Question 5.27: Show that most of the energies determined by Equation 5.71 a...

Equation 5.63 says

$q=\frac{2 \pi n}{N a}, \quad(n=0, \pm 1, \pm 2, \ldots)$       (5.63).

$q=\frac{2 \pi n}{N a} \Rightarrow q a=2 \pi \frac{n}{N}$; on page 223 we found that n = 0, 1, 2, … ,N – 1: Each value of n corresponds to a distinct state. To find the allowed energies we draw N horizontal lines on Figure 5.5, at heights $\cos q a=\cos (2 \pi n / N)$, and look for intersections with f(z). The point is that almost all of these lines come in pairs-two different n’s yielding the same value of $\cos q a$:

$\underline{N=1} \Rightarrow n=0 \Rightarrow \cos q a=1$. Nondegenerate.

$\underline{N=2} \Rightarrow n=0,1 \Rightarrow \cos q a=1,-1$.  Nondegenerate.

$\underline{N=3} \Rightarrow n=0,1,2 \Rightarrow \cos q a=1,-\frac{1}{2},-\frac{1}{2}$. The first is nondegenerate, the other two are degenerate.

$\underline{N=4} \Rightarrow n=0,1,2,3 \Rightarrow \cos q a=1,0,-1,0$. Two are nondegenerate, the others are degenerate.

Evidently they are doubly degenerate (two different n’s give same $\cos q a) \text { except when } \cos q a=\pm 1$ , i.e., at the top or bottom of a band. The Bloch factors $e^{i q a}$ lie at equal angles in the complex plane, starting with 1 (see figure below, drawn for the case N = 8); by symmetry, there is always one with negative imaginary part symmetrically opposite each one with positive imaginary part; these two have the same real part $(\cos q a)$. Only points which fall on the real axis have no twins.