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Orthogonality of Wannier functions

  1. Dec 7, 2015 #1
    I have trouble reconciling orthogonality condition for Wannier functions using both continuous and discrete k-space. I am using the definition of Wannier function and Bloch function as provided by Wikipedia (https://en.wikipedia.org/wiki/Wannier_function).

    Wannier function:
    932157a1333c5d1ae194c9c2be0c6b92.png
    Bloch function:
    a8588ba72f73bd2e8764f19937721139.png

    I can understand the orthogonality condition for Wannier functions in the discrete k-space as provided in Wikipedia:
    f1f492e4f7c73214620aa6842437c0c0.png


    But when I transform the summation over k-point to the integral representation using the relation:
    $$ \sum_{\mathbf{k}} \rightarrow \frac{N\Omega}{(2\pi)^3} \int_{BZ} d\mathbf{k} $$
    where N is the number of unit cell and ##\Omega## is the primitive cell volume, so that
    $$ \phi_{n\mathbf{R}}(\mathbf{r}) = \frac{\sqrt{N}\Omega}{(2\pi)^3} \int_{BZ} e^{-i\mathbf{k \cdot R}} \psi_{n\mathbf{k}}(\mathbf{r}) d\mathbf{k} $$
    which I cannot regain the orthogonality behaviour.

    These are my calculations:
    $$
    \langle \phi_{n\mathbf{R}}(\mathbf{r}) \vert \phi_{m\mathbf{R'}}(\mathbf{r}) \rangle = N \left( \frac{\Omega}{(2\pi)^3} \right) ^2 \int_V \int_{BZ} \int_{BZ} e^{i\mathbf{k \cdot R}} e^{-i\mathbf{k' \cdot R'}} \psi^*_{n\mathbf{k}}(\mathbf{r}) \psi_{m\mathbf{k'}}(\mathbf{r}) d\mathbf{k'} d\mathbf{k} d\mathbf{r}
    $$
    where V is the total volume included in the Born von Karman periodic boundary condition. Using
    $$
    \int_V \psi^*_{n\mathbf{k}}(\mathbf{r}) \psi_{m\mathbf{k'}}(\mathbf{r}) d\mathbf{r} = \delta_{mn}\delta(\mathbf{k-k'})
    $$
    I get
    $$
    \langle \phi_{n\mathbf{R}}(\mathbf{r}) \vert \phi_{m\mathbf{R'}}(\mathbf{r}) \rangle = N \left( \frac{\Omega}{(2\pi)^3} \right) ^2 \int_{BZ} \int_{BZ} e^{i\mathbf{k \cdot R}} e^{-i\mathbf{k' \cdot R'}} \delta_{mn} \delta(\mathbf{k-k'}) d\mathbf{k'} d\mathbf{k}
    $$
    Using (I don't know this one is correct or not)
    $$
    \int_{BZ} e^{-i\mathbf{k' \cdot R'}} \delta(\mathbf{k-k'}) d\mathbf{k'} = \frac{(2\pi)^3}{\Omega} e^{-i\mathbf{k \cdot R'}}
    $$
    then
    $$
    \langle \phi_{n\mathbf{R}}(\mathbf{r}) \vert \phi_{m\mathbf{R'}}(\mathbf{r}) \rangle = \delta_{mn} N \frac{\Omega}{(2\pi)^3} \int_{BZ} e^{i\mathbf{k \cdot (R-R')}} d\mathbf{k}
    $$
    With
    $$
    \frac{\Omega}{(2\pi)^3} \int_{BZ} e^{i\mathbf{k \cdot (R-R')}} d\mathbf{k} = \delta_{\mathbf{R,R'}}
    $$
    Finally I get
    $$
    \langle \phi_{n\mathbf{R}}(\mathbf{r}) \vert \phi_{m\mathbf{R'}}(\mathbf{r}) \rangle = N \delta_{mn} \delta_{\mathbf{R,R'}}
    $$
    which contains an additional term N.

    Anyone can point out my mistake?
     
  2. jcsd
  3. Dec 7, 2015 #2

    DrDu

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  4. Dec 7, 2015 #3
    Thanks for your reply. I am tempted to use that one because it is closer to the solution I want. But even then the final form is just
    $$
    \langle \phi_{n\mathbf{R}}(\mathbf{r}) \vert \phi_{m\mathbf{R'}}(\mathbf{r}) \rangle = N \frac{\Omega}{(2\pi)^3} \delta_{mn} \delta_{\mathbf{R,R'}}
    $$
    which differs from the discrete version.

    I figure that this question contains more fundamental aspect I might want to be sure of first, before I continue to tackle the previous question.
    Thank you.
     
  5. Dec 7, 2015 #4

    DrDu

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    Why only BZ on the RHS?
     
  6. Dec 7, 2015 #5
    It is not? It is in accordance with that on Wikipedia
    38e38263c62f4517ff8d703257591575.png
    ##\Omega## in this case is the Brillouin zone volume. Most of the references I consult automatically limit the integration region to the first Brillouin zone.

    If not then how do I limit the integration region to only the first Brillouin zone? Using the same concept as in real space like introducing ## \mathbf{k \rightarrow k+G} ## and ## \sum_{\mathbf{G}} ## ?
     
  7. Dec 7, 2015 #6

    DrDu

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    If ##k_i=2\pi n_i/L## with ##V=L^3## being the crystal volume (##V=N\Omega##), then ##\sum_\mathbf{k}=\sum_\mathbf{n}=\frac{V}{(2\pi)^3} \sum_n \frac{(2\pi)^3}{V}=\frac{V}{(2\pi)^3} \int d\mathbf{n} \frac{(2\pi)^3}{V}=\frac{V}{(2\pi)^3} \int d\mathbf{k} ##.
    If the integrand is periodic, then ##\int_V d\mathbf{k}=N\int_{BZ} d\mathbf{k}##.
     
  8. Dec 7, 2015 #7
    Ok so now ## \sum_{\mathbf{k}} \rightarrow \frac{N^2\Omega}{(2\pi)^3} \int_{BZ} d\mathbf{k} ## and
    $$
    \phi_{n\mathbf{R}}(\mathbf{r}) = \frac{N^{3/2}\Omega}{(2\pi)^3} \int_{BZ} e^{-i\mathbf{k \cdot R}} \psi_{n\mathbf{k}}(\mathbf{r}) d\mathbf{k}
    $$
    and the final form become
    $$
    \langle \phi_{n\mathbf{R}}(\mathbf{r}) \vert \phi_{m\mathbf{R'}}(\mathbf{r}) \rangle = N^3 \frac{\Omega}{(2\pi)^3} \delta_{mn} \delta_{\mathbf{R,R'}}
    $$
    which I feel weird.
     
  9. Dec 7, 2015 #8

    DrDu

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    $$
    \frac{N\Omega}{(2\pi)^3} \int_{BZ} e^{i\mathbf{k \cdot (R-R')}} d\mathbf{k} = \delta_{\mathbf{R,R'}}
    $$
     
  10. Dec 7, 2015 #9

    DrDu

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    That's also not correct, you have
    $$
    \int_V \psi^*_{n\mathbf{k}}(\mathbf{r}) \psi_{m\mathbf{k'}}(\mathbf{r}) d\mathbf{r} = \delta_{mn}\delta_{kk'}
    $$
    and you can't simply replace Kronecker by Dirac.
     
  11. Dec 7, 2015 #10
    Since ##
    \sum_\mathbf{k}=\sum_\mathbf{n}=\frac{V}{(2\pi)^3} \sum_n \frac{(2\pi)^3}{V}=\frac{V}{(2\pi)^3} \int d\mathbf{n} \frac{(2\pi)^3}{V}=\frac{V}{(2\pi)^3} \int d\mathbf{k}
    ##
    we have made a transformation from discrete k to continuous k, then why can't we use dirac delta function?
     
  12. Dec 7, 2015 #11

    DrDu

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    You can, but there will be an N dependent prefactor
     
  13. Dec 8, 2015 #12
    Sorry, I forgot to mention that the summation on k is restricted to first Brillouin zone only, in that case does ## \sum_{\mathbf{k} \in BZ} \rightarrow \frac{N\Omega}{(2\pi)^3} \int_{BZ} d\mathbf{k} ##
    apply? I mean directly restricting the integration region to BZ.

    I have some doubts on this.

    From Ashcroft and Mermin's Solid State Physics Appendix F, it is given that
    $$
    \sum_{\mathbf{k} \in BZ} e^{i\mathbf{k \cdot R}} = N \delta_{\mathbf{R},0}
    $$
    if the above transformation is true then
    $$
    \frac{\Omega}{(2\pi)^3} \int_{BZ} e^{i\mathbf{k \cdot (R-R')}} d\mathbf{k} = \delta_{\mathbf{R,R'}}
    $$
    is it?

    Care to provide some reasoning on this?
     
  14. Dec 8, 2015 #13

    DrDu

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    No, that's your job!
    Do not simply take over some equations from books or even worse, Wikipedia. They may use different conventions. Try to count the number of k states in the BZ. Transform the sums to integrals as I have shown you. Use the basic relations for delta functions and Kronecker delta.
     
  15. Dec 8, 2015 #14

    DrDu

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    Ok, this removes at least two of the dubious N's.
    Yes, I think this is correct and maybe I stepped in the same trap as you did. Namely I assumed that ##\delta(R-R')=\delta_{R,R'}##.
    So I think the problem is that you changed ##\delta_{k,k'}## to ##\delta(k-k')##.
    But ##\sum_k \delta_{k,0}=1## while ##V/(2\pi)^3 \int_{BZ} dk \delta(k)=V/(2\pi)^3##. So you should substitute ##\delta_{k,k'}\to (2\pi)^3/V \delta(k-k')##
     
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