add solutions to Sommerfeld exercises

src/4_sommerfeld_model_solutions

+# Solutions for Sommerfeld model exercises
+
+### Exercise 1: potassium
+
+1.
+
+Alkali metals mostly have a spherical fermi surface. Their energy depends only on the magnitude of the Fermi wavevector.
+
+2.
+
+Refer to the lecture notes.
+
+3.
+
+Electrons are fermions and obeys pauli exclusion principle. As electrons cannot occupy the same state, they are forced to occupy higher energy states resulting in high Fermi energy and high Fermi temperature.
+
+4.
+
+$$
+n = \frac{N}{V} = 2\frac{1}{(2\pi)^3}\frac{4}{3}\pi k_f^3
+$$
+
+5.
+
+$$
+n = \frac{\rho N_A Z}{M}
+$$
+
+where $\rho$ is the density, $N_A$ is the Avogadro's constant, $M$ is molar mass and $Z$ is the valence of potassium atom.
+
+Comparing total and free electron density, only few electrons are available for conduction which is roughly 1 free electron per potassium atom.
+
+### Exercise 2: the n-dimensional free electron model
+
+1.
+
+Distance between nearest k-points is $\frac{2\pi}{L}$ and their density across n-dimensions is $(\frac{L}{2\pi})^n$.
+
+2.
+
+$$
+g(k)dk = \left(\frac{L}{2\pi}\right)^n\int_k d\bf k
+$$
+
+Check how to convert volume to surface intergral and surface to line integral.
+
+3, 4.
+
+$$
+g(k)dk = \left(\frac{L}{2\pi}\right)^n S_{n-1}(k)dk = \left(\frac{L}{2\pi}\right)^n \frac{2\pi^{\frac{n}{2}}k^{n-1}}{\Gamma(\frac{n}{2})}dk
+$$
+
+5.
+
+$$
+2g(k)dk = g(E)dE
+$$
+$$
+g(E)=2g(k(E))\frac{dk}{dE}
+$$
+
+$$
+g(E) = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m_e}{2\pi}}\right)^n (E)^{\frac{n}{2}-1}
+$$
+
+6.
+
+$$
+N = \int_{0}^{\infty}g(E)f(E)dE = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m_e}{2\pi}}\right)^n\int_{0}^{\infty}\frac{(E)^{\frac{n}{2}-1}}{e^{\frac{E-\mu}{k_BT}}+1}dE
+$$
+
+Total energy: $E = \int_{0}^{\infty}g(E)f(E)EdE $
+
+### Exercise 3: a hypothetical material
+
+1.
+
+$$
+E = \int_{0}^{\infty}\epsilon g(\epsilon)f(\epsilon)d\epsilon = 2.10^{10}\int_{0}^{\infty}\frac{\epsilon^{\frac{3}{2}}}{e^\frac{\epsilon-5.2}{k_BT}}d\epsilon
+$$
+
+2.
+
+Substitute T=0 in the integral expression for total energy to find the ground state energy.
+
+3.
+
+$$
+\Delta E = \frac{\pi^2}{6}(k_B T)^2\frac{\partial}{\partial \epsilon}\left(\epsilon g(\epsilon)\right)\bigg|_{\epsilon=\epsilon _F}
+$$
+
+4, 6.
+
+Check the source code written in python for solving integral using midpoint rule
+
+```python
+mu = 5.2
+kB = 8.617343e-5
+T = 1000 #kelvin
+
+import numpy as np
+
+np.seterr(over='ignore')
+
+# Fermi-Dirac distribution 
+def f(E, T):
+    return 1 / (np.exp((E - mu)/(kB*T)) + 1)
+
+# Density of states
+def g(E):
+    return 2e10 * np.sqrt(E)
+
+#integration function
+def integral(E, T):
+    return f(E, T)*g(E)*E
+
+## Solve integral using midpoint rule
+elements = np.linspace(0, 1e6, 1e7) #Free to choose the number of discrete steps
+dE = np.sum(integral(elements, T))*((1e6 - 0)/len(elements)) - 0.8e10 * 5.2**(5./2)
+
+dT = 0.001
+dEplus = np.sum(integral(elements, 1000+dT))*((1e6 - 0)/len(elements)) - 0.8e10 * 5.2**(5./2)
+dEmin = np.sum(integral(elements, 1000-dT))*((1e6 - 0)/len(elements)) - 0.8e10 * 5.2**(5./2)
+
+CV = (dEplus - dEmin) / (2*dT);
+```
+
+### Exercise 4: graphene
+
+1.
+
+```python
+x = np.linspace(-1, 1, 100)
+fig, ax = plt.subplots(figsize=(7, 5))
+ax.plot(x, x, 'b')
+ax.plot(x,-x, 'b')
+
+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('$\epsilon$', fontsize=18, rotation='horizontal')
+ax.yaxis.set_label_coords(0.5,1)
+ax.xaxis.set_label_coords(1.0, 0.49)
+```
+
+2.
+
+In two dimensions: $g(k) dk = (2+2)\frac{Ak}{2\pi}dk$
+
+$$
+g(\epsilon)d\epsilon = \frac{2Ak}{\pi}dk
+$$
+
+Substituting $\epsilon(k) = \pm c\mid k\mid$,
+
+$$
+g(\epsilon) = \frac{2A\mid \epsilon \mid}{\pi c^2}
+$$
+
+3.
+
+$g(\epsilon)$ vs $\epsilon$ is a linear plot. Here, the region marked by $k_B T$ is a triangle whose area gives the number of electrons that can be excited.
+
+$$
+n_{ex} = \frac{Ak_B^2T^2}{\pi c^2}
+$$
+
+$$
+E(T) - E_0 = \frac{Ak_B^3T^3}{\pi c^2}
+$$
+
+4.
+
+$$
+C_v(T) = \frac{\partial E}{\partial T} = \frac{3Ak_B^2T^2}{\pi c^2}
+$$