 Solution

ID: past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2019/section-b/4/b/solution

Write

U = V_{g}

∣ 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) .

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

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.

 New to topics? Read the docs here!