6
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ You have to apply $\Psi_{n\vec{k}}\left(\vec{r} + \vec{R}\right) = e^{i\vec k \vec R} \Psi_{n\vec{k}}\left(\vec{r} \right)$, and $\int_V d^3r \Psi^*_{n\vec{k}}\left(\vec{r}\right) \Psi_{n'\vec{k'}}\left(\vec{r}\right) \sim \sum_{\vec R}\int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec{r} + \vec{R}\right) \Psi_{n'\vec{k'}}\left(\vec{r} + \vec{R}\right)$ $\endgroup$
    – Trimok
    Nov 21, 2013 at 19:05
  • 2
    $\begingroup$ It is, in general, highly unhelpful not to name the book that's confusing you. $\endgroup$ Nov 21, 2013 at 20:33
  • 1
    $\begingroup$ @ Emilio, it's a german textbook titled "Theoretische Festkörperphysik" from G. Czycholl in the 3rd revised version, sorry. $\endgroup$
    – user34407
    Nov 22, 2013 at 7:02
  • $\begingroup$ I had this same question. The proof and the missing steps are shown in the following book. You can follow it up. I'm providing a google book link and the name and page number of the book where the problem is solved. The book is Solid State and Quantum Theory for Optoelectronics by Michael A. Parker, page number 595, and here is the [Google book link](books.google.co.in/… $\endgroup$
    – user65318
    Nov 25, 2014 at 19:26

2 Answers 2

7
$\begingroup$

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 :

$I \sim \delta_{nn'}\delta(\vec{k} - \vec{k'})$

$\endgroup$
1
  • $\begingroup$ Great, thank you, I guess that's it! Although this means there indeed is a typo in the textbook, as the $e^{i(\vec{k'}-\vec k)\vec r}$ doesn't show up in the integral while the $\delta(\vec{k'},\vec k)$ yielding coefficient still does, right? However, thank you very much! $\endgroup$
    – user34407
    Nov 23, 2013 at 6:33
0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ I think the problem is precisely that I don't understand how to "factor out" this $\int d^3re^{i\vec r\left(\vec k - \vec{k'}\right)}$ expression out of the integral I suspect to be confronted with above. I added some information to clear my problem though, thank you! $\endgroup$
    – user34407
    Nov 22, 2013 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.