While going through the article given here (page 23), it seems that the following doublesummation is carried out

$\sum_{i=1,j=1}^{N^2} B_{ij} = \Big( \sum_{i,j=1}^{N^2-1} + \sum_{i=1,j=N^2}^{N^2} + \sum_{i=N^2,j=1}^{N^2} - \sum_{i=N^2,j=N^2}^{N^2} \Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i \rho_S F_j^\dagger$, is an operator.

Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?


When you write $\sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $\sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:

  • $i=1,\ldots,N^2$; $j=N^2$

  • $i=N^2$; $j=1,\ldots,N^2$

But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.


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