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Can't make [L_z,p^2]=0

  1. Dec 5, 2007 #1
    I think [L_z,p^2] is supposed to equal zero, but when I'm getting

    = x*(p_y)*(p_x)^2 - y*(p_x)*(p_y)^2

    and that doesn't appear to equal zero. Could someone please show a couple steps to help me out?
     
  2. jcsd
  3. Dec 5, 2007 #2

    cristo

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    What method have you used? You want to calculate the commutator [itex][L_z,p^2]=[xp_y-yp_x,p^2][/tex]. You need to expand this and then simplify.
     
  4. Dec 5, 2007 #3

    mjsd

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    isn't p stands for something like
    [tex]p_x = -i\hbar \frac{\partial}{\partial x}[/tex]
    ?
     
  5. Dec 5, 2007 #4
    These are the steps I used:

    [itex] [L_z,p^2]=[L_z,p_x^2+p_y^2+p_z^2] [/itex]
    [itex]= [L_z,p_x^2]+[L_z,p_y^2][/itex]
    [itex]= [xp_y,p_x^2]-[yp_x,p_x^2]+[xp_y,p_y^2]-[yp_x,p_y^2][/itex]
    [itex]= p_y x p_x^2-p_x y p_y^2[/itex]
     
  6. Dec 5, 2007 #5

    mjsd

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    note:
    [tex]\left[x_i,p_j\right]= i\hbar \delta_{ij}[/tex]
    look what you have done..
     
  7. Dec 5, 2007 #6

    cristo

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    No, [tex]p=\sqrt{p_x^2+p_y^2+p_z^2}[/tex]; but the OP seems to have got this part.
     
  8. Dec 5, 2007 #7
    I've used that result to eliminate some terms, but I can't see how it would eliminate the other terms. I could use it to permute the factors of the final terms I have, but that isn't too useful.
     
  9. Dec 5, 2007 #8

    mjsd

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    sorry had some network problems...
    are you sure it is p^2 not L^2?
     
  10. Dec 5, 2007 #9

    Avodyne

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    This is correct, and the 2nd and 3rd terms are zero. But in the 1st term, we can pull out the p_y (because it commutes with everything else), and similarly the p_x in the 4th term; then we have

    [tex]= p_y[x,p_x^2]-p_x[y,p_y^2][/tex]

    Then you need to evaluate the remaining commutators using

    [tex][A,BC]=[A,B]C + B[A,C][/tex]
     
  11. Dec 5, 2007 #10
    Thank you, that identity did the trick. That means that

    [itex]p_y x p_x^2 = p_x y p_y^2[/itex]

    but they don't look equal to me! Is there any way other than using the commutators that you could prove the above equality?
     
  12. Dec 5, 2007 #11

    Avodyne

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    How did you get that? It's not correct. Did you compute the commutator of x and px^2? What did you get?
     
  13. Dec 5, 2007 #12
    [itex] [x,p_x^2]=x p_x^2 - p_x^2 x = x p_x^2[/itex]

    Oh wait I see what I did wrong! I eliminated the second term because I was hitting x with [itex]p_x^2[/itex], but I forgot that these are operators acting on an unknown function, so it's more like [itex]p_x^2 ( xf ) [/itex].
    Thanks for the help.
     
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