Double Sums and Index Shuffling

	Double sums over indexed parameters are usually bad enough, but doing arithmetic with the indices is often a bit mind boggling. Lets look at this in some detail:
	We have the double sum over k'and G with both indices running from –¥ to +¥. If, for the sake of simplicity, we consider a one-dimensional case (i.e. k' now denotes just one component) we have for the double sum
		SS    =  Sk' SG Ck' · VG · exp(i · [k'+ G]) · r)
		We can write this double sum as a matrix with, e.g., constant values of the Ck' in a row and constant values of the VG in a column . Shown is the part with k' = 4 and k' = 5, and likewise G = 7, 8, 9.
		+ + + + C4 · V7 · exp(i(4 + 7)) + C4 · V8 · exp(i(4 + 8)) + C4 · V9 · exp(i(4 + 9)) + + + + + C5 · V7 · exp(i(5 + 7)) + C5 · V8 · exp(i(5 + 8)) + C5 · V9 · exp(i(5 + 9)) + + + +

	Doing the sum does not depend on which way we take through the matrix, as long as we do not drop any element.
		"Intuitively" one would tend to go horizontally and back and forth through all the terms, but we can just as well move diagonally, following the lines indicated by identical color.
		In this case, the exponent is constant, we can name it k. The sum over a diagonal now means summing over all contributions where k' + G = const = k
	We thus can rewrite the double sum by adding up all the diagonals with fixed exponents and obtain the expression used before:
		Sk' SG Ck' · VG · exp (i · [k' + G] · r)  =  Sk – GSG Ck – G · VG · exp (ikr)

With frame

Fourier transforms and Bloch's theorem

© H. Föll (Semiconductors - Script)