1. If we solve the time-independent Schrodinger equation in 1D for a periodic potential $V(x)$ with wave function $\psi(x)$ subject to a periodic boundary condition: $\psi(x) = \psi(x + Ga)$, where $a$ is the period of $V(x)$ and $G$ is a positive integer so that the lattice lenght $L = G a$, then the general form of $\psi(x)$ is given by $$ \psi_k(x) = e^{ikx} u_k(x), $$ with $u_k(x) = u_k(x+a)$ and $k = \frac{2\pi g}{Ga}, g \in \mathbb{Z}$. This is the famous Bloch theorem.


It is said that $k$ is not uniquely determined by $\psi_k(x)$ and the periodicity of $u_k(x)$. I don't see the reason why this is so.

  1. For the first Brillouin zone, the interval for momentum $k$ is often given by

$$ \left[-\frac{\pi}{a}, \frac{\pi}{a} \right). $$


Why don't we include the point $\frac{\pi}{a}$?

The textbook that I am using is The Wave Mechanics of Electrons in Metals by Stanley Raimes (page no. 198).


1 Answer 1


Say you know $\psi_n(x)$. I will show that $\psi_n(x)$ can be written as $e^{ikx}u_n(x)$ for an infinite number of different values of $k$ and $u_n$. Let's assume you know one decomposition $\psi_n(x)=e^{ikx}u_n(x)$. Bloch's theorem guarantees one such decomposition exists. Then we can also write

$$ \psi_n(x) = e^{i(k+\frac{2\pi N}{a})x} e^{-\frac{2\pi i N}{a}}u_n(x) $$ where $N\in \mathbb Z$ is any integer. If we define $\bar k\equiv k+\frac{2\pi N}{a}$, $\bar{u}_n(x)\equiv e^{-\frac{2\pi i N}{a}}u_n(x)$, then $\bar u _n(x)$ is periodic with period $a$ and $\psi_n(x)=e^{i\bar k x}\bar u_n(x)$. Thus, $\psi_n(x)$ can be decomposed in many different possible ways.

We use this freedom to demand that $k\in\left[-\frac\pi a,\frac\pi a\right)$. We don't include BOTH $\pm\frac\pi a$, since one can be turned into the other by setting $N=\pm 1$ in the derivation above.


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