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Wannier functions integration

  1. Dec 4, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-12-4_22-3-51.png

    I did not manage to get the final form of the equation. My prefactor in the final form always remain quadratic, whereas the solution shows that it is linear,

    2. Relevant equations
    w refers to wannier function, which relates to the Bloch function

    upload_2015-12-4_22-8-7.png
    ##\mathbf{R}## is this case should be zero.

    The Bloch function
    $$\psi_{n\mathbf{k}}=e^{i\mathbf{k \cdot r}}u_{n\mathbf{k}}$$, where ##u_{n\mathbf{k}}## is the cell periodic part.

    3. The attempt at a solution
    Using the given relation ##\mathbf{r\psi_{k}}##, I manage to get the following the equation
    $$\langle w \vert \mathbf{r} \vert w \rangle = \left( \frac{\Omega}{8\pi^3} \right)^2 \int_{BZ} d\mathbf{k} d\mathbf{k}' i e^{i(\mathbf{k-k'}) \cdot r} \langle u_{\mathbf{k}} \vert \nabla u_{\mathbf{k'}} \rangle $$,
    but I cannot find a way to factorize the exponential term out or to reduce the order of magnitude the prefactor.

    Any help is appreciated.
     
    Last edited: Dec 4, 2015
  2. jcsd
  3. Dec 4, 2015 #1

    DrDu

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    Science Advisor

    That's not an equation :-)
    Do you know hot to express a delta function in terms of an integral over exp ikx?
     
  4. jcsd
  5. Dec 4, 2015 #2
    You are right. I have edited the post and fixed some typos.
    Do you mean this?
    $$ (2\pi)^3 \delta({\mathbf{k-k'}}) = \int^\infty_{-\infty} e^{i(\mathbf{k-k'}) \cdot \mathbf{r}} d\mathbf{r} $$
    But I cannot figure how to factorize the term out.
     
    Last edited: Dec 4, 2015
  6. Dec 4, 2015 #3

    DrDu

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    Science Advisor

    You still didn't write down an equation. Nevertheless I suspect you forgot about the integration over r.
     
  7. Dec 4, 2015 #4
    I do not understand what you mean. If you mean the equation at the third part, I have edited it:
    $$
    \langle w \vert \mathbf{r} \vert w \rangle = \left( \frac{\Omega}{8\pi^3} \right)^2 \int_{BZ} d\mathbf{k} d\mathbf{k}' i e^{i(\mathbf{k-k'}) \cdot r} \langle u_{\mathbf{k}} \vert \nabla u_{\mathbf{k'}} \rangle
    $$

    In the original expression there is only one integration over r, which should have been included in ## \langle u_{\mathbf{k}} \vert \nabla u_{\mathbf{k'}} \rangle ##.
    Is there anything I miss?
     
  8. Dec 5, 2015 #5

    DrDu

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    Science Advisor

    Ok, but you can't pull out the exponential function from the integration over r. You have to use the periodicity of u and write the integral over all of r as a sum over the lattice times an integral over an elementary cell.
     
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