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Demonstration [L_i,x_j]= ε_ijk x_k

  1. Jun 27, 2012 #1
    Hi!
    I have to show that
    [itex] [L_i,x_j]= i \hbar \varepsilon_{ijk} x_k [/itex]

    but my result is different, I'm definitely making a mistake :confused:
    ok I wrote
    [itex] L_i = \varepsilon_{ijk} x_j p_k [/itex]
    then
    [itex] [L_i,x_l]= \varepsilon_{ijk} ( [x_j p_k , x_l] ) = \varepsilon_{ijk} ( {x_j [p_k , x_l] + [x_j , x_l] p_k } ) = \varepsilon_{ijk} ( {x_j [p_k , x_l] } ) = [/itex]

    [itex] = \varepsilon_{ijk} ( {x_j \frac{\hbar}{i} δ_{kl} } ) = \frac{\hbar}{i} \varepsilon_{ijk} {x_j } [/itex]

    can anyone tell me where I'm wrong? :frown:
    thanks anyway! :smile:
     
  2. jcsd
  3. Jun 27, 2012 #2

    tiny-tim

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    welcome to pf!

    hi ebol! welcome to pf! :smile:
    but [itex]\frac{1}{i} \varepsilon_{ijk} {x_j } = i \varepsilon_{ijk} x_k [/itex] :wink:
     
  4. Jun 27, 2012 #3
    ah!
    and why? :)
    Because the indices [itex] j [/itex] and [itex] k [/itex] commute and changes the sign?
     
  5. Jun 27, 2012 #4

    tiny-tim

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    yup! :biggrin:

    that's what ε does!​
     
  6. Jul 3, 2012 #5
    thank you very much!
    I arrived at the solution but I did not know :D
     
  7. Jul 3, 2012 #6

    tiny-tim

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    he he :biggrin:
     
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