Kronig–Penney model

Example

The Kronig–Penney model is the one-dimensional Schrödinger operator $H = -\dfrac{d^2}{dx^2} + V$ (units with $\hbar^2/2m = 1$) with an $a$-periodic potential made of square barriers. In each period, $V = V_0 > 0$ on an interval of width $b$ and $V = 0$ on the rest, of width $a - b$. For $0 < E < V_0$, put $q = \sqrt{E}$ and $\kappa = \sqrt{V_0 - E}$. Solving $-\psi'' + V\psi = E\psi$ piece by piece and imposing the Bloch condition $\psi(x + a) = e^{ika}\psi(x)$ gives the relation $$\cos(ka) = \cos\big(q(a - b)\big)\cosh(\kappa b) + \frac{\kappa^2 - q^2}{2q\kappa} \sin\big(q(a - b)\big)\sinh(\kappa b).$$ The energy $E$ lies in a band exactly when the right-hand side lies in $[-1, 1]$.

Check in the limit of thin high barriers, $b \to 0$ with $V_0 b = \alpha$ fixed (a Dirac comb $V = \alpha \sum_{n} \delta(x - na)$): then $\kappa^2 b \to \alpha$ and $\kappa b \to 0$, and the relation becomes $$\cos(ka) = \cos(qa) + \frac{\alpha}{2q} \sin(qa).$$ At $qa = \pi$ the right side equals $-1$, a band edge. Take $\alpha > 0$ and $qa$ slightly above $\pi$. Then $\cos(qa)$ is close to $-1$ and $\sin(qa) < 0$, so the right side is below $-1$ and $E = q^2$ lies in a gap. So gaps open at $qa = \pi, 2\pi, \dots$, where the free bands would cross.