## Indirect exchange interactions Fourier transforms

##\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?
 PhysOrg.com physics news on PhysOrg.com >> Atomic-scale investigations solve key puzzle of LED efficiency>> Study provides better understanding of water's freezing behavior at nanoscale>> Iron-platinum alloys could be new-generation hard drives
 Recognitions: Science Advisor 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.
 Tnx a lot.