= Logarithmic energy gain from a Peierls distortion
{title2=$\Delta\mathcal E=-\frac{4\delta^2}{\pi t}\log\frac{t}{|\delta|}+O(\delta^2/t)$}
= Peierls energy gain
{c}
{synonym}
Fill the lower <energy band> of the <dimerized tight-binding chain> with both <spin> states. Take zero on-site <energy> and measure <ground-state energy> per doubled <primitive unit cell>. <State counting in a Bloch band> and $q=ka$ give
$$
\mathcal E(\delta)=-\frac8\pi\int_0^{\pi/2}\sqrt{t^2\cos^2q+\delta^2\sin^2q}\,dq,\qquad \mathcal E(0)=-\frac{8t}\pi.
$$
Equivalently, for $|\delta|\leq t$, $\mathcal E(\delta)=-(8t/\pi)E(\sqrt{1-\delta^2/t^2})$, where $E(k)$ is the <complete elliptic integral of the second kind> in the modulus convention.
For fixed $t>0$ and $|\delta|/t\to0$, the change is
$$
\boxed{\mathcal E(\delta)-\mathcal E(0)=-\frac{4\delta^2}{\pi t}\log\frac{t}{|\delta|}+O(\delta^2/t).}
$$
The <logarithm> comes from states near the edge of the <Brillouin zone>. Put $u=\pi/2-q$. To leading order the square root there is $\sqrt{t^2u^2+\delta^2}$. For a fixed small cutoff $\eta>0$, direct integration gives
$$
\int_0^\eta\left[\sqrt{t^2u^2+\delta^2}-tu\right]du=\frac{\delta^2}{2t}\log\frac{t}{|\delta|}+O(\delta^2/t).
$$
The rest of the band, and the higher-order corrections near its edge, contribute only $O(\delta^2/t)$; multiplying by $-8/\pi$ gives the displayed <asymptotic expansion>. Here the remainder uses <Big O notation>. The negative logarithmic term exceeds any fixed quadratic elastic cost $K\delta^2$, $K>0$, for sufficiently small nonzero $\delta$. This explains the zero-<temperature> instability to a <Peierls distortion> in the ideal chain model.
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