    import matplotlib
from matplotlib import pyplot

import numpy as np

from common import draw_classic_axes, configure_plotting

configure_plotting()

pi = np.pi

Solutions for lecture 8 exercises¶

Quick warm-up exercises¶

1.

The part first part of $ω^{2}$ will always be equal or larger than the second part. Therefore $ω^{2} \geq 0$ .

2.

Be cautious with the difference in the unit cell size.

3.

Values of $k$ outside of the 1st Brillouin zone describe the same solutions.

Exercise 1: analyzing the diatomic vibrating chain¶

1.

Use the small angle approximation $\sin (x) \approx x$ to ease calculations. For the Taylor polynomial take $ω^{2} = f (x) \approx f (0) + f^{'} (0) k + \frac{1}{2} f^{″} (0) k^{2}$ (some terms vanish, computation is indeed quite tedious, but it's a 'fun' task). Calculating the group velocity yields

| v_{g} | = \sqrt{\frac{κ a^{2}}{2 (m_{1} + m_{2})}}

2.

Optical branch corresponds with (+) in the equation given in the lecture notes. Be smart: you do not have to calculate the derivatives again. Finding the Taylor polynomial andcomputing the group velocity results in

| v_{g} | = 0

3.

Density of states is given as $g (ω) = d N / d ω = d N / d k \times d k / d ω$ . We know $d N / d k = 2 L / 2 π = L / π$ since we have 1D and positive and negative $k$ -values. $d k / d ω$ can be computed using the group velocity: $d k / d ω = (d ω / d k)^{- 1} = (v_{g})^{- 1}$

Exercise 2: the Peierls transition¶

1.

A unit cell consits of exactly one $t_{1}$ and one $t_{2}$ hopping.

2.

Using the hint we find: $E ϕ_{n} = ϵ ϕ_{n} + t_{1} ψ_{n} + t_{2} ψ_{n - 1}$ $E ψ_{n} = t_{1} ϕ_{n} + t_{2} ϕ_{n + 1} + ϵ ψ_{n}$

Notice that the hopping, in this case, is without the '-'-sign!

3.

Using the Ansatz and rearranging the equation yields:

E (\begin{matrix} ϕ_{0} \\ ψ_{0} \end{matrix}) = (\begin{matrix} ϵ & t_{1} + t_{2} e^{- i k a} \\ t_{1} + t_{2} e^{i k a} & ϵ \end{matrix}) (\begin{matrix} ϕ_{0} \\ ψ_{0} \end{matrix})

4.

The dispersion is given by:

E = ϵ \pm \sqrt{t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} \cos (k a)} .

pyplot.figure()
k = np.linspace(-2*pi, 2*pi, 400)
t1 = 1;
t2 = 1.5;
pyplot.plot(k, -(t1+t2)*np.cos(k/2),'r',label='1 atom dispersion')
pyplot.plot(k[199:100:-1],-(t1+t2)*np.cos(k[0:99]/2),'r--',label='1 atom dispersion with folded Brillouin zone')
pyplot.plot(k[299:200:-1],-(t1+t2)*np.cos(k[300:399]/2),'r--')
pyplot.plot(k, np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b',label='2 atom dispersion')
pyplot.plot(k, -np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b')

pyplot.xlabel('$ka$'); pyplot.ylabel(r'$E-\epsilon$')
pyplot.xlim([-2*pi,2*pi])
pyplot.ylim([-1.1*(t1+t2),1.1*(t1+t2)])
pyplot.xticks([-2*pi, -pi, 0, pi,2*pi], [r'$-2\pi$',r'$-\pi$', 0, r'$\pi$',r'$2\pi$'])
pyplot.yticks([-t1-t2, -np.abs(t1-t2), 0, np.abs(t1-t2), t1+t2], [r'$-t_1-t_2$',r'$-|t_1-t_2|$', '0', r'$|t_1-t_2|$', r'$t_1+t_2$']);
pyplot.vlines([-pi, pi], -2*(t1+t2)*1.1,2*(t1+t2)*1.1, linestyles='dashed');
pyplot.hlines([-np.abs(t1-t2), np.abs(t1-t2)], -2*pi, 2*pi, linestyles='dashed');
pyplot.fill_between([-3*pi,3*pi], -np.abs(t1-t2), np.abs(t1-t2), color='red',alpha=0.2);

pyplot.legend(loc='lower center');

png

(Press the magic wand tool to enable the python code that created the figure to see what happends if you change $t_{1}$ and $t_{2}$ .)

Notice that the red shaded area is not a part of the Band structure anymore! Also check the wikipedia article.

5.

For the group velocity we find

v_{g} (k) = \mp \frac{t_{1} t_{2}}{ℏ} \frac{a \sin (k a)}{\sqrt{t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} \cos (k a)}},

and for the effective mass we obtain

m^{*} (k) = \mp \frac{ℏ^{2}}{a^{2} t_{1} t_{2}} {[\frac{t_{1} t_{2} \sin^{2} (k a)}{(t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} \cos (k a))^{3 / 2}} + \frac{\cos (k a)}{\sqrt{t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} \cos (k a)}}]}^{- 1} .

6.

We know $g (E) = \frac{d N}{d k} \frac{d k}{d E} = \frac{L}{π} \frac{1}{ℏ v_{g} (E)}$ with $v_{g} (k)$ from the previous subquestion. Rewriting $v_{g} (k)$ to $v_{g} (E)$ and substituting it in $g (E)$ gives

g (E) = \frac{4 L}{a π} \frac{| E - ϵ |}{\sqrt{4 t_{1}^{2} t_{2}^{2} - {[(E - ϵ)^{2} - t_{1}^{2} - t_{2}^{2}]}^{2}}} .

Graphically the density of states looks accordingly:

pyplot.subplot(1,3,1)
k = np.linspace(-2*pi, 2*pi, 400)
t1 = 1;
t2 = 1.5;
pyplot.plot(k, -(t1+t2)*np.cos(k/2),'r',label='1 atom dispersion')
pyplot.plot(k[199:100:-1],-(t1+t2)*np.cos(k[0:99]/2),'r--',label='1 atom dispersion with folded Brillouin zone')
pyplot.plot(k[299:200:-1],-(t1+t2)*np.cos(k[300:399]/2),'r--')
pyplot.plot(k, np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b',label='2 atom dispersion')
pyplot.plot(k, -np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k)),'b')

pyplot.xlabel('$ka$'); pyplot.ylabel(r'$E-\epsilon$')
pyplot.xlim([-2*pi,2*pi])
pyplot.xticks([-2*pi, -pi, 0, pi,2*pi], [r'$-2\pi$',r'$-\pi$', 0, r'$\pi$',r'$2\pi$'])
pyplot.yticks([-t1-t2, -np.abs(t1-t2), 0, np.abs(t1-t2), t1+t2], [r'$-t_1-t_2$',r'$-|t_1-t_2|$', '0', r'$|t_1-t_2|$', r'$t_1+t_2$']);

pyplot.subplot(1,3,2)
w = np.sqrt(t1**2 + t2**2+2*t1*t2*np.cos(k))
pyplot.hist(w,30, orientation='horizontal',ec='black',color='b');
pyplot.hist(-w,30, orientation='horizontal',ec='black',color='b');
pyplot.xlabel(r'$g(E)$')
pyplot.ylabel(r'$E-\epsilon$')
pyplot.yticks([],[])
pyplot.xticks([],[])

pyplot.subplot(1,3,3)
w = -(t1+t2)*np.cos(k/2)
pyplot.hist(w,60, orientation='horizontal',ec='black',color='r');
pyplot.xlabel(r'$g(E)$')
pyplot.ylabel(r'$E-\epsilon$')
pyplot.yticks([],[])
pyplot.xticks([],[])

pyplot.suptitle('Density of states for 2 atom unit cell and 1 atom unit cell');

png

Exercise 3: atomic chain with 3 different spring constants¶

1.

The unit cell should contain exactly one spring of $κ_{1}$ , $κ_{2}$ and $κ_{3}$ and exactly three atoms.

2.

The equations of motion are

\begin{aligned} m {\ddot{u}}_{n, 1} & = - κ_{1} (u_{n, 1} - u_{n, 2}) - κ_{3} (u_{n, 1} - u_{n - 1, 3}) \\ m {\ddot{u}}_{n, 2} & = - κ_{2} (u_{n, 2} - u_{n, 3}) - κ_{1} (u_{n, 2} - u_{n, 1}) \\ m {\ddot{u}}_{n, 3} & = - κ_{3} (u_{n, 3} - u_{n + 1, 1}) - κ_{2} (u_{n, 3} - u_{n, 2}) \end{aligned}

3.

Substitute the following ansatz in the equations of motion:

(\begin{matrix} u_{1, n} \\ u_{2, n} \\ u_{3, n} \end{matrix}) = e^{i ω t - i k x_{n}} (\begin{matrix} A_{1} \\ A_{2} \\ A_{3} \end{matrix})

4.

Because the spring matrix and the matrix $X$ commute, they share a common set of eigenvectors. The eigenvalues of the matrix $X$ are $λ = - 1$ with an eigenvector

(\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix})

and $λ = + 1$ with eigenvectors

(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) .

These eigenvectors can be used to calculate the eigenvalues of the spring matrix. However, be cautious! The eigenvalue $λ = + 1$ is degenerate and to find the eigenvalue of the spring matrix we should take a linear combination of the two corresponding eigenvectors. This is explained at page 3 in this document.

The eigenvalues of the spring matrix are

ω^{2} = (\begin{matrix} ω_{1}^{2} \\ ω_{2}^{2} \\ ω_{3}^{2} \end{matrix}) = \frac{1}{m} (\begin{matrix} q \\ k_{3} + \frac{3 q}{2} - \frac{\sqrt{4 k_{3}^{2} - 4 k_{3} q + 9 q^{2}}}{2} \\ k_{3} + \frac{3 q}{2} + \frac{\sqrt{4 k_{3}^{2} - 4 k_{3} q + 9 q^{2}}}{2} \end{matrix})

5.

If $κ_{1} = κ_{2} = κ_{3}$ then we have the uniform mono-atomic chain. If the length of the 3 spring constant unit cell is $a$ , then the length of the mono-atomic chain is $a / 3$ .