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    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 7 exercises

Warm up exercises

1. Check by yourself

2. $$ [ v_g ] = \frac{[E]}{[p]} = \frac{m}{s} $$

3. $$ [m^*] = \frac{[p^2]}{[E]} = kg $$

4.

$$ m^* = m_e, $$ where \(m_e\) is the free electron mass. This is expected because the free elctrons are not subject to a potential

5. If the dispersion relation is parabolic, so in the free electron model.

Exercise 1: Lattice vibrations

1. The group velocity is given by \begin{align} v_g(k)&=\frac{\partial \omega(k)}{\partial k}\ &= a \sqrt{\frac{\kappa}{m}}\cos(\frac{ka}{2}) \frac{\sin(ka/2)}{|\sin(ka/2)|}, \end{align} which can be written as $$ v_g = a \sqrt{\frac{\kappa}{m}} \begin{cases} &\cos(\frac{ka}{2}), 0<ka<\pi\ &-\cos(\frac{ka}{2}), -\pi<ka<0 \end{cases} $$

2. The density of states is \begin{align} g(\omega) &= \frac{L}{\pi} \left|\frac{1}{v_g}\right| \ &= \frac{L}{a \pi} \sqrt{\frac{m}{\kappa}}\frac{1}{\cos(ka/2)}\ &= \frac{L}{a \pi} \sqrt{\frac{m}{\kappa}}\frac{1}{\sqrt{1-\sin^2(ka/2)}}\ &=\frac{2L}{a \pi} \frac{1}{\sqrt{4\kappa / m - \omega^2}}, \end{align} where we substituted back the dispersion relation.

3.

pyplot.subplot(1,2,1)
k = np.linspace(-pi+0.01, pi-0.01, 300)
pyplot.plot(k[0:149], np.sin(k[0:149])/(np.sqrt(1-np.cos(k[0:149]))),'b');
pyplot.plot(k[150:300], np.sin(k[150:300])/(np.sqrt(1-np.cos(k[150:300]))),'b');
pyplot.xlabel(r'\(ka\)'); pyplot.ylabel('\(v(k)\)');
pyplot.xticks([-pi, 0, pi], [r'\(-\pi/2\)', 0, r'\(\pi/2\)']);
pyplot.yticks([-np.sqrt(2), 0, np.sqrt(2)], [r'\(-2\sqrt{\frac{\kappa}{m}}\)', 0, r'\(2\sqrt{\frac{\kappa}{m}}\)']);
pyplot.tight_layout();


pyplot.subplot(1,2,2)
w = np.linspace(0, 0.95, 300);
g = 1/np.sqrt(1-w**2);
pyplot.plot(w, g, 'b');
pyplot.xlabel(r'\(\omega\)'); pyplot.ylabel('\(g(w)\)');
pyplot.xticks([0, 1], [0, r'\(2\sqrt{\frac{k}{m}}\)']);
pyplot.yticks([0.5, 1], [0, r'\(\frac{L}{2\pi a}\sqrt{\frac{\kappa}{m}}\)']);
pyplot.tight_layout();

png

4.

Hint: The group velocity is given as \(v = \frac{d\omega}{dk}\), draw a coordinate system under or above the dispersion graph with \(k\) on the x-axis in which you draw \(\frac{d\omega}{dk}\). Draw a coordinate system next to the dispersion with \(g(\omega)\) on the y-axis in which you graph \(\big(\frac{d\omega}{dk}\big)^{-1}\).

Plots

Exercise 2: Vibrational heat capacity of a 1D monatomic chain

1.

For the energy we have: \(\(\langle E \rangle = \int \hbar \omega g(\omega) (n_{BE}(\hbar \omega) + \frac{1}{2})d\omega\)\) with \(g(\omega)\) being the DOS calculated in exercise 1 and \(n_{BE}(\hbar \omega) = \frac{1}{e^{\hbar\omega/k_BT}-1}\).

2.

For the heat capacity we have: \(\(C = \frac{d \langle E \rangle}{d T} = \int g(\omega) \hbar\omega \frac{d n_{BE}(\hbar \omega)}{d T}d\omega\)\)

Exercise 3: Next-nearest neighbors chain

1.

The Schrödinger equation is given by: \(|\Psi\rangle = \sum_n \phi_n |n\rangle\) such that we find $$ E\phi_n = E_0\phi_n - t\phi_{n-1} - t\phi_{n+1} - t'\phi_{n-2} - t'\phi_{n+2}$$

2.

Solving the Schrödinger equation yields dispersion: \(\(E(k) = E_0 -2t\cos(ka) -2t'\cos(2ka)\)\)

3.

\[m^* = \frac{\hbar^2}{2a^2}\frac{1}{t\cos(ka)+4t'\cos(2ka)}\]

Plot for t=t':

k1 = np.linspace(-pi, -pi/2-0.01, 300);
k2 = np.linspace(-pi/2+0.01, pi/2-0.01, 300);
k3 = np.linspace(pi/2+0.01, pi, 300);

pyplot.plot(k1, 1/(5*np.cos(k1)),'b');
pyplot.plot(k2, 1/(5*np.cos(k2)),'b');
pyplot.plot(k3, 1/(5*np.cos(k3)),'b');
pyplot.xlabel(r'$k$'); pyplot.ylabel('$m_{eff}(k)$');
pyplot.xticks([-pi,0,pi],[r'$-\pi/a$',0,r'$\pi/a$']);
pyplot.yticks([],[]);
pyplot.tight_layout();

png

4.

Plots for 2t'=t, 4t'=t and 10t'=t:

def m(k,t):
    return 1/(np.cos(k)+4*t*np.cos(2*k))

k1 = np.linspace(-1.6, -0.83, 300);
k2 = np.linspace(-0.826, 0.826, 300);
k3 = np.linspace(0.83, 1.6, 300);

pyplot.plot(k1, m(k1,2),'b');
pyplot.plot(k2, m(k2,2),'b');
pyplot.plot(k3, m(k3,2),'b',label='t=2t\'');
pyplot.xlabel('$k$'); pyplot.ylabel('$m_{eff}(k)$');
pyplot.xticks([-1.6,0,1.6],[r'$-\pi/a$',0,r'$\pi/a$']);
pyplot.yticks([0],[]);
pyplot.tight_layout();

k1 = np.linspace(-1.58, -0.81, 300);
k2 = np.linspace(-0.804, 0.804, 300);
k3 = np.linspace(0.81, 1.58, 300);

pyplot.plot(k1, m(k1,4),'r');
pyplot.plot(k2, m(k2,4),'r');
pyplot.plot(k3, m(k3,4),'r',label='t=4t\'');

k1 = np.linspace(-1.575, -0.798, 300);
k2 = np.linspace(-0.790, 0.790, 300);
k3 = np.linspace(0.798, 1.575, 300);

pyplot.plot(k1, m(k1,10),'k');
pyplot.plot(k2, m(k2,10),'k');
pyplot.plot(k3, m(k3,10),'k',label='t=10t\'');

pyplot.legend();

png