The potential has period $2\pi$, so the first Brillouin zone is $-1/2\leq k<1/2$. At its boundary the free-particle states
$$
|+\rangle=e^{ix/2},
\qquad
|-\rangle=e^{-ix/2}
$$
are degenerate, with energy
$$
E_{\rm B}=\frac{\hbar^2}{8m}.
$$
The relevant Fourier coefficient is supplied by $V_0\cos x$:
$$
\langle+|V|-\rangle=\frac{V_0}{2}.
$$
The term $-V_0\cos2x$ changes wavevector by two and has no matrix element within this degenerate pair. Thus <degenerate perturbation theory> gives the matrix
$$
\begin{pmatrix}
E_{\rm B}&V_0/2\\
V_0/2&E_{\rm B}
\end{pmatrix}
$$
and the two band-edge energies
$$
E_\pm=E_{\rm B}\pm\frac{V_0}{2}
+O\!\left(\frac{V_0^2}{E_{\rm R}}\right),
\qquad E_{\rm R}=\frac{\hbar^2}{2m}.
$$
The <nearly-free electron model> therefore predicts the lowest <band gap>
$$
\boxed{
\Delta_{\rm weak}
=V_0+O\!\left(\frac{V_0^2}{E_{\rm R}}\right)
}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 35C / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/35c/b/i/solution