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Indirect exchange interactions Fourier transforms

  1. Oct 21, 2012 #1
    ##\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. jcsd
  3. Oct 21, 2012 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  4. Oct 22, 2012 #3
    Tnx a lot.
     
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