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enter image description here \begin{equation} \sum_{n}\left[\left(\frac{2 \pi n}{a}+k_{x_{0}}\right)^{2}+k_{y}^{2}\right] a_{n} \delta\left(\frac{2 \pi p}{a}-\frac{2 \pi n}{a}\right)=k_{0}^{2} \sum_{m} \sum_{m} a_{n} b_{m} \delta\left(\frac{2 \pi p}{a}-\frac{2 \pi m}{a}-\frac{2 \pi n}{a}\right) \end{equation}(2.20) \begin{equation} \left[\left(\frac{2 \pi n}{a}+k_{x_{0}}\right)^{2}+k_{y}^{2}\right] a_{n}=k_{0}^{2} \sum_{m} b_{n-m} a_{m} \end{equation}(2.21) where $\delta$ is the delta-function.

I try to simplify the eq2.20, but I can obtain the eq2.21, can you give me some advise？ thanks

yun shi
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1 Answers1

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The author is just changing the dummy indices. I guess $p$ is also a dummy variable (he probably uses it in a sum somewhere before) otherwise this is wrong.

On the left side from the $\delta$ you get $n=p$ and in (2.21) he made the change of dummy index $p\rightarrow n$.

On the right side, first your sum is $\sum\limits_m \sum\limits_n$ and not $\sum\limits_m \sum\limits_m$ (you made a small mistake). Then, from the $\delta$ you have $m=p-n$ giving a term $a_nb_{p-n}$ then, he changes the index $n\rightarrow m$ giving $a_m b_{p-m}$ and finally as on the left side, he made $p\rightarrow n$ giving $a_mb_{n-m}$...this is quite a messy notation.

E. Bellec
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