Question

One-dimensional chain of N identical atoms separated by a distance , has the potential and the function wave can be written as:

$\psi = \alpha e^{i (kx)} + \beta e^{i (k - 2\pi / a) x}$

Replace this wave function in the Schroedinger equation
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$

Multiply the resulting equation
by $\exp{(-ikx)}$ and integrate over the entire chain, and
by $\exp{[-i (k - 2\pi / a) x]}$ and integrate over the entire chain.

If it is requested that the two resulting equations have a non-trivial solution for $\alpha$ and $\beta$ , prove that the energy for this wave function can be written as:

$E=\frac{\hbar^2 k^2}{2m}+\frac{\hbar^2\pi}{ma}\left \{ \left ( \frac{\pi}{a}-k \right )\pm\left [ \left ( \frac{\pi}{a}-k \right )^2 + \left ( \frac{amV_1}{2\pi\hbar^2} \right )^2\right ]^{1/2} \right \}$

What is the value of the energy gap for $k = \pi / a$ ? Show that the shape of the scattering curve for away from the boundary of the first zone is that of the free particle.

Answer 1

"given potential V = V_1 cos (2 pi x/a) psi = de^ikx + Beta e^I (k - 2pi/a)x implies normalization condition =