# 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
S S | = | 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)