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Commutativity Equation Of Hamilton and Position Operators

  1. Sep 18, 2007 #1
    How can we show [tex]\left[\hat{H},\hat{x}\right]=\frac{-i\hbar}{m} \hat{p_{x}}[/tex] ?
     
  2. jcsd
  3. Sep 18, 2007 #2
    is this regular ?

    [tex]\left[\hat{p_{x}}^2,\hat{x}\right]=\left[\hat{p_{x}},\hat{x}\right]\hat{p_{x}}+\hat{p_{x}}\left[\hat{x},\hat{p_{x}}\right][/tex] ( from leibniz rule)
     
  4. Sep 18, 2007 #3

    cristo

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    Well, do you know the form of the Hamiltonian?
     
  5. Sep 18, 2007 #4
    [tex][p^2,x] = ppx - xpp = p[p,x]-pxp - [x,p]p + pxp[/tex]
     
  6. Sep 18, 2007 #5

    cristo

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    Yes, that is correct.
     
  7. Sep 18, 2007 #6
    if we know [tex]\left[\hat{p}_x,\hat{x}\right]=-i\hbar[/tex] we can show this equation. Thanks my friends for your helps.
     
  8. Sep 18, 2007 #7

    dextercioby

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    So how does the Hamiltonian look like ?
     
  9. Sep 19, 2007 #8
    Yes, but you will have to start with the Hamiltonian operator first,
    [tex]\hat H = {\hat {p}^2 \over {2m}} + V(\vec r)}[/tex]
    and arrive at your result.
     
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