-
Bowy La Riviere authoredBowy La Riviere authored
search:
exclude: true
Solutions for lecture 12 exercises
from matplotlib import pyplot as plt
import numpy as np
from math import pi
Exercise 1: 3D Fermi surfaces
Question 1.
Well described: (close to) spherical.
Question 2.
K is more spherical, hence 'more' free electron model. Li is less spherical, hence 'more' nearly free electron model. Take a look at Au, and see whether you can link this to what you learned in lecture 11.
Question 3.
Yes. Cubic -> unit cell contains one atom -> monovalent -> half filled band -> metal.
Question 4.
With Solid State knowledge: Na has 1 valence electron, Cl has 7. Therefore, a unit cell has an even number of electrons -> insulating.
Empirical: Salt is transparent, Fermi level must be inside a large bandgap -> insulating.
Exercise 2: Tight binding in 2D
Question 1.
E \phi_{n,m} = \varepsilon_0\phi_{n,m}-t_1 \left(\phi_{n-1,m}+\phi_{n+1,m}\right) -t_2 \left(\phi_{n,m-1}+\phi_{n,m+1}\right)
Question 2.
\psi_n(\mathbf{r}) = u_n(\mathbf{r})e^{i\mathbf{k}\cdot\mathbf{r}} \quad \leftrightarrow \quad \phi_{n,m} = \phi_0 e^{i(k_x n a_x + k_y m a_y)}
Question 3.
E = \varepsilon_0 -2t_1 \cos(k_x a_x) -2t_2 \cos(k_y a_y)
Question 4 and 5.
Monovalent -> half filled bands -> rectangle rotated 45 degrees.
Much less than 1 electron per unit cell -> almost empty bands -> elliptical.
def dispersion2D(N=100, kmax=pi, e0=2):
# Define matrices with wavevector values
kx = np.tile(np.linspace(-kmax, kmax, N),(N,1))
ky = np.transpose(kx)
# Plot dispersion
plt.figure(figsize=(6,5))
plt.contourf(kx, ky, e0-np.cos(kx)-np.cos(ky))
# Making things look ok
cbar = plt.colorbar(ticks=[])
cbar.set_label('$E$', fontsize=20, rotation=0, labelpad=15)
plt.xlabel('$k_x$', fontsize=20)
plt.ylabel('$k_y$', fontsize=20)
plt.xticks((-pi, 0 , pi),('$-\pi/a$','$0$','$\pi/a$'), fontsize=17)
plt.yticks((-pi, 0 , pi),('$-\pi/a$','$0$','$\pi/a$'), fontsize=17)
dispersion2D()
Exercise 3: Nearly-free electron model in 2D
Question 1.
Construct the Hamiltonian with basis vectors (\pi/a,0) and (-\pi/a,0), eigenvalues are
E=\frac{\hbar^2}{2m} \left(\frac{\pi}{a}\right)^2 \pm \left|V_{10}\right|.
Question 2.
Four in total: (\pm\pi/a,\pm\pi/a).