Question

The tight binding energy dispersion (E-k) relation for electrons in a one-dimensional array of atoms having lattice constant 'a' and total length 'L' is :

E=E0−β−2γcoska

Where E0,β and γ are constants and k is the wave vector.

The density of states of electrons (including spin degeneracy) in the band is given by?

Answer 1

Given:

$$E(k)=E_0-\beta-2\gamma\cos ka$$

The density of states

$$D(E)=\frac{1}{\pi}\frac{dE(k)}{dk}=\frac{1}{\pi}a*2\gamma\sin ka$$

$$\sin ka=\sqrt{1-\left(\frac{E_0-\beta-E}{2\gamma}\right)^2}$$

So

$$D(E)=\frac{2a}{\pi}\sqrt{1-\left(\frac{E_0-\beta-E}{2\gamma}\right)^2}$$