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Simple question: Do L^2 and r^2 commute?

  1. Dec 15, 2009 #1
    Hi

    The subject says it all. I'm wondering if [tex][L^2,r^2] = 0[/tex] is true?

    regards
    Frímannn
     
  2. jcsd
  3. Dec 15, 2009 #2

    dextercioby

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    What do you think ? On what variables does [itex] L^2 [/itex] depend ?
     
  4. Dec 15, 2009 #3
    Well we have that:

    [tex]\underline{\hat L }^2 = \hat L_1^2+\hat L_2^2+\hat L_3^2
    [/tex]

    [tex]\hat L_1 = \hat x_2 \hat p_3 - \hat x_3 \hat p_2
    [/tex]

    [tex]\hat L_2 = \hat x_3 \hat p_3 - \hat x_1 \hat p_3
    [/tex]

    [tex]\hat L_3 = \hat x_1 \hat p_2 - \hat x_2 \hat p_1
    [/tex]

    [tex]\underline{r }^2 = \hat x_1^2+\hat x_2^2+\hat x_3^2
    [/tex]

    and

    [tex][x_i,p_j] = i\hbar\delta_{i,j} [/tex]

    [tex][L_i,p_j] = 0, i = j [/tex]

    [tex][L_i,p_j] \ne 0, i \ne j [/tex]

    [tex][L_i,x_j] = 0, i = j [/tex]

    [tex][L_i,x_j] \ne 0, i \ne j [/tex]

    So it seems to me that they don't commute. But I'we been told otherwize so I'm trying to figure it out :)
     
    Last edited: Dec 15, 2009
  5. Dec 15, 2009 #4
    Are you sure about that? I get [tex][x_1,L_2] = i\hbar x_3[/tex]
     
  6. Dec 15, 2009 #5
    Sorry that was a mistake. That should have been a [tex]\ne[/tex]. I fixed the original post.
     
  7. Dec 16, 2009 #6

    dextercioby

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    I'll give you a further hint: use spherical coordinates.
     
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