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Is there a closed-form expression for the Floquet transform of the derivative of a function $f$ (analog to the well-kown property of the Fourier transform)?

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Dear Carlo Thank you for your quick reply. It's indeed true that the derivative of the Floquet transform equals the Floquet tranform of the derivative. I should have pointed it out more precisely, but I am actually interested in the fact if the Floquet tranform of the derivative of a function $f(r)$ can be expressed in terms of the Floquet tranform of the function $f(r)$. So, is there a relation between $(U \frac{\partial }{\partial r}f)(r)$ and $(U f)(r)$ (like there is for the Fourier transform, i.e. $(F \frac{\partial }{\partial r}f)(r)=i\omega (F f)(r)$)? Many thanks in advance ! Jeff –  user14903 May 5 '11 at 18:21
    
Since Answers should be answers to the original question, this should have been a comment to Carlo's answer... –  Dirk May 5 '11 at 18:26
    
Aren't we asking for the impossible here? How could the derivative be expressed in terms of the function? –  Michael Renardy May 5 '11 at 18:28
    
I have to second Micheal's objection here-unless we're misunderstanding the question. –  Andrew L Jun 2 '11 at 20:47
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1 Answer

If I'm not mistaken, the definition of the Floquet transform is

$(Uf)(r)=\sum_{R\in L}e^{ik\cdot R}f(r-R)$,

where the sum runs over vectors $R$ on a $d$-dimensional periodic lattice $L$. The vector $k$ is fixed. Now I just take the derivative of both sides with respect to $r$,

$\frac{\partial}{\partial r}(Uf)(r)=(U\frac{\partial}{\partial r}f)(r)$.

In words: the derivative of the Floquet transform is the Floquet transform of the derivative.

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