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    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, \(1/T \rightarrow \infty\). The heat capacity is then given as:

    \[ C \overset{\mathrm{low \: T}}{\approx} 9Nk_{\mathrm{B}}\left(\frac{T}{T_{D}}\right)^3\int_0^{\infty}\frac{x^4{\mathrm{e}}^x}{({\mathrm{e}}^x-1)^2}{\mathrm{d}}x. \]

  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. \[ \int k_x k_y \rightarrow \int_{0}^{2\pi} \mathrm{d} \theta \int_{0}^{\infty} k \mathrm{d} k = 2\pi \int_{0}^{\infty} k \mathrm{d} k \]

  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();

png

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

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

Exercise 2: Debye model in 2D.

  1. See lecture notes. 2.

    \[\begin{align} E &= \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega \\ &= \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + C. \end{align}\]

  2. High temperature implies \(\beta \rightarrow 0\), hence \(E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + C\), and then \(C = \frac{k_BL^2\omega^2_D}{2\pi v^2} = 2Nk_B\). 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 \(\beta \rightarrow \infty\), hence \(E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + C = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + C\). Finally \(C = \frac{6\zeta(3)k^3_BL^2}{\pi v^2\hbar^2}T^2\). We used the fact that \(\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx = 2\zeta(3)\) 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 \(C = 3Nk_B\).
  2. In the low temperature limit we have that \(C \sim \frac{2\pi^2k_B^4L^3}{15\hbar^3}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)T^3\). We used that \(\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}\).

Exercise 4: Anisotropic sound velocities.

\[ E = 3\left(\frac{L}{2\pi}\right)^3\int d^3k\hbar\omega(\mathbf{k})\left(n_B(\beta\hbar\omega(\mathbf{k})) + \frac{1}{2}\right) = 3\left(\frac{L}{2\pi}\right)^3\frac{1}{v_xv_yv_z}\int d^3\kappa\frac{\hbar\kappa}{e^{\beta\hbar\kappa} - 1} + C, \]

where we used the substitutions \(\kappa_x = k_xv_x,\kappa_y = k_yv_y, \kappa_z = k_zv_z\). Finally

\[ E = \frac{3\hbar L^3}{2\pi^2}\frac{1}{v_xv_yv_z}\int_0^{\kappa_D} d\kappa\frac{\kappa^3}{e^{\beta\hbar\kappa} - 1} + C = \frac{3L^3}{2\pi^2\hbar^3\beta^4}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1} + C, \]

hence \(C = \frac{\partial E}{\partial T} = \frac{6k_B^4L^3T^3}{\pi^2\hbar^3}\frac{1}{v_xv_yv_z}\int_0^{\beta\hbar\kappa_D} dx\frac{x^3}{e^{x} - 1}\). We see that the result is similar to the one with the linear dispersion, the only difference is the factor \(1/v_xv_yv_z\) instead of \(1/v^3\).