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Relating the angular momentum with lapacian

  1. Sep 16, 2010 #1
    The angular momentum operator is given by
    [tex]\bold{L}=\bold{r}\times-i\hbar \bold{\nabla}[/tex]


    How do we compute
    [tex]\bold{L}\cdot \bold{L}=(\bold{r}\times-i\hbar \bold{\nabla})\cdot(\bold{r}\times-i\hbar \bold{\nabla})[/tex]??? so that we can get a relation of L^2 with the lapacian operator


    i found this in a lecture note and it gave this as the first line (i might have inserted the factors -ihbar wrong)
    [tex]L^2=-\bold{r} \cdot(-i\hbar\bold{\nabla }\times (\bold{r} \times -i\hbar\bold{\nabla })[/tex] ???? (is it correct and can you help me proof it)
     
    Last edited: Sep 16, 2010
  2. jcsd
  3. Sep 17, 2010 #2
    I did a bit of work and this is what i got,
    can someone tell me if this is right and is there another way that will make this shorter!?
    [tex] \bold{L}\cdot\bold{L}=-\hbar^{2}(\epsilon_{ijk}r_{j}\nabla_{k})(\epsilon_{ilm}r_{m}\nabla_{m})[/tex]
    [tex]-\frac{L^{2}}{\hbar^{2}}=(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})r_{j}\nabla_{k}\, r_{l}\nabla_{m}[/tex]
    [tex]\\=r_{j}\nabla_{k}r_{j}\,\nabla_{k}-r_{j}\nabla_{k}\,r_{k}\nabla_{j}[/tex]
    [tex]\\=r_{j}\delta_{kj}\nabla_{k}+r_{j}r_{j}\nabla_{k}\nabla_{k}-(r_{j}\delta_{kk}\nabla_{j}+r_{j}r_{k}\nabla_{k}\nabla_{j})[/tex]
    [tex]\\=-2r_{j}\nabla_{j}+r_{j}r_{j}\nabla_{k}\nabla_{k}-r_{j}r_{k}\nabla_{j}\nabla_{k}[/tex][tex]\\=-2r_{j}\nabla_{j}+r_{j}r_{j}\nabla_{k}\nabla_{k}-r_{j}(\nabla_{j}\,r_{k}-\delta_{ij})\nabla_{k}[/tex]
    [tex]\\=-r_{j}\nabla_{j}+r_{j}r_{j}\nabla_{k}\nabla_{k}-r_{j}\nabla_{j}\,r_{k}\nabla_{k}[/tex]
    [tex]\\=-\bold{r}\cdot\nabla+r^{2}\nabla^{2}-(\bold{r}\cdot\nabla)(\bold{r}\cdot\nabla) [/tex]
     
  4. Sep 18, 2010 #3
    See Landau, Vol 3, 3rd edition, section 28 "eigenfunctions of the angular momentum". For a mathematical derivation, see "Lie Groups, Lie Algebras, and Representations" by Brian C Hall.
     
  5. Sep 18, 2010 #4
    thank you so much. However i didnt find much in Landau, under which section will i find the mathematical derivation in your second reference by hall
     
  6. Sep 19, 2010 #5
    I am sad that I don't have a copy of Hall, so I can't give you the reference. But I studied that book with the guidance of my mentor, and I remember the proof in there was really nice. The thing is, most people just transform the laplacian to spherical polar coordinates and wave their hands to say that's the angular momentum eigenfunctions. But in Hall, there is no hand waving, and in order to really appreciate the proof you kind of have to work through what comes before. You could do a lot worse than to study this book from the beginning. Lie groups/algebra's are the most important part of math for physics (they underlie the relationship between conservation laws and symmetry, as well as the fact that the lie algebra elements ARE the quantum particles) and this is far and away the best book on lie groups for beginners, imo. I skimmed through basically every lie group book in the math/physics libraries and this is the one I liked the most.
     
  7. Sep 19, 2010 #6

    Meir Achuz

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    I got your answer by a different method, but I can't say my method is shorter.
     
  8. Sep 19, 2010 #7
    We may use the method introduced by Feynman, that will make it shorter.

    See Feynman Vol2, derivation of Poynting's Vector. There he describes a handy method to use the operator like a normal vector, and everythig else follows directly........
     
  9. Sep 20, 2010 #8

    Meir Achuz

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    That was the method I used. It avoids all the indices, but is still a bit complicated.
     
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