Integrating Wannier Functions: Simplifying the Prefactor Equation

  • #1
Sheng
11
0

Homework Statement


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,

Homework 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.

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:
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  • #2
Sheng said:

The Attempt at a Solution


Using the given relation ##\mathbf{r\psi_{k}}##, I manage to get the following the equation
$$\left( \frac{\Omega}{8\pi^3} \right)^2 \int_{BZ} d\mathbf{k} d\mathbf{k}' i e^{i(\mathbf{k-k'}) \cdot r} \langle u_{\mathbf{k}} \vert \nabla u_{\mathbf{k}} \rangle $$,
That's not an equation :-)
Do you know hot to express a delta function in terms of an integral over exp ikx?
 
  • #3
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:
  • #4
You still didn't write down an equation. Nevertheless I suspect you forgot about the integration over r.
 
  • #5
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?
 
  • #6
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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