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Coulomb potential as an operator

  1. Feb 9, 2016 #1

    ShayanJ

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    I want to calculate the commutator ##{\Large [p_i,\frac{x_j}{r}]}## but I have no idea how I should work with the operator ##{\Large\frac{x_j}{r} }##.
    Is it ## x_j \frac 1 r ## or ## \frac 1 r x_j ##? Or these two are equal?
    How can I calculate ##{\Large [p_i,\frac 1 r]}##?
    Thanks
     
  2. jcsd
  3. Feb 9, 2016 #2
    My first thought is that the momentum operator act as a derivative on a function of the operator r
     
  4. Feb 9, 2016 #3

    A. Neumaier

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    Since the ##x_j## commute and ##r^2=\sum x_j^2##, ##x## and ##r## commute. You can differentiate the relation defining ##r## to find the partial derivatives, and then use ##[p_k,f(x)]=-i\hbar \partial f(x)/\partial x_k##. (Don't use ##i## as index if it appears as ##\sqrt{-1}##, too.)
     
    Last edited: Feb 10, 2016
  5. Feb 9, 2016 #4

    ShayanJ

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    I've been able to prove ## [p_k,f(x)]=-i\hbar \frac{df(x)}{dx_k} ## only for the case of ## f(x) ## being a polynomial in ## x_k ##. But here ## \frac 1 r ## doesn't satisfy this condition. Is there a general proof?
     
    Last edited: Feb 9, 2016
  6. Feb 9, 2016 #5

    blue_leaf77

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    Actually the differential operator should be of partial one.
    The commutator suggested by Neumaier is applicable to any smooth function ##f(x)##, not only polynomials.
     
  7. Feb 9, 2016 #6

    ShayanJ

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    ## f(x) ## is a function of positon operator, not a smooth function of a real variable!
     
  8. Feb 9, 2016 #7

    blue_leaf77

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    Who said that ##f(x)## is a function of real variable?
     
  9. Feb 9, 2016 #8

    ShayanJ

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    I got the impression that you meant it was trivial since you didn't provide any proof, so I thought that's what you mean. Do you know where can I find a proof?
     
  10. Feb 9, 2016 #9

    blue_leaf77

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    Consider
    $$
    [p,x^n] = -i\hbar nx^{n-1}
    $$
    and the Taylor expansion of ##f(x)##.
     
  11. Feb 9, 2016 #10

    ShayanJ

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    A while ago I asked about this and micromass didn't seem to agree with you.(See here!)
    Actually that's why I didn't think about using the formula in the first place.
     
  12. Feb 9, 2016 #11

    blue_leaf77

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    It seems that Taylor expansion of a function operator cannot be taken lightly, I have actually also addressed exactly the same problem on this regard almost a year ago There I asked if a function ##f(x)=(1+x)^{-1}## can be expanded in power terms, as the same function with ##x## real numbers can only have Taylor expansion for a bounded x ##-1<x<1##.
     
  13. Feb 9, 2016 #12

    ShayanJ

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    Is there any proof that doesn't use Taylor expansion of operator functions?
     
  14. Feb 9, 2016 #13

    strangerep

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    Well, it starts from Taylor expansion of ordinary functions. But yes, one can extend to very general (complex analytic) functions.

    To get started, restrict to the case where an (ordinary, polynomial) operator ##A## has an inverse ##A^{-1}##. I.e., ##A A^{-1} = 1##. Then take the commutator with ##p## on both sides, and use the Leibniz rule on the LHS. What do you find? :oldwink:


    [And thank you to @A. Neumaier for teaching me this generalized material ages ago.]
     
  15. Feb 9, 2016 #14

    ShayanJ

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    For any operator B, I find ## [B,A]A^{-1}=A[A^{-1},B] ##! So?
     
  16. Feb 9, 2016 #15

    strangerep

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    Take it further...

    When ##A = A(x)## (ordinary polynomial), and ##B = p##, and you've already found that ##[p, A(x)] = A'(x)## (modulo factors of ##i## and ##\hbar##, depending on your conventions),... then what is $$[p, A^{-1}(x)] ~=~ ?$$
     
  17. Feb 9, 2016 #16

    ShayanJ

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    I don't see how ## [p,A]A^{-1}=A[A^{-1},p] ## implies ##[p,A(x)]\propto A'(x)##!
     
  18. Feb 9, 2016 #17

    strangerep

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    You already had that back in post #4. Now use that to find ##[p,A^{-1}] = ?##
     
  19. Feb 9, 2016 #18

    ShayanJ

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    ## [p,A(x)^{-1}]=i\hbar A(x)^{-1} A'(x) A^{-1}(x)##
     
  20. Feb 9, 2016 #19

    strangerep

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    You keep stopping too soon.

    Let ##F(x) := A^{-1}(x)##. From the above, what is ##[p, F(x)] = \; ?##
     
  21. Feb 9, 2016 #20
    Can you be sure that A^-1 commutes with A' ?
     
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