Positive-energy solutions. These are the same as before, except that α (and hence also β) is now a negative number.
Negative-energy solutions. On 0 < x < a we have
\frac{d^{2} \psi}{d x^{2}}=\kappa^{2} \psi, \quad \text { where } \quad \kappa \equiv \frac{\sqrt{-2 m E}}{\hbar} \Rightarrow \psi(x)=A \sinh \kappa x+B \cosh \kappa x .
According to Bloch’s theorem the solution on -a < x < 0 is
\psi(x)=e^{-i q a}[A \sinh \kappa(x+a)+B \cosh \kappa(x+a)] .
Continuity at x = 0 ⇒
B=e^{-i q a}[A \sinh \kappa a+B \cosh \kappa a], \quad \text { or } \quad A \sinh \kappa a=B\left[e^{i q a}-\cosh \kappa a\right] [\star] .
The discontinuity in ψ′ (Eq. 2.128)⇒
\kappa A-e^{-i q a} \kappa[A \cosh \kappa a+B \sinh \kappa a]=\frac{2 m \alpha}{\hbar^{2}} B, \text { or } A\left[1-e^{-i q a} \cosh \kappa a\right]=B\left[\frac{2 m \alpha}{\hbar^{2} \kappa}+e^{-i q a} \sinh \kappa a\right] [\blacklozenge] .
\Delta\left(\frac{d \psi}{d x}\right)=-\frac{2 m \alpha}{\hbar^{2}} \psi(0) (2.128).
Plugging F into \star and \blacklozenge cancelling B:
\left(e^{i q a}-\cosh \kappa a\right)\left(1-e^{-i q a} \cosh \kappa a\right)=\frac{2 m \alpha}{\hbar^{2} \kappa} \sinh \kappa a+e^{-i q a} \sinh ^{2} \kappa a .
e^{i q a}-2 \cosh \kappa a+e^{-i q a} \cosh ^{2} \kappa a-e^{-i q a} \sinh ^{2} \kappa a=\frac{2 m \alpha}{\hbar^{2} \kappa} \sinh \kappa a .
e^{i q a}+e^{-i q a}=2 \cosh \kappa a+\frac{2 m a}{\hbar^{2} \kappa} \sinh \kappa a , \cos q a=\cosh \kappa a+\frac{m \alpha}{\hbar^{2} \kappa} \sinh \kappa a .
This is the analog to Eq. 5.71. As before, we let \beta \equiv m \alpha a / \hbar^{2} (but remember it’s now a negative number), and this time we define z \equiv-\kappa a extending Eq. 5.72 to negative z, where it represents negative-energy solutions. In this region we define
\cos (q a)=\cos (k a)+\frac{m \alpha}{\hbar^{2} k} \sin (k a) (5.71).
z \equiv k a, \quad \text { and } \quad \beta \equiv \frac{m \alpha a}{\hbar^{2}} (5.72).
f(z)=\cosh z+\beta \frac{\sinh z}{z} [\star \star] .
In the figure below I have plotted f(z) for β = -1.5, using Eq. 5.73 for positive z and \star \star for negative z. As before, allowed energies are restricted to the range -1 \leq f(z) \leq 1 , and occur at intersections of f(z) with the N horizontal lines \cos q a=\cos (2 \pi n / N a) \text {, with } n=0,1,2 \ldots N-1 . Evidently the first band (partly negative, and partly positive) contains N states, as do all the higher bands.
