Bloch wave function orthonormality? there is this text book that is giving me a hard time for a while now:
It shows that Bloch wave functions can be written as
$$\Psi_{n\vec{k}}\left(\vec{r}\right) = \frac{1}{\sqrt{V}}e^{i\vec k \vec r}u_{n\vec k}\left(\vec r\right),$$
which is fine to me. It also states that the Bloch factors $u_{n\vec k}\left(\vec r\right)$ may be orthonormalized on the (primitive) unit cell volume $V_{UC}$:
$$\frac{1}{V_{UC}}\int_{V_{UC}}d^3r u^*_{n\vec k}\left(\vec r\right) u_{n'\vec k}\left(\vec r\right) = \delta_{n n'}$$.
However, and here starts my problem, it then concludes that therefore, the Bloch functions $\Psi_{n\vec{k}}\left(\vec{r}\right)$ fulfill
$$\int_V d^3r \Psi^*_{n\vec{k}}\left(\vec{r}\right) \Psi_{n'\vec{k'}}\left(\vec{r}\right)=$$
$$=\frac 1 V \sum_\vec{R} \int_{V_{UC}\left(\vec R\right)}d^3r e^{-i\vec k \left(\vec R + \vec r\right)} u^*_{n\vec k}\left(\vec R + \vec r\right) e^{i\vec {k'} \left(\vec R + \vec r\right)} u_{n'\vec{k'}}\left(\vec R + \vec r\right)=$$
$$=\frac 1 N \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \frac{1}{V_{UC}}\int_{V_{UC}} d^3r u^*_{n\vec{k}}\left(\vec r\right) u_{n'\vec{k'}}\left(\vec r\right)=$$
$$=\delta_{\vec{k'}\vec{k}}\delta_{n'n}$$
with lattice vectors $\vec R$ and crystal volume $V = N V_{UC}$.
But I just don't get the last two lines. I mean, the integral in the second last line actually should read $\int_{V_{UC}} d^3r e^{i\left(\vec{k'}-\vec{k}\right)\vec r}u^*_{n\vec{k}}\left(\vec r\right) u_{n'\vec{k'}}\left(\vec r\right)$, shouldn't it? And, if that is true and I didn't miss something important already, I can't understand how that would yield these two $\delta$s ...
I'm almost sure I missed something, but I just desperately keep fail getting it, so any help would be greatly appreciated!
EDIT
Thank you very much for your reactions! However, it seems I failed to state my problem clear enough, so I figured it might be best to tell you what my approach so far was step by step so may be someone can see where I actually go wrong:
Starting with
$$\int_V d^3r \Psi^*_{n\vec{k}}\left(\vec{r}\right) \Psi_{n'\vec{k'}}\left(\vec{r}\right),$$
I partitioned the integration domain $V=NV_{CU}$, thus getting a sum of integrations over the unit cell volume, yielding
$$\sum_{\vec R}\int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec{r} + \vec R \right) \Psi_{n'\vec{k'}}\left(\vec{r} + \vec R\right),$$
which happens to be exactly the second line when exploiting $\Psi_{n\vec{k}}\left(\vec{r}\right) = \frac{1}{\sqrt{V}}e^{i\vec k \vec r}u_{n\vec k}\left(\vec r\right)$. I then proceeded using $\Psi\left( \vec r + \vec R\right)=e^{i\vec k \vec R}\Psi\left(\vec r\right)$. But this yields
$$ \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right)$$
which disagrees with the third line in the book where the integral is over the Bloch factors $u_{n\vec k}\left(r\right)$ only.
However even assuming this is just a typo (which I'm not so sure of ...), I would be confronted with the integral $\int_{V_{UC}} d^3r e^{i\left(\vec {k'} - \vec k\right)\vec r} u^*_{n\vec{k}}\left(\vec r\right) u_{n'\vec{k'}}\left(\vec r\right)$ and I can't see how those two $\delta$s would arise from that either.
Thank you all again for your reactions and I hope I know actually stated my problem clearly.
 A: Let 
$I \sim \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right)$
The term $\sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R}$ gives you a $\sim \delta(\vec{k} - \vec{k'})$ term.
Now, you have : $\Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right) \delta(\vec{k} - \vec{k'}) \sim e^{i(\vec k'- \vec k).\vec r} u^*_{n\vec k}\left(\vec r\right) u_{n'\vec k'}\left(\vec r\right)\delta(\vec{k} - \vec{k'})$
Now, with the $\delta(\vec{k} - \vec{k'})$ term, $e^{i(\vec k'- \vec k).\vec r}$ becomes 1.
So you have : 
$\Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right) \delta(\vec{k} - \vec{k'}) = u^*_{n\vec k}\left(\vec r\right) u_{n'\vec k}\left(\vec r\right)\delta(\vec{k} - \vec{k'})$
where we have replaced $k'$ by $k$, in the indice of $u_{n'\vec k}$.
So, finally, after integration on $r$, we get : 
$I \sim \delta_{nn'}\delta(\vec{k} - \vec{k'})$
A: It's because of the integral:
$\int d^3r e^{i\vec{r}\cdot\vec{k}} = \delta^{(3)}(\vec{k})$
When you combine the two exponential factors you get (for the first case of $\vec{R}=0$):
$\int d^3r e^{i\vec{r}\cdot(\vec{k}-\vec{k}')}$
Which using the above result is just $\delta^{(3)}(\vec{k}-\vec{k}')$. In your last line where you have a Kronecker delta with a $\vec{k}$ and a $\vec{k}'$ that is just shorthand for continuous 3D delta function.  You can generalise that argument for each $\vec{R}$ in your sum fairly easily.
The delta function involving $n$'s comes from your second line about the orthonormality of the $u$'s.
Does that help?
Jack
