Orthogonality of Bloch factors in $k$ When I see the derivation of Bloch functions $\psi_{n\mathbf{k}}(\mathbf{r})=\mathrm{e}^{\mathrm{i}\mathbf{k}\mathbf{r}}u_{n\mathbf{k}}(\mathbf{r})$, the eigenfunctions of electrons in a periodic lattice and their properties, I always see a relation for the orthogonality of Bloch factors $u_{n\mathbf{k}}(\mathbf{r})$ with respect to the band index $n$:
$$\int u_{n\mathbf{k}}(\mathbf{r}) u_{n'\mathbf{k}}(\mathbf{r}) \ \mathrm{d^3}r=\delta_{nn'}$$
This relation always appears with respect to the same $\mathbf{k}$ for both factors. I wonder how this relation would look for different $\mathbf{k}$ values, $\mathbf{k}$ and $\mathbf{k}'$, while the band indices are also different. I.e. how would the general case of the scalar product of two Bloch factors look like?
 A: For a fixed $\textbf{k}$, a Bloch factor $u_{n\textbf{k}}$ satisfies the following eigenvalue equation:
\begin{equation}
\frac{(\hat{\textbf{p}} + \hbar \textbf{k})^2}{2m}u_{n\textbf{k}}(\textbf{r}) = E_{n\textbf{k}} u_{n\textbf{k}}(\textbf{r})
\end{equation}
subject to a periodic boundary condition over a unit cell. Note that Bloch factors corresponding to different Bloch wave vectors ($\textbf{k}$'s) constitute separate eigensystems. Therefore, it is meaningless to consider an orthonormality relation between Bloch factors unless they have the same $\textbf{k}$. All we can say is that for each $\textbf{k}$ separately, the relation
\begin{equation}
\int_{\Omega} d^3 \textbf{r}\, u_{n\textbf{k}}^*(\textbf{r})u_{n'\textbf{k}}(\textbf{r})  = \delta_{nn'}
\end{equation}
holds. ($\Omega$ denotes a unit cell.) 
Nevertheless, an orthonormality relation between Bloch wave functions $\psi_{n\textbf{k}}(\textbf{r}) = e^{i\textbf{k}\cdot\textbf{r}}u_{n\textbf{k}}(\textbf{r})$ still exists, and it is defined as an integral over the entire space:
\begin{equation}
\begin{split}
\int d^3\text{r}\, \psi_{n\textbf{k}}^*(\textbf{r})\, \psi_{n'\textbf{k}'}(\textbf{r}) &= \int d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}} \\
&= \sum_{\textbf{R}} e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{R}}\int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}}\\
&=\frac{(2\pi)^3}{V_{\Omega}}\,\delta^{(3)}(\textbf{k}-\textbf{k}') \int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}'}(\textbf{r})\, e^{-i(\textbf{k}-\textbf{k}')\cdot\textbf{r}}\\
&= \frac{(2\pi)^3}{V_{\Omega}}\,\delta^{(3)}(\textbf{k}-\textbf{k}') \int_{\Omega} d^3\text{r}\, u_{n\textbf{k}}^*(\textbf{r})\, u_{n'\textbf{k}}(\textbf{r})\\
&=\frac{(2\pi)^3}{V_{\Omega}}\,\delta_{nn'}\,\delta^{(3)}(\textbf{k}-\textbf{k}').
\end{split}
\end{equation}
Here, $\textbf{R}$ denotes a lattice vector of the crystal, and $V_{\Omega}$ the volume of each unit cell. Also, the property $u_{n\textbf{k}}(\textbf{r}) = u_{n\textbf{k}}(\textbf{r} + \textbf{R})$ was used to obtain the second line.
A: I don't think there is any similar orthonormality condition on the $u_{n,k}$ for different $k$'s. In the limit when the potential $V(\mathrm{r}) \to 0$, the Bloch wavefunctions are essentially plane waves so that, in $1\mathrm{D}$ and up to a normalization factor:
$$u_{n,k}(\mathrm{r}) = \exp \left(i \left( k + (-1)^n \mathrm{sign}(k)  \mathrm{ceil} \left(\frac{n}{2} \right)\frac{2 \pi}{a} \right) r\right)$$ (please check the expression, I could easily have made an error as band mapping is a nightmare).
The important result is that, while we still have orthonormality condition at fixed $k$:
$$\int u_{n\mathbf{k}}^*(\mathbf{r}) u_{n'\mathbf{k}}(\mathbf{r}) \ \mathrm{d^3}r=\delta_{nn'},$$
We don't have a similar condition for different $k$. For instance, staying in $1\mathrm{D}$ and looking at the lowest band ($n=0$), we should have something like:
$$\int u_{0\mathbf{k}}^*(\mathbf{r}) u_{0\mathbf{k'}}(\mathbf{r}) \ \mathrm{d^3}r=\mathrm{sinc}\left((k-k')a \right).$$
(But you can also have a non-zero overlap between $u_{n,k}$ and $u_{n',k'}$ for $n \neq n'$ if $k \neq k'$.)
In the other limiting case, when $V(r)$ is infinitely strong, the bands are flat and $u_{n,k}(\mathrm{r}) = 1$ up to a phase and a normalization factor. In this case you will have something looking like:
$$\int u_{n\mathbf{k}}^*(\mathbf{r}) u_{n'\mathbf{k'}}(\mathbf{r}) \ \mathrm{d^3}r=\delta_{nn'}.$$
I hope this helps.
PS : what I've written when integrating two $u_{n,k}(\mathbf{r})$ is still a simplification, because it is possible to add a global phase depending on $\mathbf{k}$ for all $n$ i.e setting $u'_{n,\mathbf{k}}(\mathbf{r}) = \exp (i \theta(\mathbf{k})) u_{n,\mathbf{k}}(\mathbf{r})$ without any physically significant change. This would change the previous results by an extra phase factor of $\exp (i(\theta(\mathbf{k}) - \theta(\mathbf{k'})))$ but this does not change the main argument. If anything, it should convince you that comparing two $u_{n,k}$ with different values of $k$ is not so significant most of the time.
