Replace 4_sommerfeld_model_solutions.md

docs/4_sommerfeld_model_solutions.md

@@ -130,7 +130,66 @@ $$
 C_v(T) = \frac{\partial E}{\partial T} = \frac{3L^2k_B^3T^2}{\pi c^2}
 $$
 
+### Exercise 4: Two energy bands
 
+#### Question 1.
+```python
+import numpy as np
+import matplotlib.pyplot as plt
+x = np.linspace(-1, 1, 100)
+fig, ax = plt.subplots(figsize=(7, 5))
+ax.plot(x, x**2, 'b')
+ax.plot(x, x**2 + -0.2, 'b')
+ax.axhline(y=0.5, color='r', linestyle='-')
+
+ax.spines['left'].set_position('center')
+ax.spines['bottom'].set_position(('data', 0.0))
+
+# Eliminate upper and right axes
+ax.spines['right'].set_color('none')
+ax.spines['top'].set_color('none')
+
+ax.set_xticks([])
+ax.set_yticks([])
+
+ax.set_xlabel(r'$\mid \vec k \mid$', fontsize=14)
+ax.set_ylabel(r'$\varepsilon$', fontsize=18, rotation='horizontal')
+ax.yaxis.set_label_coords(0.5,1)
+ax.xaxis.set_label_coords(1.0, 0.49)
+```
+#### Question 2. 
+We compute the density of states separately for each band and add the result. Note that for the first band there will be no states below $\varepsilon <0$ and for the second band no states below $\varepsilon<\varepsilon_0$.
+\begin{align}
+N_{1,states} &= 2_s (\frac{L}{2\pi})^2 2\pi \int k \textrm{d}k \\
+			 &= 2_s (\frac{L}{2\pi})^2 2\pi \int  sqrt{\frac{\varepsilon}{A}} \frac{1}{2}\sqrt{\frac{1}{\varepsilon A}} \textrm{d}\varepsilon\\
+g_1(\varepsilon) = 2_s (\frac{L}{2\p})^2 \frac{\pi}{A}\\
+N_{2,states} &= 2_s (\frac{L}{2\pi})^2 2\pi \int k \textrm{d}k \\
+			 &= 2_s (\frac{L}{2\pi})^2 2\pi \int  sqrt{\frac{\varepsilon-varepsilon_0}{A}} \frac{1}{2}\sqrt{\frac{1}{varepsilon-varepsilon_0 A}} \textrm{d}\varepsilon\\
+g_2(\varepsilon) &= 2_s (\frac{L}{2\pi})^2 \frac{\pi}{A}
+\end{align}
+$$
+g(\varepsilon) =\begin{cases}
+ 2_s (\frac{L}{2\pi  })^2 \frac{\pi}{A} &\text{for}&0<\varepsilon<\varepsilon_0 \\
+ 2_s (\frac{L}{2\pi  })^2 \frac{2\pi}{A} &\text{for}&\varepsilon>\varepsilon_0 
+\end{cases} 
+$$
+#### Question 3. 
+Add $T = 0$ the chemical potential $\mu$ is equal to the Fermi energy $E_F$ and the Fermi-Dirac distributionis the step function
+$$
+N &= \int_0^\infty g(\varepsilon) n_F(\beta(\varepsilon-E_F)) \textrm{d}\varepsilon \\
+  &= \int_0^{\varepsilon_0} g(\varepsilon)  \textrm{d}\varepsilon + \int^{E_F}_{\varepsilon_0} g(\varepsilon)  \textrm{d}\varepsilon
+  &= 2_s (\frac{L}{2\pi})^2 \frac{\pi}{A} \varepsilon_0 + 2_s (\frac{L}{2\pi})^2 \frac{2\pi}{A}  (E_F-\varepsilon_0)
+  &=  2_s (\frac{L}{2\pi})^2 \frac{\pi}{A} (2E_F -\varepsilon_0)
+$$
+#### Question 4. 
+$$
+N = \int_\varepsilon_a^\infty g(\varepsilon) n_F(\beta(\varepsilon-E_F)) \textrm{d}\varepsilon 
+$$
+#### Question 5. 
+In this case the Fermi-Dirac can be approximated by $n_F(\beta(\varepsilon-E_F)) \approx e^{-\beta (\varepsilon-E_F)}$, and working out the integral we obtain
+$$
+N = 2_s (\frac{L}{2\pi})^2 \frac{2\pi}{A} \int_{varepsilon_a}^\infty  e^{-\beta (\varepsilon-E_F)} \textrm{d}\varepsilon = 2_s (\frac{L}{2\pi})^2 \frac{2\pi}{A} k_B T e^{-\beta (\varepsilon_a-E_F)}
+$$
 ## extra exercises
 ### Exercise 1: the n-dimensional free electron model