Past exam of the quantum condensed matter course of the University of Cambridge / 2019 / Section B / 4 / b

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- Solution b

Solution

Write

U = V_{g}

for the coupling Fourier coefficient, with

∣ U ∣ = ∣ V_{0} ∣

for the supplied cosine potential. Projecting the Schrödinger equation onto the two retained plane waves gives

(ϵ_{q + g /2} U^{*} U ϵ_{q - g /2}) (A B) = E (A B) .

The diagonal potential average is zero. The characteristic equation is

(E - ϵ_{q + g /2}) (E - ϵ_{q - g /2}) - ∣ U ∣^{2} = 0,

so the nearly-free electron dispersion near a one-dimensional band gap is

E_{\pm} (q) = \frac{ϵ _{q + g /2} + ϵ _{q - g /2}}{2} \pm \frac{1}{2} (ϵ_{q + g /2} - ϵ_{q - g /2})^{2} + 4 V_{0}^{2} .

Only the coupling magnitude enters the energies; its sign changes the standing-wave eigenvectors. The two-wave truncation is controlled near this crossing when

∣ V_{0} ∣ ≪ ϵ_{g /2}

, since the other plane waves are separated by much larger kinetic energies.

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