Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 33D / Solution

Assume $\lambda >0$ for the attractive wells. Write $E=-{\hslash }^2{\kappa }^2/(2m)$ with $\kappa >0$. Between wells the solution is a linear combination of $e^{\kappa x}$ and $e^{-\kappa x}$. Across a well it is continuous and its derivative jumps by $-2\lambda \psi$. Free propagation by $a$ and the jump have matricesTheir product has determinant one and trace $2[\cosh \kappa a-(\lambda /\kappa )\sinh \kappa a]$. Bloch theorem $e^{\pm ika}$ therefore give the negative-energy band of an attractive delta comb equationNegative energies are allowed precisely where the right side lies in $[-1,1]$. A cell wavefunction can be written explicitly ascontinued by Bloch theorem; the band equation enforces its derivative jump.

For $\lambda a\gg 1$, the isolated well has $\kappa =\lambda$ and energy $E_0=-{\hslash }^2{\lambda }^2/(2m)$. Writing $\kappa =\lambda +\delta$, the band equation gives $\delta \sim 2\lambda e^{-\lambda a}\cos ka$. HenceThis is the nearest-neighbour tight-binding model. Its hopping magnitude follows independently from isolated wavefunctions ${\varphi }_n=\sqrt{\lambda }{e}^{-\lambda |x-na|}$: the neighbouring delta well contributes $⟨{\varphi }_0,(H-E_0){\varphi }_1⟩\simeq -({\hslash }^2\lambda /m){\varphi }_0(0){\varphi }_1(0)=-t$. The subtraction removes the overlap contribution of $E_0$. More distant contributions are exponentially smaller. For repulsive or absent wells, $\lambda \le 0$, the quadratic form is nonnegative and there are no negative-energy bands.