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I have to work with the following sum: $\sum_{ijk}\omega u_iu_j \delta_{ik}\delta_{kj}$ where $\omega$ is a constant in $\mathbb{C}$.

Is the answer: $$\omega \sum_k \sum_i u_i \delta_{ik}\sum_j u_j\delta_{kj}=\omega\sum_k u_k^2$$ or: $$\sum_{ij}u_iu_j \left(\sum_k \omega\delta_{ik}\delta_{kj}\right)=\sum_{ij}u_iu_j \left(2\omega-\omega\delta{ij}\right)=2\omega\sum_{ij}u_iu_j-\omega\sum_iu_i^2$$.

This is a pretty basic question, but I don't see where my logic is wrong here... If it is too obvious I can delete the question once my slow mind gets it... Anyway, thanks for your help!

Edit:

For the second line of calculation here was my reasoning:

In the $\sum_k \omega\delta_{ik}\delta_{kj}$ there are only two non-zero terms, namely when $k=i$ and when $k=j$ therefore $\sum_k \omega\delta_{ik}\delta_{kj}=2\omega$. However if $i=j$ then there is only one non-zero term. Therefore to consider this case I have to add $-\omega\delta_{ij}$.

Which yields: $\sum_k \omega\delta_{ik}\delta_{kj}=2\omega-\omega\delta_{ij}$.

Matt
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    Could you develop how you get the first equality of the second line of computations? – D. Thomine Jul 27 '19 at 12:18
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    It's the first one, not the second one. I don't follow the logic behind the second one, unfortunately. – Theo Bendit Jul 27 '19 at 12:19
  • Yes I think it is the first one indeed... By developing the second line of computations I think I saw that was wrong. In the second line of computation the sum should be then $\sum_{i\neq j}$ which is thus the same as the first one... – Matt Jul 27 '19 at 12:28
  • I would like someone to confirm before I delete or mark it as solved. – Matt Jul 27 '19 at 12:29
  • The mistake is more elementary, give me one second. – D. Thomine Jul 27 '19 at 12:32

1 Answers1

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When you manipulate this expression:

$$\sum_{i, j} u_i u_j \left( \sum_k \omega \delta_{ik} \delta_{kj} \right),$$

you first (rightfully) consider the sum over $k$:

$$\sum_k \omega \delta_{ik} \delta_{kj}.$$

A product of real numbers is non-zero if and only if each of the terms is non-zero. Here, if $\delta_{ik}=0$ or $\delta_{kj} = 0$, then $\omega \delta_{ik} \delta_{kj} = 0$; if $\delta_{ik}=\delta_{kj} = 1$, then $\omega \delta_{ik} \delta_{kj} = \omega$. Said otherwise:

  • If $i \neq k$ or $j \neq k$, then $\omega \delta_{ik} \delta_{kj} = 0$;

  • If $i=j=k$, then $\omega \delta_{ik} \delta_{kj} = \omega$.

If $i \neq j$, the second condition is never satisfied, so all of the terms are zero. If $i=j$, then there is only one non-zero term, corresponding to $k=i(=j)$. Hence:

$$\sum_k \omega \delta_{ik} \delta_{kj} = \omega \delta_{ij}.$$

Finally,

$$\sum_{i, j} u_i u_j \left( \sum_k \omega \delta_{ik} \delta_{kj} \right) = \sum_{i, j} u_i u_j \omega \delta_{ij} = \omega \sum_i u_i^2,$$

which is the same answer as with the first method.

D. Thomine
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