# Orthogonality of Wannier functions

1. Dec 7, 2015

### Sheng

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:
Bloch function:

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

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. Dec 7, 2015

3. Dec 7, 2015

### Sheng

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.

4. Dec 7, 2015

### DrDu

Why only BZ on the RHS?

5. Dec 7, 2015

### Sheng

It is not? It is in accordance with that on Wikipedia
$\Omega$ 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 $\sum_{\mathbf{G}}$ ?

6. Dec 7, 2015

### DrDu

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 $\int_V d\mathbf{k}=N\int_{BZ} d\mathbf{k}$.

7. Dec 7, 2015

### Sheng

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.

8. Dec 7, 2015

### DrDu

$$\frac{N\Omega}{(2\pi)^3} \int_{BZ} e^{i\mathbf{k \cdot (R-R')}} d\mathbf{k} = \delta_{\mathbf{R,R'}}$$

9. Dec 7, 2015

### DrDu

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.

10. Dec 7, 2015

### Sheng

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?

11. Dec 7, 2015

### DrDu

You can, but there will be an N dependent prefactor

12. Dec 8, 2015

### Sheng

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?

13. Dec 8, 2015

### DrDu

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 $\delta_{k,k'}\to (2\pi)^3/V \delta(k-k')$