Why energy bands are flat at the edges of the 1BZ?

Question

I’m currently studying for an exam concerning Low Dimensional Systems. The professor asked at the exam this question: why energy bands are flat at the edges? I was not present at the exam session, so I don’t know the specific context of the background, but I believe it was talking about the nearly free electron model and the bands resulting after the opening of the gap. Could you help me? I know that the answer involves periodicity and standing waves, but I don’t know much more. Thanks in advance!

EDIT: to add more context, I’m in the situation where I’m applying the nearly free electron model and I want to study the energy of two degenerate states satisfying the von Laue condition (k = k’+K), where k and k’ are two points of the reciprocal lattice and K is a vector belonging to the reciprocal lattice. The resulting wavefunction is the sum of two stationary waves (a sine and a cosine), but I don’t know how this result leads to the flat edges of the band at the edges of the First Brillouin Zone.

Answer 1

This is due to Bragg reflection at the zone boundary. I'll stick to a 1D treatment for simplicity in my answer

We know that the potential is periodic, i.e.

\begin{equation} V(x) = V(x + a) \end{equation}

resulting in a periodic solution for the wavefunction via the Bloch theorem

\begin{equation} \psi_{n,k}(x) = e^{ikx}u_{n,k}(x) \end{equation}

for $ -\pi/a < k < \pi/a$.

The Bragg condition is $\boldsymbol k ^{\prime} = \boldsymbol k + \boldsymbol G$ where $\boldsymbol G$ is a lattice vector. We find in our 1D Brillouin zone that this is satisfied by the pair of momenta $\pm \pi/a$, therefore a wave travelling right will be reflected left (and vice-versa) due to the Bragg condition. Rephrased, the eigenstates at $k = \pm \pi/a$ will couple

\begin{equation} \begin{split} \phi_{\pm} &= \psi_{\pi/a} \pm \psi_{-\pi/a} \\ &\propto e^{i\pi x/a} \pm e^{-i\pi x/a} \end{split} \end{equation}

resulting in standing waves $\phi_{+} \propto \cos{\pi x/a}$ and $\phi_{-} \propto \sin{\pi x/a}$.

Now, why does this imply that the energy dispersion should be flat near the zone boundaries, i.e. why does $|\frac{\partial E_{n,k}}{\partial k}|_{k=\pm \pi/a} = 0$? Well the solutions at the BZ boundary do not depend on $k$, therefore the first derivative is $0$ and the dispersion goes flat.

A slightly more hand wavey explanation is that $v_{g} = \frac{1}{\hbar}\frac{\partial E_{n,k}}{\partial k}$ where $v_{g}$ is the group velocity. A standing wave is necessarily not a travelling wave and therefore has $v_{g} = 0$ meaning that the dispersion is flat at those points.