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I can't find a satsifying explanaition for the fact that the components of the $\vec k$ vector only take discretized values $$k_i = 2\pi n_i/L_i \qquad i\in {x,y,z}$$ with $L_i$ being the periodicity in the corresponding direction. I know Bloch's theorem, i.e. $\Psi_{\vec k + \vec K}(\vec r) = \Psi_{\vec k}(\vec r) $ where $\vec K$ is a vector of the reciprocal lattice and the wave functions being Bloch functions $$\Psi_{\vec k}(\vec r) = u_{\vec k}(\vec r) e^{i\vec k \cdot \vec r}.$$

I believe I have the necessary "puzzle blocks" to solve the question, I just can't put them together properly. Please give me other suggestions/ways to derive the discretized values of $k$ in momentum space.

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  • $\begingroup$ Periodic boundary conditions $\endgroup$
    – hft
    Nov 9, 2022 at 19:06
  • $\begingroup$ (enforced at the "edges" ($0$ and $L$) of the entire crystal. Your textbook should discuss this quite early on) $\endgroup$
    – hft
    Nov 9, 2022 at 19:07
  • $\begingroup$ No I checked three textbooks and all of them just put the formua... $\endgroup$ Nov 9, 2022 at 19:09
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    $\begingroup$ Which three textbooks did you check? $\endgroup$
    – hft
    Nov 9, 2022 at 19:10
  • $\begingroup$ Recall that atom positions in a crystal lattice are discrete, so their Fourier transform needs to be as well. $\endgroup$
    – Jon Custer
    Nov 9, 2022 at 19:10

1 Answer 1

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I believe I found the proper way to do it with Bloch's theorem and due to the other replies:

\begin{align*} \Psi_{\vec k}(\vec r+L_i \vec e_i) &= \Psi_{\vec k}(\vec r)\qquad \forall i\in{x,y,z} \\ e^{i\vec k \cdot \vec r} &= e^{i\vec k \cdot \left(\vec r + L_i \vec e_i\right)} \\ e^{i\vec k \cdot \vec r} &=e^{i\vec k \cdot \vec r} e^{ik _iL_i} \\ 1 & = e^{ik _iL_i} \end{align*} $ \Rightarrow k _iL_i = n_i2\pi \Leftrightarrow k_i = \frac{n_i 2\pi}{L_i} \qquad n_i \in \mathbb{Z} $

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