Past exam of the condensed matter physics course of the University of Oxford / 2022 / 2 / b / iii

Table of contents
- Solution iii

Solution

For

ϵ \to 0^{+}

, the Taylor expansions needed in the exact energy density are

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.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version