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Homework Help: Commutator Relation [L_i,L_j]

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Find ##[L_i,L_j]##.

    2. Relevant equations

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

    3. The attempt at a solution

    [tex][L_i,L_j] = \epsilon_{ijk}\epsilon_{jlm} [x_jp_k,x_lp_m][/tex]
    [tex] = \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}\right)[x_jp_k,x_lp_m][/tex]
    [tex] = [x_jp_k,x_jp_k] - [x_jp_k,p_kx_j][/tex]
    [tex] = p_kx_jx_jp_k - x_jp_kp_kx_j[/tex]

    Since ##x_jp_k = p_kx_j## for j≠k, we swap the two in the second term:

    [tex] = p_kx_jx_jp_k - x_jp_kx_jp_k[/tex]
    [tex] = [p_k,x_j]x_jp_k[/tex]
    [tex] = -i\hbar \delta_{jk}x_jp_k[/tex]

    I think the answer is something like ##i\hbar L_k##.
     
  2. jcsd
  3. Mar 21, 2014 #2

    TSny

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    You've used j as a dummy summation index on the right, which leads to trouble because on the left j is a fixed index. You'll need to use a dummy index other than j when you express ##L_i##.
     
  4. Mar 21, 2014 #3
    [tex][L_i, L_l] = \epsilon_{ijk}\epsilon_{lmn}[x_jp_k,x_mp_n][/tex]
    [tex] = (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km})[x_jp_k, x_mp_n][/tex]
    [tex] = -[x_jp_k, x_kp_j] [/tex]
    [tex] = -\left( [x_jp_k,x_k]p_j + x_k[x_jp_k,p_j]\right) [/tex]
    [tex] = -\left( x_j[p_kx_k]p_j + x_k[x_j,p_j]p_k\right)[/tex]
    [tex] = i\hbar \left( x_jp_j - x_kp_k\right)[/tex]
    [tex] = 0[/tex]
     
    Last edited: Mar 21, 2014
  5. Mar 21, 2014 #4

    TSny

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    [tex]\epsilon_{ijk}\epsilon_{lmn} \neq (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km})[/tex]
    (Note that indices ##i## and ##l## appear on the left but not on the right.)
     
  6. Mar 21, 2014 #5
    Ah that's right, for that identity to work they must have the same leftmost letter.
     
  7. Mar 21, 2014 #6

    TSny

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    Yes. Then you would be summing over the leftmost index of each Levi-Civita symbol. But, in your case there is no summation (yet).
     
  8. Mar 21, 2014 #7
    Ok I've given it another go!

    [tex][x_jp_k, x_mp_n] = [x_jp_k, x_m]p_n + x_m[x_jp_k, p_n][/tex]
    [tex] = \left( x_j[p_k, x_m]p_n + x_m[x_j, p_n]p_k\right) [/tex]
    [tex] = -i\hbar \left( \delta_{km}x_jp_n - \delta_{jn}x_mp_k\right] [/tex]

    Now applying ##\epsilon_{ijk}\epsilon_{lmn}##:

    [tex] = -i\hbar \left( \epsilon_{ijk}\epsilon_{lkn}x_jp_k - \epsilon_{ijk}\epsilon_{lmj}x_mp_k\right)[/tex]

    [tex] = -i\hbar \left( -\epsilon_{kij}\epsilon_{kln}x_jp_n + \epsilon_{jik}\epsilon_{jlm}x_mp_k\right)[/tex]

    [tex] = i\hbar \left[ (\delta_{il}\delta_{jn} - \delta_{in}\delta_{jl})x_jp_n - (\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})x_mp_k\right][/tex]

    Now, ##i=m=n## and ##j=k=l##:

    [tex] = i\hbar \left( x_jp_j - x_jp_i -x_jp_j+x_ip_j\right) [/tex]

    [tex] = i\hbar (x_ip_j - x_jp_i)[/tex]

    [tex][L_i,L_j] = i\hbar \epsilon_{kij}x_ip_j[/tex]

    [tex] [L_i,L_j] = i\hbar L_k[/tex]
     
  9. Mar 21, 2014 #8

    TSny

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    OK. Looks good.

    From your post #3 you are evaluating ##[L_i, L_l]##. So the indices ##i## and ##l## are fixed.
    Have another go at simplifying [tex] i\hbar \left[ (\delta_{il}\delta_{jn} - \delta_{in}\delta_{jl})x_jp_n - (\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})x_mp_k\right][/tex]
     
  10. Mar 22, 2014 #9
    [tex] i\hbar \left[ (\delta_{il}\delta_{jn} - \delta_{in}\delta_{jl})x_jp_n - (\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})x_mp_k\right][/tex]

    [tex] = i\hbar [\delta_{im}\delta_{kl}x_mp_k - \delta_{in}\delta_{jl}x_jp_n][/tex]

    [tex] = i\hbar [(\delta_{im}\delta_{kl} - \delta_{ik}\delta_{ml})x_mp_k][/tex]
     
  11. Mar 22, 2014 #10

    TSny

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    OK. You can keep going and simplify further. Note that you are summing over m, k, [STRIKE]j[/STRIKE], and [STRIKE]n[/STRIKE].

    [EDIT: Sorry. I see you've reduced it to just summing over m and k]
     
    Last edited: Mar 22, 2014
  12. Mar 22, 2014 #11
    I have simplified it till it's a sum over m and k only. But that doesn't reduce to 2 epsilons
     
  13. Mar 22, 2014 #12

    TSny

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    OK. What is your final expression after summing over m and k?
     
    Last edited: Mar 22, 2014
  14. Mar 22, 2014 #13
    the deltas don't seem to combine to give an epsilon
     
  15. Mar 22, 2014 #14

    TSny

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    We'll see how to get an epsilon factor later. For now, what expression do you get after summing over m and k?
     
  16. Mar 22, 2014 #15
    Actually, check out post #7. I solved it already by expanding. I thought you meant that there was a way of making use of the identity ##(\epsilon_{1}\epsilon_{2} = \delta\delta - \delta\delta)## to get the epsilon more explicitly..

    In post #7, the epsilon was drawn out by observation of a cross product ##x_ip_j - p_ix_j##
     
  17. Mar 22, 2014 #16

    TSny

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    Yes, in post 7 you are getting the right type of result. It's confusing, because you started out post 7 as a continuation of post 3 where you were calculating ##[L_i, L_l]##, but somehow ended up with an expression for ##[L_i, L_j]##.

    But if you now feel that you understand all the steps to get from ##[L_i, L_j]## to ## i \hbar (x_i p_j - p_i x_j )##, you are essentially done.

    If you let ##k## denote the direction perpendicular to the ##i##-##j ## plane,then you have shown that ##[L_i, L_j] = i \hbar (x_i p_j - p_i x_j) = i \hbar L_k##

    Can you show that this is equivalent to ##[L_i, L_j] = i \hbar \epsilon_{ijm} L_m## with summation over ##m##?
     
  18. Mar 22, 2014 #17
    It's ##[L_i, L_l] = i\hbar (x_ip_l - p_ix_l) = i\hbar L_m## I'm still missing an ##\epsilon## though..
     
    Last edited: Mar 22, 2014
  19. Mar 22, 2014 #18

    TSny

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    OK, that's correct with the understanding that ##m## refers to the index that defines the direction perpendicular to the ##i##-##l## plane and that the indices ##i, l## and ##m## are in cyclic order.

    The nice thing is that the one expression [tex][L_i, L_j] = i \hbar \sum_{k=1}^{3} \epsilon_{ijk}L_k = i \hbar \epsilon_{ijk}L_k [/tex] automatically takes care of everything. In the last expression the summation over ##k## is assumed (using the Einstein summation convention for repeated indices).
     
  20. Mar 22, 2014 #19
    I seem to be missing an ##\epsilon## in my final answer..
     
  21. Mar 22, 2014 #20

    TSny

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    We have ##[L_i, L_j] = i \hbar (x_ip_j - x_jp_i)##.

    The right hand side contains ##(x_ip_j - x_jp_i)##, which is the component of angular momentum that is perpendicular to the ##i##-##j## plane. A convenient way to express this component of angular momentum is ##\epsilon_{ijk}L_k## (where summation over the index ##k## is assumed). So the epsilon is "put in by hand" as a way of expressing the angular momentum component that is perpendicular to the ##i##-##j## plane.

    Thus, we can write

    ##[L_i, L_j] = i \hbar \epsilon_{ijk}L_k##

    As an example, if ##i = 1## and ##j = 2##, then ##[L_1, L_2] = i \hbar \epsilon_{12k}L_k = i \hbar L_3##.
     
  22. Mar 23, 2014 #21
    Yeah, but matematically ##(x_ip_j - x_jp_i)## simply means ##\vec{x}##x##\vec{p}##. So by putting an epsilon sign, it becomes ##\epsilon_{ijk}x_jp_k## which is simply ##L_k##. By putting another epsilon, won't you get an extra factor of epsilon?
     
  23. Mar 23, 2014 #22

    TSny

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    ##(x_ip_j - x_jp_i) = (\vec{x} \times \vec{p})_k## where ##k## refers to the component of the cross product that is perpendicular to the i-j plane.

    I'm not following. ##\epsilon_{ijk}x_jp_k## represents the ##i^{th}## component of the cross product ## \vec{x} \times \vec{p}##. (Note the repeated indices j and k in your expression, which means you are summing over j and k.)

    I don't see how you would get two epsilon factors. For fixed i and j, ##(x_ip_j - x_jp_i) = \epsilon_{ijk}L_k## with just one epsilon factor (and summation over the index k).

    In summary, you have for fixed i and j, [tex][L_i, L_j] = i \hbar (x_ip_j - x_jp_i) = i\hbar \epsilon_{ijk}L_k[/tex]
     
  24. Mar 23, 2014 #23
    Because ##L_k## is defined as ##\epsilon_{kij}x_ip_j ##
     
    Last edited: Mar 23, 2014
  25. Mar 25, 2014 #24
    any input?
     
  26. Mar 25, 2014 #25

    Fredrik

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    About what? Are you still talking about this:

    You want to find an equality that holds for all ##i,l\in\{1,2,3\}## such that ##i\neq l##. If m=1, then ##[L_i,L_l]=i\hbar L_m## can't possibly hold for all ##i,l\in\{1,2,3\}## such that ##i\neq l##. Same thing if m=2, or if m=3.

    The right-hand side (and every intermediate step) must depend on i and j, and nothing else.

    Yes, we have ##L_k=\epsilon_{kij} x_i p_j##. Now what does that imply about ##\epsilon_{ijk}L_k##?

    Depending on the values of i and j, ##x_ip_j - x_jp_i## is either equal to 0, or to one of the components of ##\vec x\times\vec p##. Which component that is depends on the values of i and j. If i≠j, then it's the kth component, where k is the unique element of {1,2,3} such that ##\epsilon_{ijk}\neq 0##.
     
    Last edited: Mar 25, 2014
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