Particle in a one-dimensional lattice

A quantum particle in a one-dimensional lattice experiences a periodic potential. Bloch states organize its spectrum into bands and gaps; the Kronig-Penney model realizes this behavior with an idealized array of barriers or narrow delta barriers.

The Kronig-Penney model is an exactly solvable quantum mechanics model showing how a periodic potential produces energy bands and band gaps. R. de L. Kronig and W. G. Penney began with rectangular potential barriers and then took a narrow-barrier limit. This makes the relation between repeated quantum scattering and electronic band structure explicit.^[ref]

In the delta potential version, a particle of mass $m$ moves inThe lattice spacing $a$ has units of length; the strength $V_0$ has units of energy times length. The wavefunction is continuous, but integrating the Schrödinger equation across a barrier givesBetween barriers the wavefunction is a combination of plane waves. Bloch's theorem requires translation by one cell to multiply it by a phase $e^{iqa}$, where $q$ is the crystal wavenumber. Matching across one cell yields the Floquet discriminant of the delta-comb Kronig-Penney model, which decides whether a real $q$ is possible. With $V_0=0$ the band gaps close and the free particle dispersion is recovered; finite barriers open forbidden intervals through coherent repeated scattering.

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.

Let $E={\hslash }^2{k}^2/(2m)>0$ for the Kronig-Penney model, and propagate the column $y=(\psi ,{\psi }^{\prime }{)}^T$ by one lattice cell. Free propagation through distance $a$ and passage across one delta potential respectively give the propagation and jump matricesFor a cell beginning just after a barrier, $M=DP$ takes it to just after the next barrier. Both matrices have determinant one, so their product has reciprocal eigenvalues. Multiplication gives the half-trace, or Floquet discriminant,A Bloch state has Floquet matrix eigenvalues $e^{\pm iqa}$, hence $\cos (qa)=\Delta (E)$. Real crystal wavenumber $q$ therefore requires $|\Delta (E)|\le 1$. The values $\Delta =\pm 1$ give periodic or antiperiodic band-edge conditions. When $|\Delta |>1$, the reciprocal eigenvalues have unequal magnitudes and one cell solution grows while the other decays, producing a band gap in the infinite periodic system. The displayed positive-energy formula continues to negative energy by replacing trigonometric functions with their hyperbolic counterparts; the limit at $k=0$ uses $\sin (ka)/k\to a$.

Take lattice spacing $a>0$ and well strength $\lambda >0$ in the attractive Kronig-Penney modelIntegrating the Schrödinger equation across one Dirac delta distribution gives continuity of the wavefunction and ${\psi }^{\prime }(n{a}^{+})-{\psi }^{\prime }(n{a}^{-})=-2\lambda \psi (na)$. Free propagation between wells and this jump determine the Floquet matrix. The periodic matching method is the thin-barrier construction of the original paper; the attractive sign here follows the displayed potential.^[ref]

For $E={\hslash }^2{k}^2/(2m)>0$, real crystal wavenumber $q$ requiresFor $E=-{\hslash }^2{\mu }^2/(2m)<0$, substitute $k=i\mu$. The condition becomesThe equalities belong to the allowed energy bands: they give periodic or antiperiodic Bloch states at the edges. Strict inequalities describe their interiors. At $E=0$, take the limit separately: the half-trace is $1-\lambda a$, so zero belongs to the spectrum exactly when $0<\lambda a\le 2$.

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