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Homework Help: Evaluate expression with permutaion symbol and Kronecker delta

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following expression:

    2. Relevant equations
    [itex]\delta_{ij}[/itex] = [itex][i = j][/itex]

    3. The attempt at a solution
    I don't have a solution attempt to this one yet, because somehow I completely missed out on what the permutation thing has to do with anything.

    This is the second expression given on this homework assigment. The first one was a little easier, which I did work out, and came up with the solution. I'm going to show you guys this first problem so you know I at least know a little of what I'm doing..

    Evaluate expression:

    I used my knowlege of the Kronecker delta to say that:
    [itex]\delta_{ij}\delta_{ji} = \delta_{ii} = \delta_{jj}[/itex]

    Then using my knowledge of the trace of an n x n matrix (since I'm only dealing with square matrices), the trace of an n x n matrix is just n. So the final solution to the expression I found to be:
    [itex]\sum_{i}\sum_{k}\delta_{ij}\delta_{ji} = \sum_{i}\delta_{ii} = tr(I_{i}) = i[/itex]

    So I do have some of the knowledge I'm expected to have, but I really have no idea how to progress further, with the [itex]\epsilon_{ijk}[/itex] thrown in there. Any help is greatly appreciated. Thanks

    edit: I should have mentioned that I do at least know what the permutation symbol is.. it is valued at 0 if any of the i,j,k are the same, it is valued at +1 if the indicies are in cyclic order (123,231,312), -1 if they are are not in cyclic order
  2. jcsd
  3. Sep 9, 2012 #2


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    Did you mean j instead of k on the second summation?

    This isn't quite right. If i≠j, the first quantity is 0, but ##\delta_{ii}## is always equal to 1. Without the summations, the best you can say is [itex]\delta_{ij}\delta_{ji}=\delta_{ij}[/itex]. When you do the summation over j, for example, then you get
    $$\sum_i \sum_j \delta_{ij} = \sum_i \delta_{ii}.$$ Only the terms where i=j survive.
    You mean tr(In)=n, right?

    Perform the summation over k in the expression above. What do you end up with?
  4. Sep 9, 2012 #3
    Pardon my issues with the BB code and Latex, I've pretty much never used it before a week ago. Yes, you were able to pick out my errors quite nicely.

    As for my main question, I've tried to invision what is going on. It seems this question is inherently trivial because [itex]\epsilon_{ijk}\delta_{jk}[/itex] appears to always evaluate to be zero..

    If j and k are not equal, then the delta expression goes to zero, which brings the entire expression to zero with it. If j and k ARE equal, then the epsilon portion goes to zero, which brings the whole expression to zero with it.

    So I guess that expression should be evaluated as zero for all values of i,j, and k between 1 and 3 (since only 1,2,3 are applicable with the permutation symbol)
    Last edited: Sep 9, 2012
  5. Sep 9, 2012 #4


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    Right. If you perform the summation over k, the delta function picks out terms where k=j, so you end up with ##\sum \epsilon_{ijj}##, which, as you said, is 0.
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