 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2022/2/b/iii/solution

For

ϵ \to 0^{+}

4 + ϵ^{2} = 2 + \frac{ϵ ^{2}}{4} + O (ϵ^{4}), ln \frac{2 + 4 + ϵ ^{2}}{ϵ} = ln \frac{4}{ϵ} + O (ϵ^{2}) .

Substitution yields

U (ϵ) = \frac{E _{ZB}}{a} [\frac{2}{3} - \frac{ϵ ^{2}}{4} - \frac{ϵ ^{2}}{2} ln \frac{4}{ϵ} + O (ϵ^{4})] = \frac{E _{ZB}}{a} [\frac{2}{3} - \frac{ϵ ^{2}}{4} (1 - 2 ln \frac{ϵ}{4}) + O (ϵ^{4})] .

Therefore

α = 1, β = 2.

Here Big O notation describes the neglected small-coupling terms, and the logarithm is natural. The nonanalytic

ϵ^{2} ln ϵ

term comes from the formerly degenerate quantum states near the Brillouin zone boundary; a uniform nondegenerate perturbation theory expansion of the integrand would fail there. This is the same mechanism as the Peierls energy gain.

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