Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 35C / b / i

Table of contents
- Solution i

Solution

The potential has period

2 π

, so the first Brillouin zone is

- 1/2 \leq k < 1/2

. At its boundary the free-particle states

∣ + ⟩ = e^{i x /2}, ∣ - ⟩ = e^{- i x /2}

are degenerate, with energy

E_{B} = \frac{ℏ ^{2}}{8 m} .

The relevant Fourier coefficient is supplied by

V_{0} cos x

:

⟨ + ∣ V ∣ - ⟩ = \frac{V _{0}}{2} .

The term

- V_{0} cos 2 x

changes wavevector by two and has no matrix element within this degenerate pair. Thus degenerate perturbation theory gives the matrix

(E_{B} V_{0} /2 V_{0} /2 E_{B})

and the two band-edge energies

E_{\pm} = E_{B} \pm \frac{V _{0}}{2} + O (\frac{V _{0}^{2}}{E _{R}}), E_{R} = \frac{ℏ ^{2}}{2 m} .

The nearly-free electron model therefore predicts the lowest band gap

Δ_{weak} = V_{0} + O (\frac{V _{0}^{2}}{E _{R}}) .

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