1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deriving commutator for angular momentum components

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data

    Prove that

    ## [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##

    using Einstein summation convention.

    I think I have achieved the solution but I am not sure of my last steps, since this is one of my first excersises using this convention.

    2. Relevant equations

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

    3. The attempt at a solution

    Well, using (3) and (4)

    ## [L_a,L_b] = [\epsilon_{auv} x_u p_v, L_b] = \epsilon_{auv} x_u [p_v, L_b] + \epsilon_{auv} [x_u , L_b] p_v ##

    using (1) and (2), I get:

    ## \epsilon_{auv} x_u (- i \hbar \epsilon_{bvw}p_w)+\epsilon_{auv} x_u ((- i \hbar \epsilon_{but} x_t) p_v = - i \hbar (\epsilon_{auv} \epsilon_{bvw} x_u p_w + \epsilon_{auv} \epsilon_{but} x_t p_v )##

    And using (5) and (6) :

    ## - i \hbar (\epsilon_{vau} \epsilon_{vwb} x_u p_w + \epsilon_{uva} \epsilon_{utb} x_t p_v ) = - i \hbar (x_b p_a - x_u x_u + x_t x_t - x_a p_b) ##

    This expression reduces to

    ## i \hbar (x_a p_b - x_b p_a) ##

    I see this is the expression for the "c" component of the angular momentum but I would like to get the final expression using Einstein's notation and the Levi-Civita, so I guess this could be written as

    ## i \hbar \epsilon_{cab} x_a p_b ##

    Using (3) I get :

    ## i \hbar L_c ##

    Which is different to what I want to prove.

    ## [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##

    Then my assumption is that the Levi Civita symbol in the original equation makes explicit that c is different from a and b and that

    ## [L_b,L_a] = - [L_a,L_b] ##

    So would it be correct to do the following in order to arrive to the solution?

    ## i \hbar \epsilon_{cab} x_a p_b = i \hbar \epsilon_{abc} \epsilon_{cab} x_a p_b = i \hbar \epsilon_{abc} L_c##
     
  2. jcsd
  3. Feb 8, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    These two expressions are not equivalent. The first expression makes no reference to index c, whereas the second expression does explicitly reference c. Also, the first expression contains fixed values of a and b. But the second expression is a summation over all values of a and b.

    I will let you think about this a bit more.
     
  4. Feb 8, 2015 #3
    I knew I was guessing... :eek:
    Thanks for pointing that out!

    Now I am very uncomfortable with the previous steps (and with this notation in general, for now at least).:)

    I was trying to avoid summation symbols but I think I have to. When I wrote

    Am I really doing
    ## - i \hbar ( \sum_u \sum_v \sum_w \epsilon_{vau} \epsilon_{vwb} x_u p_w + \sum_u \sum_v \sum_t \epsilon_{uva} \epsilon_{utb} x_t p_v ) = \\ =- i \hbar ( \sum_u \sum_v \sum_w (\delta_{aw}\delta_{ub}-\delta_{ab}\delta_{uw})x_u p_w + \sum_u \sum_v \sum_t (\delta_{vt}\delta_{ab}-\delta_{vb}\delta_{at}) x_t p_v ) = \\ = - i \hbar ( \sum_v x_b p_a - \sum_v \sum_u x_u p_u + \sum_u \sum_v x_v p_v - \sum_u x_a p_b) = - i \hbar ( \sum_v x_b p_a- \sum_u x_a p_b)= - i \hbar (3 x_b p_a- 3 x_a p_b) ##

    ?
     
  5. Feb 8, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    In going from ##\sum_u \sum_v \sum_w \epsilon_{vau} \epsilon_{vwb} x_u p_w ## to ##\sum_u \sum_v \sum_w (\delta_{aw}\delta_{ub}-\delta_{ab}\delta_{uw})x_u p_w ##, should there still be a sum over ##v##?

    Note, your original work looked good to me all the way to the result of ##i \hbar (x_a p_b- x_b p_a) ## (except for a couple of typos where you had ##x_u x_u## instead of ##x_u p_u## and ##x_t x_t## instead of ##x_t p_t##). The problem was when you said that ##i \hbar (x_a p_b- x_b p_a) = i\hbar \epsilon_{cab} x_a p_b ##.
     
  6. Feb 8, 2015 #5
    True. I was a bit in panic. Those 3's made no sense at all. I'll think a bit more about it.
     
  7. Feb 9, 2015 #6
    Well I think I've got it by exploring all possible values:

    ##(x_a p_b - x_b p_a) = \left\{\begin{array}{cc}0,&\mbox{ if } a=b \\L_3,&\mbox{ if } (a,b) =(1,2) \\ -L_3,&\mbox{ if } (a,b) =(2,1)\\ L_2,&\mbox{ if } (a,b) =(1,3)\\ -L_2,&\mbox{ if } (a,b) =(3,1) \\ L_1,&\mbox{ if } (a,b) =(2,3) \\ -L_1,&\mbox{ if } (a,b) =(3,2) \end{array}\right . = \epsilon_{abc} L_c##

    However, I still wonder if I have done it correctly, and if I have, if there is a more elegant algebraic solution for this (with no numbers).

    Moreover, as I think about it I am not sure if I even understand the original identity:

    ## [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##

    Is here a summation over all possible values of c but none over a and b because they are fixed on the left hand side?

    My apologies but as I said before this is the first time I use Einstein's convention.
     
  8. Feb 9, 2015 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think it is correct. You just have to realize that ##x_a p_b - x_b p_b## is equivalent to ##\epsilon_{abc} L_c##.

    Yes, that's right. For example ## [L_1,L_2] = i \hbar \epsilon_{12c} L_c ## where c is summed over the values 1, 2, 3. So, the right hand side reduces to## i \hbar L_3##.
     
  9. Feb 9, 2015 #8
    I think I understand Einstein's convention a little more. I will keep working on this topic: ## [L_i, L^2] = 0## next. I hope I manage to prove this alone.

    Thanks for your help!
     
  10. Feb 19, 2015 #9
    I can't get this one correct either... I'll post a new thread anyway. Thanks a lot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Deriving commutator for angular momentum components
Loading...