Show that most of the energies determined by Equation 5.71 are doubly degenerate. What are the exceptional cases? Hint: Try it for N = 1,2,3,4,.., to see how it goes. What are the possible values of \cos (q a) in each case?
\cos (q a)=\cos (k a)+\frac{m \alpha}{\hbar^{2} k} \sin (k a) (5.71).
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.
