that way. http://en.wikipedia.org/wiki/Spherical_coordinates
     
  8. Jun 22, 2008 #7

    reilly

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    Yes, this will work. Look at the last term, with df/dr.
    The term dr/dz = z/ sqrt(x*x+Y*y+z*z), or [tex] \cos \theta[/tex].

    The theta term requires [tex]\frac{\partial \theta}{\partial z}[/tex]
    with z= r [tex] \cos \theta[/tex], from which you can compute the desired formula in spherical coords(hold r constant), and so on This is pretty much the standard way to do the problem -- grind it out.
    Regards,
    Reilly Atkinson
     
    Last edited: Jun 22, 2008
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