= Solution
{c}
For a long chain of length $L$ with periodic boundary conditions, each <spin> has $L\,dk/(2\pi)$ <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 density>
$$
\mathcal U(\epsilon)=\frac1\pi\int_{\text{one zone}}E_-(k)\,dk
=\frac{E_{\rm ZB}}a\int_{-1}^1
\left[1+y^2-\sqrt{4y^2+\epsilon^2}\right]dy.
$$
The integral of $1+y^2$ is $8/3$. For $\epsilon>0$, the even square-root integral is
$$
\int_{-1}^1\sqrt{4y^2+\epsilon^2}\,dy
=\sqrt{4+\epsilon^2}+\frac{\epsilon^2}{2}
\ln\frac{2+\sqrt{4+\epsilon^2}}{\epsilon}.
$$
For example, this follows by setting $u=2y$, applying the supplied antiderivative, and evaluating both endpoints; the lower-end <logarithm> contributes $\ln\epsilon$. Hence
$$
\boxed{\mathcal U(\epsilon)=\frac{E_{\rm ZB}}a
\left[\frac83-\sqrt{4+\epsilon^2}
-\frac{\epsilon^2}{2}\ln\frac{2+\sqrt{4+\epsilon^2}}{\epsilon}\right].}
$$
The continuous zero-coupling limit is $\mathcal U(0)=2E_{\rm ZB}/(3a)$. This agrees with directly integrating the one-dimensional <free electron model> up to $k_F=\pi/a$ at two <electrons> per <unit cell>.
Back to article page
