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I am working on a problem in statistical mechanics involving a double sum of two dirac-delta functions. I am not sure how to $$ \text{relate} \quad \sum_{i=1}^{N} \sum_{j=1}^{N} \delta (r - r_{i}) \delta (r' - r_{j}) \quad \text{to} \quad \sum_{i=1}^{N} \sum_{\substack{j=1 \\ j \neq i}}^{N} \delta (r - r_{i}) \delta (r' - r_{j}) $$ My attempt is to take out an $i$ term from the inner sum: $$ \sum_{i=1}^{N} \sum_{j=1}^{N} \delta (r - r_{i}) \delta (r' - r_{j}) \to \sum_{i=1}^{N} \sum_{\substack{j=1 \\ j \neq i}}^{N} \delta (r - r_{i}) \delta (r' - r_{j}) + \sum_{i=1}^{N} \delta (r - r_{i}) \delta (r' - r_{i}) $$ However, the given answer is $$ \sum_{i=1}^{N} \sum_{j=1}^{N} \delta (r - r_{i}) \delta (r' - r_{j}) = \sum_{i=1}^{N} \sum_{\substack{j=1 \\ j \neq i}}^{N} \delta (r - r_{i}) \delta (r' - r_{j}) + \delta (r' - r) \sum_{i=1}^{N} \delta (r - r_{i}) $$ How did the second delta function change from $\delta(r'-r_{j})$ into $\delta(r' - r)$?

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  • $\begingroup$ @MishaLavrov It is Dirac delta. The particle density is given by $\rho (r) = \sum_{i=1}^{N} \delta (r - r_{i})$. $\endgroup$ – ferris Apr 25 at 3:01
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Since $\delta(x-y)f(x)=\delta(x-y)f(y)$, $\delta(r-r_i)\delta(r'-r_i)=\delta(r-r_i)\delta(r'-r)$. Therefore, $$\sum_{i=1}^N \delta (r-r_i)\delta (r'-r_i) = \sum_{i=1}^N \delta (r-r_i)\delta (r'-r) = \delta(r'-r)\sum_{i=1}^N \delta(r-r_i)$$

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