from matplotlib import pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable

import numpy as np
from scipy.optimize import curve_fit
from scipy.integrate import quad

from common import draw_classic_axes, configure_plotting

configure_plotting()
  

Solutions for lecture 2 exercises

warm-up exercises

1. For low T, $1/T\to \mathrm{\infty }$. The heat capacity is then given as:

   $$C\stackrel{\mathrm{low}\mathrm{T}}{\approx }9N{k}_{\mathrm{B}}{\left(\frac{T}{T_D}\right)}^3{\int }_0^{\mathrm{\infty }}\frac{x^4{\mathrm{e}}^x}{({\mathrm{e}}^x-1{)}^2}\mathrm{d}x.$$

2. See plot below (shown for $T_{D,1}<T_{D,2}$)

3. The polarization is related to the direction of the amplitudes of the waves with respect to the direction of the wave. In 3D, there are only 3 different amplitude directions possible.
4. $$\int k_x{k}_y\to {\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^{\mathrm{\infty }}k\mathrm{d}k=2\pi {\int }_0^{\mathrm{\infty }}k\mathrm{d}k$$

5. The Debye frequency ${\omega }_D$.

6. The wavelength is of the order of the interatomic spacing:

   $$\lambda =(\frac{4}{3}\pi {)}^{1/3}a.$$

fig, ax = plt.subplots()
T = np.linspace(0.1, 3)
T_D = [1,2]
ax.plot(T, (T/T_D[0])**3, 'b-', label = r'$T_{D,1}$')
ax.plot(T, (T/T_D[1])**3, 'r-', label = r'$T_{D,2}$')
ax.set_ylim([0,3])
ax.set_xlim([0,3])
ax.set_xlabel('$T$')
ax.set_xticks([0])
ax.set_xticklabels(['$0$'])
ax.set_ylabel('$C$')
ax.set_yticks([0])
ax.set_yticklabels(['$0$'])
ax.legend();

Exercise 1: Debye model: concepts.

1. $k=\frac{4\pi }{L}$ and $k=-\frac{4\pi }{L}$.

3. The number of states per $k$ or per frequency. 4.

   $$g(\omega )=\frac{dN}{d\omega }=\frac{dN}{dk}\frac{dk}{d\omega }=\frac{1}{v}\frac{dN}{dk}.$$

   We assume that in $d$ dimensions there are $d$ polarizations.

For 1D we have that $N=\frac{L}{2\pi }{\int }_{-k}^{k}dk$, hence $g(\omega )=\frac{L}{\pi v}$.

For 2D we have that $N=2{\left(\frac{L}{2\pi }\right)}^2\int d^{2}k=2{\left(\frac{L}{2\pi }\right)}^2\int 2\pi kdk$, hence $g(\omega )=\frac{L^2\omega }{\pi v^2}$.

For 3D we have that $N=3{\left(\frac{L}{2\pi }\right)}^3\int d^{3}k=3{\left(\frac{L}{2\pi }\right)}^3\int 4\pi k^{2}dk$, hence $g(\omega )=\frac{3L^3{\omega }^2}{2{\pi }^2{v}^3}$.

Exercise 2: Debye model in 2D.

1. See lecture notes. 2.

   $$\begin{array}{rl}E& ={\int }_0^{{\omega }_D}g(\omega )\hslash \omega (\frac{1}{e^{\beta \hslash \omega }-1}+\frac{1}{2})d\omega \\ & =\frac{L^2}{\pi v^2{\hslash }^2{\beta }^3}{\int }_0^{\beta \hslash {\omega }_D}\frac{x^2}{e^x-1}dx+C.\end{array}$$

2. High temperature implies $\beta \to 0$, hence $E=\frac{L^2}{\pi v^2{\hslash }^2{\beta }^3}\frac{(\beta \hslash {\omega }_D{)}^2}{2}+C$, and then $C=\frac{k_B{L}^2{\omega }_D^2}{2\pi v^2}=2N{k}_B$. We've used the value for ${\omega }_D$ calculated from $2N={\int }_0^{{\omega }_D}g(\omega )d\omega$.

3. In the low temperature limit we have that $\beta \to \mathrm{\infty }$, hence $E\approx \frac{L^2}{\pi v^2{\hslash }^2{\beta }^3}{\int }_0^{\mathrm{\infty }}\frac{x^2}{e^x-1}dx+C=\frac{2\zeta (3)L^2}{\pi v^2{\hslash }^2{\beta }^3}+C$. Finally $C=\frac{6\zeta (3)k_B^3{L}^2}{\pi v^2{\hslash }^2}{T}^2$. We used the fact that ${\int }_0^{\mathrm{\infty }}\frac{x^2}{e^x-1}dx=2\zeta (3)$ where $\zeta$ is the Riemann zeta function.

Exercise 3: Different phonon modes.

1.

\begin{gather}
g(\omega) = \sum_{\text{polarizations}}\frac{dN}{dk}\frac{dk}{d\omega} = \left(\frac{L}{2\pi}\right)^3\sum_{\text{polarizations}}4\pi k^2\frac{dk}{d\omega} = \frac{L^3}{2\pi^2}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\omega^2,\\
E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^3}{2\pi^2\hbar^3\beta^4}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx + C.
\end{gather}

1. Note that we can get ${\omega }_D$ from $3N={\int }_0^{{\omega }_D}g(\omega )$ so everything cancels as usual and we are left with the Dulong-Petit law $C=3N{k}_B$.
2. In the low temperature limit we have that $C\sim \frac{2{\pi }^2{k}_B^4{L}^3}{15{\hslash }^3}(\frac{2}{v_{\perp }^3}+\frac{1}{v_{\parallel }^3})T^3$. We used that ${\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x-1}dx=\frac{{\pi }^4}{15}$.

Exercise 4: Anisotropic sound velocities.

$$E=3{\left(\frac{L}{2\pi }\right)}^3\int d^{3}k\hslash \omega (\mathbf{k})(n_B(\beta \hslash \omega (\mathbf{k}))+\frac{1}{2})=3{\left(\frac{L}{2\pi }\right)}^3\frac{1}{v_x{v}_y{v}_z}\int d^3\kappa \frac{\hslash \kappa }{e^{\beta \hslash \kappa }-1}+C,$$

where we used the substitutions ${\kappa }_x=k_x{v}_x,{\kappa }_y=k_y{v}_y,{\kappa }_z=k_z{v}_z$. Finally

$$E=\frac{3\hslash L^3}{2{\pi }^2}\frac{1}{v_x{v}_y{v}_z}{\int }_0^{{\kappa }_D}d\kappa \frac{{\kappa }^3}{e^{\beta \hslash \kappa }-1}+C=\frac{3L^3}{2{\pi }^2{\hslash }^3{\beta }^4}\frac{1}{v_x{v}_y{v}_z}{\int }_0^{\beta \hslash {\kappa }_D}dx\frac{x^3}{e^x-1}+C,$$

hence $C=\frac{\partial E}{\partial T}=\frac{6k_B^4{L}^3{T}^3}{{\pi }^2{\hslash }^3}\frac{1}{v_x{v}_y{v}_z}{\int }_0^{\beta \hslash {\kappa }_D}dx\frac{x^3}{e^x-1}$. We see that the result is similar to the one with the linear dispersion, the only difference is the factor $1/v_x{v}_y{v}_z$ instead of $1/v^3$.