Kronig-Penney model

The delta-comb Kronig-Penney potential is

For $V(x)=\alpha {\sum }_j\delta (x-ja)$ with $\alpha >0$, put $P=m\alpha a/{\hslash }^2$ and $x=ka$. The Floquet discriminant of the delta-comb Kronig-Penney model is $f(x)=\cos x+P\sin x/x$. Extended Bloch states require $|f(x)|\le 1$. Since $f(0)=1+P>1$, there is a positive lower band edge; a repulsive potential does not support negative-energy states in this energy convention.

The $j$th positive-energy band ends at $x=j\pi$. Its lower edge lies in $((j-1)\pi ,j\pi )$ and solves $f(x)=(-1{)}^{j-1}$. A finite periodic ring selects only the discrete allowed values of crystal momentum within these bands. Nonzero band gaps do not by themselves imply an insulator: the band filling must also place the Fermi energy in a gap.

For $E={\hslash }^2{k}^2/(2m)$ and $\gamma =m{V}_0/({\hslash }^{2}k)$,Thus the allowed bands satisfy $|\cos (ka)+\gamma \sin (ka)|\le 1$.

Forthe wavefunction is continuous across every lattice point and obeysAt negative energy $E=-{\hslash }^2{\mu }^2/(2m)$, an allowed energy band satisfiesAt positive energy $E={\hslash }^2{k}^2/(2m)$, it satisfies

With well strength $\lambda$ and spacing $a$, negative-energy Bloch states obey $\tanh y\le 2y/(\lambda a)\le \coth y$, where $y=\kappa a/2$. There is one negative band: it reaches zero when $\lambda a\le 2$ and is separated from zero when $\lambda a>2$.

If one symmetric cell has reflection and transmission amplitudes $r,t$, thenAt a periodic or antiperiodic band edge, with $\mathrm{tr}M=2\sigma$ and $\sigma =\pm 1$, setting $z=e^{-ika}$ gives