Updated solutions and added solutions to the warm-up exercises

src/2_debye_model_solutions.md

+---
+jupyter:
+  jupytext:
+    text_representation:
+      extension: .md
+      format_name: markdown
+      format_version: '1.0'
+      jupytext_version: 0.8.6
+  kernelspec:
+    display_name: Python 3
+    language: python
+    name: python3
+---
+
+```python tags=["initialize"]
+from matplotlib import pyplot
+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, $\beta \rightarrow \infty$. The heat capacity is then given as:
+$$
+C \overset{\mathrm{low \: T}}{\approx} 9Nk_{\mathrm{B}}\left(\frac{T}{T_{D}}\right)^3\int_0^{T_{D}/T}\frac{x^4{\mathrm{e}}^x}{({\mathrm{e}}^x-1)^2}{\mathrm{d}}x.
+$$
+2. 
+```python
+fig, ax = plt.subplots()
+T = np.linspace(0.1, 3)
+T_D = [1,2]
+ax.plot(omega, (T/T_D[0])**3, 'b-', label = r'$T_{D,1}$')
+ax.plot(omega, (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.set_title('$T_{D,1}<T_{D,2}$')
+ax.legend();
+```
+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. $$
+\int k_x k_y \rightarrow \int_limits_{0}^{2\pi} \mathrm{d} \theta \int_limits_{0}^{\infty} k \mathrm{d} k = 2\pi \int_limits_{0}^{\infty} k \mathrm{d} k 
+$$
+5. The Debye frequency $\omega_D$.
+6. The wavelength is of the order of the interatomic spacing:
+$$
+\lambda = \sqrt{\frac{4}{3}\pi} a.
+$$
 
 ### Exercise 1: Debye model: concepts.
 
-1, 2, 3. 
-
-See the lecture notes.
-
-4.
-
+1. $k = \frac{L}{4\pi}$ and $k = -\frac{L}{4\pi}$.  
+2. 
+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 $g(\omega) = \frac{L}{\pi v}$.
@@ -22,29 +79,21 @@ For 3D we have that $N = 3\left(\frac{L}{2\pi}\right)^3\int d^3k = 3\left(\frac{
 
 ###  Exercise 2: Debye model in 2D.
 
-1.
-
-See the lecture notes.
-
-2.
-
+1. See lecture notes.
+2. 
 $$
-E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant}.
+\begin{align}
+E &= \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega \\
+&= \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\beta\hbar\omega_D}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant}.
+\end{align}
 $$
-
-3.
-
-High temperature implies $\beta \rightarrow 0$, hence $E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + T \text{ independent constant}$, 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 $2N = \int_{0}^{\omega_D}g(\omega)d\omega$.
-
-4.
-
-In the low temperature limit we have that $\beta \rightarrow \infty$, hence $E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant} = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + T \text{ independent constant}$. 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.
+3. High temperature implies $\beta \rightarrow 0$, hence $E = \frac{L^2}{\pi v^2\hbar^2\beta^3}\frac{(\beta\hbar\omega_D)^2}{2} + T \text{ independent constant}$, 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 $2N = \int_{0}^{\omega_D}g(\omega)d\omega$.
+4. In the low temperature limit we have that $\beta \rightarrow \infty$, hence $E \approx \frac{L^2}{\pi v^2\hbar^2\beta^3}\int_{0}^{\infty}\frac{x^2}{e^{x} - 1}dx + T \text{ independent constant} = \frac{2\zeta(3)L^2}{\pi v^2\hbar^2\beta^3} + T \text{ independent constant}$. 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.
-
+1. 
 $$
 g(\omega) = \sum_{\text{polarizations}}\frac{dN}{dk}\frac{dk}{d\omega} = \left(\frac{L}{2\pi}\right)^3\sum_{\text{polarizations}}4\pi k^2\frac{dk}{d\omega} = \frac{L^3}{2\pi^2}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\omega^2
 $$
@@ -53,13 +102,9 @@ $$
 E = \int_{0}^{\omega_D}g(\omega)\hbar\omega\left(\frac{1}{e^{\beta\hbar\omega} - 1} + \frac{1}{2}\right)d\omega = \frac{L^3}{2\pi^2\hbar^3\beta^4}\left(\frac{2}{v_\perp^3} + \frac{1}{v_\parallel^3}\right)\int_{0}^{\beta\hbar\omega_D}\frac{x^3}{e^{x} - 1}dx + T \text{ independent constant}.
 $$
 
-2.
-
-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 $C = 3Nk_B$.
-
-3.
+2. 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 $C = 3Nk_B$.
 
-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 $\int_{0}^{\infty}\frac{x^3}{e^{x} - 1}dx = \frac{\pi^4}{15}$.
+3. 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.