Dispersion relation near Brillouin zones - Periodic potentials

Question

I am trying to wrap my head around periodic potentials and weak periodic potentials from the reduced zone schemes. From the definition of $\psi_k$: $$ \psi_k(x)=\sum_G C_{k-G}e^{i(k-G)x} $$

I understand how $\psi_k=\psi_{k+ng}$ when $g$ is the smallest reciprocal lattice vector. As the sum is infinite a simple relabeling demonstrates the wave function periodicity in $k-space$ and therefore we see that the same periodicity is carried to the energy dispersion relation $E_k=E_{k+ng}$. But if that is true then, any wavevector that differs only by a reciprocal lattice vector (be it $+g$, $+2g$, $+3g$, etc.) then the energy will be the same! So no matter what the wave vector is the are going to be infinite bloch functions with the same energy and the wave function should be a superposition of every single one of them. In the free electron the wave function should be: $$ \psi=\sum_ne^{i(k+ng)x} $$

so the brillouin zone boundary carries nothing special... I know this isn't true, but the whole concept of bands, periodicity in k-space and periodic dispersion relations is killing me, I can't understand what is so special at the zone boundaries so as to not consider other terms in the superposition. Please just keep in one dimension, this is already hard as it is.

Additional: This obstacle came originally from a problem in Oxford Solid State Basics where it is asked to explain why the wave function at the 1st BZ boundary the wavefunction should be: $$ |\psi\rangle=A|k\rangle+B|k+G\rangle $$

I have no idea why this is the case only at the zone boundary and not at all wave vectors $k$ since he dispersion relation is periodic.

Answer 1

The states that differ by a reciprocal lattice vector are the same state (i.e., not different states with the same energy). This is part of the statement of the Bloch theorem.

This fact is often utilized for various tricks in mapping and discussing the Brillouin zone - which is sometimes considered as "extended" and sometimes is "wrapped" to the interval $[-G/2, G/2]$. The problem cited in the question deals with this case: at the center of the Brillouin zone the states can be considered plane-wave-like, i.e. we can neglect the existence of the periodic potential. By wrapping the plane waves with spectrum $$\epsilon(k) = \frac{\hbar^2k^2}{2m}, k \in [-\infty, +\infty]$$ to the first Brillouin zone we obtain a series of bands $$\epsilon_n(k) = \frac{\hbar^2(k+nG)^2}{2m}, k\in\left[-\frac{G}{2}, +\frac{G}{2}\right], n = 0, \pm 1, \pm 2,....$$ At the edges of the Brillouin zone the existence of the periodic potential cannot be neglected, since the pairs of bands have nearly the same energy, e.g., for $k=G/2 - q$, where $0<q\ll G/2$: $$\epsilon_n\left(\frac{G}{2} - q\right) = \epsilon\left(\frac{G}{2} - q + nG\right) \approx \epsilon\left(\frac{G}{2} + q + nG\right) = \epsilon\left(-\frac{G}{2} - q - nG\right) = \epsilon\left(\frac{G}{2} - q - (n+1)G\right)= \epsilon_{-(n+1)}\left(\frac{G}{2} - q\right),$$ that is $$\epsilon_n(k) \approx\epsilon_{-n}(k-G) = \epsilon_{-(n+1)}(k).$$ This near degeneracy is then resolve using the degenerate perturbation theory, i.e. solving exactly for the degenerate pair of states, which in this case are $|k+nG\rangle$ and $|k-(n+1)G\rangle$. (My notation may differ from the one in the problem, so you have to work it out yourself.)

Answer 2

Based on the argument on your comment, that is not true. For non-interacting electrons in the jellium model, $H_{jel}=T_{jel}(K.E.)$. Therefore, $$H_{jel}\psi_{\vec{k},\sigma}=\frac{\hbar^2\vec{k}^2}{2m}\psi_{\vec{k},\sigma}$$ $$\psi_{\vec{k},\sigma}=\frac{1}{\sqrt{V}}e^{i\vec{k}\cdot\vec{r}}\chi_{\sigma}$$ So, one can see that the eigenvalue is not periodic. Hence you can't say $E_{k}=E_{k+ng}$ since the nature of energy is quadratic.