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I No r dependence in L operator?

  1. Feb 20, 2017 #1
    In classical mechanics, angular momentum, L = r x p, depends of r. For a given momentum p, the bigger r is, the bigger is the angular momentum. Event in spherical coordinates, r still appears in the classical angular momentum.

    However, the angular momentum operator in QM has no r dependence, it's only a function of angles phi and theta. While I can follow the derivation of the operator, I find it surprising that the r dependency vanishes. How can we explain this physically ?

    80822983L.png
     
  2. jcsd
  3. Feb 20, 2017 #2

    BvU

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  4. Feb 20, 2017 #3
    Thank you. Is there a specific part adressing my question ? This seem to be quite complete course on QM angular momentum, but after a quick search all I have found regarding r, is around the end when it says they haven't adressed r yet. But they refer to the case of the hydrogen atom I think, which has a hamiltonian with a r dependent potential.
     
  5. Feb 20, 2017 #4

    BvU

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    First page
     
  6. Feb 20, 2017 #5
    After reading it I don't see where is it ? It shows how we derive the QM version of angular momentum from the classical picture. This doesn't bother me. And even if the maths of this derivation also works out for the angular momentum operator, I just wonder, on a conceptual level, why the r dependence vanishes.
     
  7. Feb 20, 2017 #6

    Nugatory

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    The ##x##, ##y##, and ##z## that appear in the first page are the components of ##\vec{r}##. So it's there.
     
  8. Feb 20, 2017 #7
    Right, I should have been more precise : why r vanishes in the spherical coordinate version of L ?
     
  9. Feb 20, 2017 #8

    BvU

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    Math, math and math. The transition from cartesian to spherical makes it look like a magic disappearance, I concede.

    http://quantummechanics.ucsd.edu/ph130a/130_notes/node216.html

    My hunch is the dependence sits in the wavefunction itself, but I quit trying to form such mental pictures a long time ago.
    I do feel with you, so let's ask @Nugatory :smile: whether this is only seemingly counter-intuitive, or whether there's some other insight we overlook ...
     
    Last edited: Feb 20, 2017
  10. Feb 20, 2017 #9
    Exactly. Good to know I am not alone. :D I came accross the same link BTW searching an answer on Google.
     
  11. Feb 20, 2017 #10
    What adds to the weirdness is that angular momentum in classical mechanics, even if expressed in spherical coordinates, does have a r dependency :

    853522classicalLspherical.png

    Source (page 11).
     
  12. Feb 20, 2017 #11

    blue_leaf77

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    Since it has been confirmed by many experiments that such formulation of angular momentum operator does work, I think I will just accept it as granted. Besides that, angular momentum is defined as the generator of rotation in space in quantum mechanics, so I think it makes sense that it should not depend on r.
     
  13. Feb 20, 2017 #12

    PeterDonis

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    The quantum operator has no ##r## dependence, but that does not mean the actual measured values of angular momentum have no ##r## dependence. The latter is what you would expect to be analogous to the classical angular momentum.
     
  14. Feb 20, 2017 #13
    Right ! So, as @BvU suggested, the r dependence would be in the wavefunction ? But...The eigenstates of the angular momentum operator are spherical harmonics, which are not functions of r, but of theta and phi only.
     
  15. Feb 20, 2017 #14

    PeterDonis

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    No, they are spherical harmonics combined with a function of ##r## only.
     
  16. Feb 20, 2017 #15
    Do you mean if the particle is in a r-dependent potential ? The courses I read present the eigenstates of L as being solely spherical harmonics. After a quick search on Google, same thing.
     
  17. Feb 20, 2017 #16

    BvU

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    central potential, e.g. hydrogen atom allows separation into radial and angular eigenfunctions. L is 'on the angular side' :smile:
     
  18. Feb 20, 2017 #17
    Yes I have seen that separation in the case of the hydrogen atom. But I'm trying to consider L just for itself, outside any specific potential considerations. Maybe this doesn't make sense then ?
     
  19. Feb 20, 2017 #18
    Well, I suppose I am considering the case of a free particle then. And it is true that even in classical mechanics, the angular momentum of a free particle is a constant, so not really dependent of r. Mmh, is this a viable answer ? I'm not sure at all. :D
     
  20. Feb 20, 2017 #19

    PeterDonis

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    No, it doesn't. Any quantum system will have some Hamiltonian, and therefore some potential. That determines the energy eigenstates, and it's meaningless to talk about the action of any operator, including L, without some specification of the states it's acting on.
     
  21. Feb 20, 2017 #20
    What about the free particle case ?
     
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