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

For a long chain of length $L$ with periodic boundary conditions, each spin has $Ldk/(2\pi )$ quantum states in a wavenumber interval $dk$. Including two spin states, a full energy band contains $2L/a$ electrons. The state counting in a Bloch band therefore gives the energy densityThe integral of $1+y^2$ is $8/3$. For $ϵ>0$, the even square-root integral isFor example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end logarithm contributes $\ln ϵ$. HenceThe continuous zero-coupling limit is $\mathcal{U}(0)=2E_{\mathrm{ZB}}/(3a)$. This agrees with directly integrating the one-dimensional free electron model up to $k_F=\pi /a$ at two electrons per unit cell.

At every $y$ and every $ϵ>0$,so the lower energy band lies strictly below its zero-coupling counterpart. Its state counting in a Bloch band and complete occupation are unchanged. Subtracting the two energy densities directly givesThusEquality holds at $ϵ=0$. This is the downward displacement of the filled lower energy band in the sketch. The upward displacement of the empty upper energy band contributes nothing to the zero-temperature electronic energy density.

For $ϵ\to 0^{+}$, the Taylor expansions needed in the exact energy density areSubstitution yieldsThereforeHere 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.