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The Bloch theorem states that a non-interacting electronic state of a confined system with periodic boundary conditions (and a periodic potential) is represented as $\psi_{\mathbf{k}}(\mathbf{r}) = u_{\mathbf{k}}(\mathbf{r}) e^{i{\mathbf{k}}r}$. In the picture of a free-electron confined in a periodic box, the momentum values are quantized as $k_i=\frac{2\pi n_i}{L}~\mathrm{for}~i\in {x,y,z}$. So in principle, also in Density Functional Theory (DFT), the $\mathbf{k}$-vector should be quantized. However, when we plot the result from DFT band calculations, we always see continuous bands, what is the reason for that?

Edit: I want to stress that I understand that if we make the unit cell very large, there will be a continuous amount of states, but that is not the case in a conventional periodic DFT calculation, where our cell size is finite...

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  • $\begingroup$ From your comment to the answer below I see that you now understand that the periodicity of the wave function in the periodic box model differs from the Bloch waves we actually have in infinite crystals. It would be nice if you write an answer to your own question in which you explicitly describe the difference. I guess you are by far not the only person who gets irritated by this. I actually wonder where this periodic box model is useful. $\endgroup$ Jun 15 '20 at 23:00
  • $\begingroup$ 2 months is long for no upvoted answers, maybe copy and paste into materials.stackexchange.com $\endgroup$ Jul 23 '20 at 18:18
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Your understanding is correct and your answer is right there. In the first Brillouin zone, the 'K-points' are discrete and quantized, according to the equation you mentioned in the question. For a crystal, as one might infer, you would have plenty of 'K-points' due to the periodic boundary condition (and how much bigger the system is compared to say, an isolated molecule). The E vs K diagram (electronic band structure) plotted in the 1st BZ therefore looks continuous because of high number of K points.

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  • $\begingroup$ Yeah but this was not my question. It is clear to me that physically if we would choose periodic boundary conditions with a huge crystal, the states will be continuous. But in DFT, we have small unit cell sizes, so that the periodic boundary conditions contain only 1 or 2 lattices. This means that the k will naturally be coarsely quantized and far from a continuum. This is of course some artificial artifact of choosing a small unit cell size in DFT. But I am wondering what enables us to solve this problem then. $\endgroup$
    – Guiste
    May 4 '20 at 7:03
  • $\begingroup$ What do you mean we have small unit cell sizes in DFT? System sizes vary all the time. Periodic conditions (in context of DFT) are usually discussed only in 'k-space' and not real space. If you are interested in how bands are calculated in DFT for very big systems (for small systems, you can imagine how bands tend to look continuous due to many K points), it's a matter of discretion. Some calculations are done at just one K point - the rest is 'folded' into the 1st BZ - You can look up 'Brillouin zone folding'. Or, if you had the computational power, you could obviously sample more K points. $\endgroup$
    – Xivi76
    May 4 '20 at 9:02
  • $\begingroup$ I realized that I confused the periodicity of the electronic wave function with that of the potential. The wave function has a larger periodicity of macroscopic length scales, so that in principle the k-space is almost continuous, while the periodic potential has the periodicity of a lattice cell. $\endgroup$
    – Guiste
    May 6 '20 at 12:09

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