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Determining the commutation relation of operators - Einstein summation notation

  1. Oct 10, 2012 #1
    Determining the commutation relation of operators -- Einstein summation notation

    1. The problem statement, all variables and given/known data

    Determine the commutator [itex] [L_i, C_j] [/itex].

    2. Relevant equations

    [itex]L_i = \epsilon_{ijk}r_j p_k[/itex]
    [itex]C_i = \epsilon_{ijk}A_j B_k[/itex]
    [itex] [L_i, A_j] = i \hbar \epsilon_{ijk} A_k[/itex]
    [itex] [L_i, B_j] = i \hbar \epsilon_{ijk} B_k[/itex]

    3. The attempt at a solution

    To be clear, the goal of this procedure is to become familiar with Einstein summation notation. That said, I've broken open [itex]C_j = \epsilon_{jmn}A_m B_n[/itex] and expanded the commutator accordingly. My problem is opening up [itex]L_i[/itex] in a meaningful way that gives me a nice identity with four deltas. Will I have to expand two epsilons with 6 different indices into deltas, or is there any way to get the epsilons to share an index that will give me the result (which I know to be [itex] [L_i, C_j] = i \hbar \epsilon_{ijk}C_k[/itex])?
     
  2. jcsd
  3. Oct 10, 2012 #2
    Re: Determining the commutation relation of operators -- Einstein summation notation

    Well, since the two free indices i and j are expressedly independent you can't set i=j (it's more interesting to calculate the commutator for two specific terms than the sum anyway). And the other four indices are already contracted, so you can't contract them with either i or j. You will just have to use all six indices.
     
  4. Oct 10, 2012 #3
    Re: Determining the commutation relation of operators -- Einstein summation notation

    That's super ugly. We really only talked about expanding epsilons in terms of deltas when there were four differing indices, not six. Is there a method to do this problem that requires using only four indices?
     
  5. Oct 10, 2012 #4
    Re: Determining the commutation relation of operators -- Einstein summation notation

    It's not pretty, I agree, but then again the Levi-Civita symbol is what you use to hide the ugly things. It's possible to group the terms somewhat, but you can't remove degrees of freedom unless you now that one or more of the vector components happen to be zero... Just be happy that you don't work in four dimensions.
     
  6. Oct 10, 2012 #5
    Re: Determining the commutation relation of operators -- Einstein summation notation

    Noted. Thanks for the help.
     
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