# Why is the $\vec k$ space in solid state physics discretized?

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.

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

\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*} $$\Rightarrow k _iL_i = n_i2\pi \Leftrightarrow k_i = \frac{n_i 2\pi}{L_i} \qquad n_i \in \mathbb{Z}$$