ISEGORIA / MATH ENCYCLOPEDIA
Solid-state physics: waves in a crystal
Tight-binding bands, band gaps from a periodic potential, and Brillouin zones.
Before you begin: Quantum mechanics and Fourier analysis
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Bands from hopping
An electron hops along a chain whose bonds alternate between strengths t₁ and t₂. Each Bloch wavevector k gives two energies, so the levels form two bands. Make the bonds unequal and a gap opens at the zone edge. Drag the gold point along the band to see the wave on the chain.
Worked example. At \(ka=\pm\pi\), \(E_\pm=\pm|t_1-t_2|\), so the gap is \(2|t_1-t_2|\). With \(t_1=t_2\) the gap closes and the chain is a single cosine band folded in half.
Watch out. The model keeps only nearest-neighbour hopping and one orbital per site. Real band structures need more orbitals and longer-range terms.
What is the group velocity at the zone edge?
Zero whenever \(t_1\ne t_2\): the bands are flat there, and the wave is a standing wave.
2. Band gaps from a periodic potential
A free electron can have any energy. Put it in a crystal of square barriers and whole intervals of energy become forbidden. The left view shows the allowed bands as strips over the potential; the right view shows the same bands as E(k). Raise the barriers and the bands narrow while the gaps widen.
Worked example. Units with \(\hbar^2/2m=1\) and lattice constant \(a=b+c=1\). An energy is allowed exactly when the right side lies in \([-1,1]\); above the barrier \(\beta\) becomes imaginary and \(\cosh,\sinh\) turn into \(\cos,\sin\).
Watch out. Even above the barrier top, gaps persist: every periodic potential opens gaps at the zone boundaries, though high gaps become narrow.
Where in k do the gaps open?
At the zone boundary \(ka=\pm\pi\) (and at \(k=0\) for higher bands), where Bragg reflection mixes \(k\) with \(k-2\pi/a\).
3. Reciprocal lattice and Brillouin zones
Drag the second lattice vector. The reciprocal lattice rotates and stretches with it, and the Brillouin zones are recomputed: the n-th zone is the set of wavevectors for which the origin is the n-th nearest reciprocal lattice point. The zones fragment into shards, yet every zone has exactly the same area.
Worked example. At a 60° angle with equal lengths the lattice is hexagonal and the first zone is a regular hexagon; at 90° it is a square.
Watch out. Zone areas are computed by counting pixels in the plotted window, so high zones that leave the window are reported only approximately.
Why do all zones have equal area?
Translating the pieces of any zone by reciprocal lattice vectors reassembles exactly one copy of the first zone.