Evaluating expression involving permutation symbol and Kronecker delta

[xWaffle]

Homework Statement

Evaluate the following expression:
$\sum_{j}\sum_{k}\epsilon_{ijk}\delta_{jk}$

Homework Equations

$\delta_{ij}$ = $[i = j]$

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.
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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:
$\sum_{i}\sum_{k}\delta_{ij}\delta_{ji}$

I used my knowledge of the Kronecker delta to say that:
$\delta_{ij}\delta_{ji} = \delta_{ii} = \delta_{jj}$

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:
$\sum_{i}\sum_{k}\delta_{ij}\delta_{ji} = \sum_{i}\delta_{ii} = tr(I_{i}) = i$
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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 $\epsilon_{ijk}$ thrown in there. Any help is greatly appreciated. Thanks

 

[anathema]

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Thank you for sharing your attempt at the first expression and your concerns about the second one. It seems like you have a good understanding of the Kronecker delta and the trace of a matrix. To evaluate the second expression, you can use the same approach as you did for the first one.

First, we can use the property of the Kronecker delta, \delta_{ij} = [i = j], to simplify the expression to:
\sum_{j}\sum_{k}\epsilon_{ijk}\delta_{jk} = \sum_{j}\sum_{k}\epsilon_{ijk}[j = k]

Next, we can use the definition of the Levi-Civita symbol, \epsilon_{ijk}, to further simplify the expression. The Levi-Civita symbol is defined as:
\epsilon_{ijk} = \begin{cases}
1, & \text{if } (i,j,k) \text{ is an even permutation of } (1,2,3) \\
-1, & \text{if } (i,j,k) \text{ is an odd permutation of } (1,2,3) \\
0, & \text{otherwise}
\end{cases}

Using this definition, we can see that the only non-zero terms in the summation occur when i, j, and k are all different. This means that the only possible permutations of (i,j,k) are (1,2,3) and (3,1,2), which are both even permutations. Therefore, we can simplify the expression to:
\sum_{j}\sum_{k}\epsilon_{ijk}[j = k] = \sum_{j}\sum_{k}1[j = k]

Finally, using the property of the Kronecker delta again, we can simplify the expression to:
\sum_{j}\sum_{k}1[j = k] = \sum_{j}\delta_{jj} = \sum_{j}1 = 3

So the final solution to the expression is 3. I hope this helps and clarifies how the permutation and Kronecker delta come into play. Keep up the great work!