# Indirect exchange interactions Fourier transforms

1. Oct 21, 2012

### LagrangeEuler

$\hat{c}_{i\sigma}=\frac{1}{\sqrt{N}}\sum_{\bf{q}}e^{i\bf{q}\cdot \bf{R}_i}\hat{c}_{\bf{k}\sigma}$

$\hat{c}^+_{i\sigma}=\frac{1}{\sqrt{N}}\sum_{\bf{q}}e^{-i\bf{q}\cdot \bf{R}_i}\hat{c}^+_{\bf{k}\sigma}$

Then

$-J\sum_{i}\hat{S}_i^z\hat{c}^+_{i\sigma}\hat{c}_{i \sigma}$

in $\bf{k}$ space is equal

$-J\sum_{i}\hat{S}_i^z\frac{1}{\sqrt{N}}\sum_{\bf{q}}e^{-i\bf{q}\cdot \bf{R}_i}\hat{c}^+_{\bf{q}\sigma}\frac{1}{\sqrt{N}}\sum_{\bf{k}}e^{i\bf{k}\cdot \bf{R}_i}\hat{c}_{\bf{k}\sigma}=$
$=-\frac{J}{N}\sum_{i}\sum_{\bf{q},\bf{k}}e^{-i\bf{q}\cdot \bf{R}_i}e^{i\bf{k}\cdot \bf{R}_i}\hat{S}_i^z\hat{c}^+_{\bf{q}\sigma}\hat{c}_{\bf{k} \sigma}$

and from that we get
$=-\frac{J}{N}\sum_{i}\sum_{\bf{q},\bf{k}}e^{-i\bf{q}\cdot \bf{R}_i}\hat{S}_i^z\hat{c}^+_{\bf{q}+\bf{k} \sigma}\hat{c}_{\bf{k} \sigma}$

Can you explain me this last step?

2. Oct 21, 2012

### mathman

You didn't show the range for q. If it is infinite, then (q-k) got changed to q in the exponent, and q then got changed to (q+k) for the subscript.

3. Oct 22, 2012

Tnx a lot.