# Nearly Free Electron Model and the Reduced Zone Scheme

When for example studying the vibrational modes of a one dimensional diatomic chain we find that the dispersion relation $\omega(k)$ is periodic in the one dimensional reciprocal lattice vector $\frac{2\pi}{a}$, and so the dispersion relation can all be displayed in the first Brillouin zone in the reduced zone scheme.

When studying free electrons perturbed by a weak periodic potential we find that the dispersion relation of the free electrons $E=\frac{\hbar^2k^2}{2m}$ develops gaps at the Brillouin zone boundaries. However this (almost) parabolic dispersion does not have a functional dependence that is periodic (i.e for a single value of $k$ there is only one possible energy). Books seem to suggest we can display this in a reduced zone scheme though - why is this?

Edit: Possibly related to Bloch's theorem? I guess Bloch's theorem says that we can always reduce things into the first Brillouin zone - however if this is the case why doesn't it naturally drop out of the calculations rather than us having to force it?

Preamble:

The dispersion relation for a periodic potential is always periodic itself. That is, if $E_k$ is an energy eigenvalue for wavevector $k$ in a lattice with cell size $a$, then $$E_{k + G} = E_k, \;\;\;\forall k,\;G = \frac{2\pi}{a}n\;\;\text{(G vector of the reciprocal lattice)}$$ which follows directly from the translational symmetry of the Hamiltonian. To see this, let $\psi(x)$ be an eigenfunction of the Hamiltonian. Translational symmetry means that (Bloch's theorem) $$\psi(x+a) = e^{ika}\psi(x)$$ for some wavevector $k$, and that any eigenfunction behaving as above under translation corresponds to the same eigenvalue $E_k$. So the corresponding eigenfunctions are also indexed by $k$ (and other quantum numbers, e.g. spin), $\psi_k(x)$.

Now let $\psi_{k+G}(x)$ be an eigenfunction corresponding to energy eigenvalue $E_{k+G}$. Then by translational symmetry again, $$\psi_{k+G}(x+a) = e^{i(k+G)a}\psi_{k+G}(x) = e^{ika}\psi_{k+G}(x)$$ which means that $\psi_{k+G}(x)$ must correspond to eigenvalue $E_k$, and so $E_{k+G} = E_k$. As a side note, from time reversal symmetry it also follows that $E_k = E_{-k}$.

The first Brillouin zone singles out a first period in $E_k$, by setting $-\pi \le ka \le \pi$. If there are $N$ particles in the lattice, then there are $N$ distinct values of $k$ in the first Brillouin zone. It can be shown that relations $E_{k+G} = E_k$, $E_k = E_{-k}$ imply $\frac{dE_k}{dk} = 0$ at $k=\frac{\pi}{a}n$, $\;n=0,\pm1,\pm2,\dots$, that is, $E_k$ has extrema in the center and on the edges of Brillouin zones, where the dispersion relation can be approximated as a parabolic one, $E_k \sim \pm k^2$. In the nearly free electron model, this parabolic approximation becomes, around the center of the first Brillouin zone, $E_k = \frac{\hbar^2 k^2}{2m}$.

The reason why this periodicity is not explicit in expressions such as $E_k = \frac{\hbar^2 k^2}{2m}$ is that the complexity of energy levels, or energy bands, in the periodic system can be described in 3 different schemes. The representation described above, using $E_k = E_{k+G}$ and showing all energy levels (bands) in all regions of the wavevector space, is the so-called periodic zone scheme. The representation using $E_k = \frac{\hbar^2 k^2}{2m}$ shows all bands in the first Brillouin zone only and is known as the reduced zone scheme. There is also the extended zone scheme, shown in the other answer, which shows different bands in different Brillouin zones, with discontinuities at zone edges. 