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Angular momentum commutator

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove that

    ## [L_a,L_b L_b] =0 ##

    using Einstein summation convention.


    2. Relevant equations

    ## (1) [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##
    ## (2) \epsilon_{abc} \epsilon_{auv} = \delta_{bu} \delta_{cv}- \delta_{bv} \delta_{cu}##
    ## (3) \epsilon_{abc} = \epsilon_{bca} = \epsilon_{cab} ##
    ## (4) L_a = \epsilon_{abc} x_b p_c ##
    ## (5) [AB,C] = A [B,C] + [A,C] B ##
    ## (6) [x_a,p_b] = i \hbar \delta_{ab}##

    3. The attempt at a solution
    Well, using (1) and (5):
    ##[L_a,L_b L_b] = [L_a,L_b ] L_b+L_b [L_a,L_b] = i \hbar \epsilon_{abc} L_c L_b + L_b i \hbar \epsilon_{abd} L_d = \\
    = i \hbar \epsilon_{abc} (L_b L_c +[L_c,L_b ] ) + i \hbar \epsilon_{abd} L_b L_d = i \hbar (\epsilon_{abc} L_b L_c + \epsilon_{abc} i \hbar \epsilon_{cbe} L_e + \epsilon_{abd} L_b L_d ) ##

    Then, using (2) and (3), and putting the two equivalent terms together, I get:

    ##i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar \epsilon_{bca} \epsilon_{bec} L_e ) =
    i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar (\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) =2 i \hbar \epsilon_{abc} L_b L_c ##

    This does not look well to me... Anyway, I have continued by using (4):

    ## 2 i \hbar \epsilon_{abc} \epsilon_{buv} x_u p_v \epsilon_{ckj} x_k p_j ##

    Then (2) and (6):

    ## 2 i \hbar (\epsilon_{ckj} x_c p_a x_k p_j - \epsilon_{ckj} x_a p_c x_k p_j) = 2 i \hbar \epsilon_{ckj} [x_c, p_a] x_k p_j = 2 i \hbar \epsilon_{ckj} i \hbar \delta{ca} x_k p_j = \\ = -2 \hbar^2 \epsilon_{akj} x_k p_j = -2 i \hbar L_a ##

    I cannot find my mistake(s)... and my result is absurd.
     
  2. jcsd
  3. Feb 19, 2015 #2

    TSny

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    Homework Helper
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    In the first term on the far right you have two summation indices: b and c. In the second term you have summation indices b and d.

    Summation indices are "dummy" indices. You can rename them if you wish. What happens if in the second term you re-label the b index as c and re-label the d index as b?
     
    Last edited: Feb 20, 2015
  4. Feb 19, 2015 #3

    TSny

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    Note that ##(\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) \neq 0##.

    What is value of ##\delta_{cc}##? Remember, repeated subscripts are summed.
     
    Last edited: Feb 20, 2015
  5. Feb 22, 2015 #4
    So ## i\hbar \epsilon_{abc} L_c L_b+i \hbar \epsilon_{abd} L_b L_d = i\hbar \epsilon_{abc} L_c L_b+i \hbar \epsilon_{acb} L_c L_b = 0##

    So simple now... b is a dummy index! It is still difficult for me to realise what is "fixed" and what is a dummy index without writing down the summation symbols... Now I see that ## L_b L_b ## stands for the square of the modulus in classic mechanics: I lost this somewhere in my work.

    I think it is 3.

    Ok so the first term is actually, ## \delta_{ae}L_e = L_a## and the second term is ## 3 \delta_{ae}L_e = 3 L_a##
    so:

    ## i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar (\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) =2 i \hbar \epsilon_{abc} L_b L_c + 2 \hbar^2 L _ a ##

    And according to my previous attemp (in which I corrected the last step):
    ##2 i \hbar \epsilon_{abc} L_b L_c + 2 \hbar^2 L _ a = -2 \hbar^2 L_a + 2 \hbar^2 L _ a = 0##

    I think I got it... Thank you once again!
     
    Last edited: Feb 22, 2015
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