Integrating Wannier Functions: Simplifying the Prefactor Equation

[Sheng]

Homework Statement

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

$\mathbf{R}$ 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.

 

[DrDu]

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?

 

[Sheng]

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.

 

[DrDu]

You still didn't write down an equation. Nevertheless I suspect you forgot about the integration over r.

 

[Sheng]

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?

 

[DrDu]

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.