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Quantum Angular momentum Question

  1. Feb 9, 2008 #1

    malawi_glenn

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    [SOLVED] Quantum Angular momentum Question

    1. The problem statement, all variables and given/known data

    Show that [tex] \dfrac{{\vec{p}} ^2 }{2m} = \dfrac{{\vec{l} }^2}{2mr^2} - \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r}) [/tex] is rotational invariant under the rotation generated by: [tex] \vec{j} = \vec{l} + \vec{s} [/tex] , s is intrinic spin.

    2. Relevant equations

    [H,J] = 0 and/or [H,J^2] = 0


    3. The attempt at a solution

    I think that the second part, the radial operator [tex] \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r}) [/tex] commutes with the angular momentas, since it is just a function of radial coordinate, whereas angular momenta depends on the angles (direction) Is that correct?
     
    Last edited: Feb 9, 2008
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  3. Feb 9, 2008 #2

    kdv

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    Yes, any function of r i strivially invariant under rotation since [tex] [f(r), L_i] = 0 [/tex] because the [tex] L_i [/tex] depend only on the angles, as you say.
     
  4. Feb 9, 2008 #3

    malawi_glenn

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    Just as I thought then, Didn't wanna search or work out the differential form of J before I was sure :) Thanx!
     
  5. Feb 9, 2008 #4

    kdv

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    You're welcome.

    They are given in spherical coordinates toward the middle of the page at xbeams.chem.yale.edu/~batista/vvv/node16.html
     
  6. Feb 9, 2008 #5

    malawi_glenn

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    Thanx!

    Was wondering if you know if spin also have differential form? and j (total angular mom.). Or if you must work em out using group theory?
     
  7. Feb 9, 2008 #6

    kdv

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    You have to work using group theory. The differential form approach only produces the integer angular momentum representations. It's only by working with the abstract formalism of commutator and operators that one can generate all the spin representation including the half integer ones. For the spatial angular momentum calculations, one has the choice of working with explicit spatial wavefunctions or with matrices and column vectors, etc. For spin, one must work with the matrix representations.
     
  8. Feb 9, 2008 #7

    malawi_glenn

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    Ok I got it :)

    So how can I argue that f(r) and s commutes? same as with L, that s only depends on direction?
     
  9. Feb 9, 2008 #8

    kdv

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    They commute but the reason is that s acts on a totally different space, so they commute trivially. The total Hilbert space is a direct product of the spin space and the Hilbert space of spatial wavefunctions. Any operator acting in one of the space commutes with any operator acting in the other space.
     
  10. Feb 9, 2008 #9

    malawi_glenn

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    yeah, of course.. I have done angular momentum in QM now in 6h.. maybe shall go and cook some food ;) this one was so obviuos, I should be ashamed..

    Thanx again :) Next time I'll help you
     
  11. Feb 9, 2008 #10

    kdv

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    I am glad I could help!
    I know the feeling of not seeing simple things after having worked for several hours.
    I am sure you'll help me at some point with something in differential geometry/GR/supersymmetry/string theory/etc etc :-)
     
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