# $k$-interval for First Brillouin Zone

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.

Question:

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

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

Question:

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).

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.