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In my notes I have that

$$\sum_m a_m \delta_{nm}=a_1\delta_{n1}+a_2\delta_{n2}+a_3\delta_{n3}+\cdots=a_n\tag{A}$$

Is this really correct?

I thought that for the Kronecker delta the first index must match the summation index. For this reason, I thought that the expression $(\mathrm{A})$ should written as

$$\sum_\color{blue}{m} a_m \delta_{\color{blue}{m}n}=a_1\delta_{1n}+a_2\delta_{2n}+a_3\delta_{3n}+\cdots=a_n\tag{B}$$

Just to make my argument clear, I have made the color of the indices match.

After looking at this page on the Kronecker delta I know that it is okay to write $$\sum_ja_j\delta_{ij}=a_i\tag{1}$$ or $$\sum_ia_i\delta_{ij}=a_j\tag{2}$$

Expression $(1)$ matches $(\mathrm{A})$ and expression $(2)$ matches $(\mathrm{B})$


So does this mean that switching the order of the indices of the Kronecker delta (within a summation) has no effect on the result (the RHS), or am I missing something?

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The Kronecker symbol $\delta_{ij}$ is used to say $\delta_{ii}=1$ and $\delta_{ij}=0$ if $i \neq j$.

This is independent of any summation. Kronecker symbol can be used in general. That being said, the evaluation of formula $(A)$ is perfectly correct as the only term that is not vanishing is when index $m$ is equal to $n$.

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    $\begingroup$ So is it okay to switch the order of the indices or not (the question I'm asking)? $\endgroup$ – BLAZE Sep 10 '18 at 19:06
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    $\begingroup$ I updated the answer. In the specific case of sum $(A)$, you can indeed switch the indices. $\endgroup$ – mathcounterexamples.net Sep 10 '18 at 19:10
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By its definition, $\delta_{i,j}=\delta_{j,i},$ [one can drop the commas as you have] one can switch the indices anywhere, in a sum or otherwise.

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