polished Sommerfeld ex.

docs/4_sommerfeld_model.md

@@ -485,7 +485,7 @@ Your result should be linear with $|\varepsilon|$.
 ### Exercise 4: Two energy bands
 An 'energy band' is a range of energies within which there are states available for the particles in the system. An energy band is therefore closely related to the dispersion relation. For instance, for the free electron dispersion $\varepsilon = \hbar^2k^2/2m$, the available states (=the energy band) lie between zero and infinite energy. As we will encounter more often in our course, many materials have multiple, possibly overlapping energy bands, which are each described by their own dispersion. Here we analyze a system with two, partially overlapping, energy bands. These bands are each described by a free-electron dispersion, but they are  offset with repect to each other in energy. The goal is to calculate the density of states and electron occupation of the system.
 
-The dispersion of energy band 1 is $\varepsilon_1(\mathbf{k}) = Ak^2$ and that of band 2 is $\varepsilon_2(\mathbf{k}) = Ak^2 + \varepsilon_0$, with $A = \hbar^2/2m$. The Fermi energy is $E_F$, and we assume $E_F \gg \varepsilon_0$. We consider a _two-dimensional_ system.
+The dispersion of energy band 1 is $\varepsilon_1(\mathbf{k}) = \tfrac{\hbar^2k^2}{2m}$ and that of band 2 is $\varepsilon_2(\mathbf{k}) = \tfrac{\hbar^2k^2}{2m} + \varepsilon_0$. The Fermi energy is $E_F$, and we assume $E_F \gg \varepsilon_0$. We consider a _two-dimensional_ system.
 
 1. Sketch the two dispersions in one plot. Indicate the Fermi energy.
 2. Calculate the density of states and sketch it as a function of energy. Hint: the total density of states is obtained by adding the density of states associated with the individual bands.

docs/4_sommerfeld_model_solutions.md

@@ -2,6 +2,17 @@
 search:
   exclude: true
 ---
+
+```python tags=["initialize"]
+from matplotlib import pyplot
+
+import numpy as np
+
+from common import draw_classic_axes, configure_plotting
+
+configure_plotting()
+```
+
 # Solutions for exercises lecture 4: Sommerfeld model
 
 ### Warm-up questions
@@ -137,28 +148,18 @@ $$
 ### Exercise 4: Two energy bands
 
 ```python
-import numpy as np
-import matplotlib.pyplot as plt
-x = np.linspace(-1, 1, 100)
-fig, ax = plt.subplots(figsize=(7, 5))
+fig, ax = pyplot.subplots()
+x = np.linspace(-3, 3, 100)
 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.plot(x, x**2 + 1, 'b')
+ax.axhline(y=2, color='r', linestyle='-')
 ax.set_xticks([])
-ax.set_yticks([])
-
-ax.set_xlabel(r'$ k $', fontsize=14)
-ax.set_ylabel(r'$\varepsilon$', fontsize=18, rotation='horizontal')
-ax.yaxis.set_label_coords(0.5,1)
-ax.xaxis.set_label_coords(1.0, 0.49)
+ax.set_yticks([0,1,2]);
+ax.set_yticklabels([r'$0$',r'$\varepsilon_0$',r"$E_F$"])
+ax.set_ylabel(r"$\varepsilon$")
+ax.set_xlabel(r"$k$")
+ax.set_ylim(-.1, 4)
+draw_classic_axes(ax, xlabeloffset=.2)
 ```
 
 1. We have sketched the two bands in the plot above. The Fermi energy is indicated by the red line.
@@ -173,14 +174,14 @@ ax.xaxis.set_label_coords(1.0, 0.49)
     N = \int_0^{E_F}g(\varepsilon)d\varepsilon = A\varepsilon_0 + 2A(E_F-\varepsilon_0)
     $$
 
-4. The number of electrons between $\varepsilon_a<\varepsilon<\infty$ is
+4. The total number of electrons between $\varepsilon_a<\varepsilon<\infty$ is
 
     $$
-    N = \int_{\varepsilon_a}^\infty g(\varepsilon) \frac{1}{e^{\beta (\varepsilon-E_F)}+1} \textrm{d}\varepsilon
+    N(\varepsilon_a<\varepsilon<\infty) = \int_{\varepsilon_a}^\infty g(\varepsilon) \frac{1}{e^{\beta (\varepsilon-E_F)}+1} \textrm{d}\varepsilon
     $$
 
 5. Because $\varepsilon_a-E_F \gg k_B T$, the Fermi-Dirac can be approximated by the Boltzmann distribution: $n_F \approx e^{-\beta (\varepsilon-E_F)}$. Working out the integral we obtain
 
     $$
-    N = \int_{\varepsilon_a}^\infty 2A  e^{-\beta (\varepsilon-E_F)} \textrm{d}\varepsilon = 2A k_B T e^{-\beta (\varepsilon_a-E_F)}
+    N(\varepsilon_a<\varepsilon<\infty)  = \int_{\varepsilon_a}^\infty 2A  e^{-\beta (\varepsilon-E_F)} \textrm{d}\varepsilon = 2A k_B T e^{-\beta (\varepsilon_a-E_F)}
     $$