    from matplotlib import pyplot as plt 
import numpy as np
from math import pi

Solutions for lecture 12 exercises¶

Exercise 1: 3D Fermi surfaces¶

Subquestion 1¶

Well described: (close to) spherical.

Subquestion 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.

Subquestion 3¶

Yes. Cubic -> unit cell contains one atom -> monovalent -> half filled band -> metal.

Subquestion 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¶

Subquestion 1¶

E ϕ_{n, m} = ε_{0} ϕ_{n, m} - t_{1} (ϕ_{n - 1, m} + ϕ_{n + 1, m}) - t_{2} (ϕ_{n, m - 1} + ϕ_{n, m + 1})

Subquestion 2¶

ψ_{n} (r) = u_{n} (r) e^{i k \cdot r} \leftrightarrow ϕ_{n, m} = ϕ_{0} e^{i (k_{x} n a_{x} + k_{y} m a_{y})}

Subquestion 3¶

E = ε_{0} - 2 t_{1} \cos (k_{x} a_{x}) - 2 t_{2} \cos (k_{y} a_{y})

Subquestion 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()

png

Exercise 3: Nearly-free electron model in 2D¶

Subquestion 1¶

Construct the Hamiltonian with basis vectors $(π / a, 0)$ and $(- π / a, 0)$ , eigenvalues are

E = \frac{ℏ^{2}}{2 m} {(\frac{π}{a})}^{2} \pm | V_{10} | .

Subquestion 2¶

Four in total: $(\pm π / a, \pm π / a)$ .

Subquestion 3¶

Define a basis, e.g. $\begin{aligned} | 0 ⟩ & = (π / a, π / a) | 1 ⟩ & = (π / a, - π / a) | 2 ⟩ & = (- π / a, - π / a) | 3 ⟩ & = (- π / a, π / a) \end{aligned}$ The Hamiltonian becomes

\hat{H} = (\begin{matrix} ε_{0} & V_{10} & V_{11} & V_{10} \\ V_{10} & ε_{0} & V_{10} & V_{11} \\ V_{11} & V_{10} & ε_{0} & V_{10} \\ V_{10} & V_{11} & V_{10} & ε_{0} \end{matrix})

Subquestion 4¶

Using the symmetry of the matrix, we try a few eigenvectors: $v_{\pm} = (\begin{matrix} 1 \pm 1 1 \pm 1 \end{matrix})$ $v_{α} = (\begin{matrix} α 1 - α - 1 \end{matrix})$

E_{\pm} = ε_{0} + V_{11} \pm 2 V_{10} and E_{α} = ε_{0} - V_{11}