Lecture 4 solutions

Adjustments to the solutions of the 4th lecture.

src/4_sommerfeld_model_solutions.md

-# Solutions for Sommerfeld model exercises
+# Solutions for exercises lecture 4: Sommerfeld model
 
-### Exercise 1: potassium
+### Warm-up questions
+1. See lecture notes
+
+2. 
+$$
+E = \int \limits_0^{\infty} g(\epsilon) n_{FD}(\beta (\epsilon - \mu)) \epsilon \mathrm{d} \epsilon
+$$
 
-1.
+3. ?
 
-Alkali metals mostly have a spherical fermi surface. Their energy depends only on the magnitude of the Fermi wavevector.
+4. Thermal smearing is too significant and we can not accurately approximate the fraction of the excited electron through triangles anymore. Thus the Sommerfeld expansion breaks down.
 
-2.
+5. Electrons.
 
-Refer to the lecture notes.
+### Exercise 1: potassium
 
-3.
+1. Alkali metals mostly have a spherical Fermi surface. Their energy depends only on the magnitude of the Fermi wavevector.
 
-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.
+2. Refer to the lecture notes.
 
-4.
+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.
-
+5. 
 $$
-n = \frac{\rho N_A Z}{M}
+n = \frac{\rho N_A Z}{M},
 $$
-
 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.
+1. Distance between nearest k-points is $\frac{2\pi}{L}$ and their density across n-dimensions is $(\frac{L}{2\pi})^n$.
 
+2. 
 $$
-g_{1D}(k)dk = \left(\frac{L}{2\pi}\right) 2 .dk
+g_{1D}(k)dk = \left(\frac{L}{2\pi}\right) 2 \mathrm{d} k
 $$
-
-Factor 2 is due to positive and negative k-points having equal energy.
-
+The factor 2 is due to positive and negative $k$-values having equal enery
 $$
-g_{2D}(k)dk = \left(\frac{L}{2\pi}\right)^2 2\pi k . dk
+g_{2D}(k)dk = \left(\frac{L}{2\pi}\right)^2 2\pi k \mathrm{d} k
 $$
-
 $$
-g_{3D}(k)dk = \left(\frac{L}{2\pi}\right)^3 4\pi k^2 . dk
+g_{3D}(k)dk = \left(\frac{L}{2\pi}\right)^3 4\pi k^2 \mathrm{d} k
 $$
 
-$4\pi k^2$ is the volume of spherical shell enclosed between k and k + dk.
-
-3, 4.
-
+3. 
 $$
-g(k)dk = \left(\frac{L}{2\pi}\right)^n S_{n-1}(k)dk = \left(\frac{L}{2\pi}\right)^n \frac{2\pi^{\frac{n}{2}}k^{n-1}}{\Gamma(\frac{n}{2})}dk
+\begin{align}
+g(k)dk &= \left(\frac{L}{2\pi}\right)^n S_{n-1}(k)dk \\
+&= \left(\frac{L}{2\pi}\right)^n \frac{2\pi^{\frac{n}{2}}k^{n-1}}{\Gamma(\frac{n}{2})}dk
+\end{align}
 $$
 
-5.
+4. See hint of the question
 
+5. 
 $$
 2g(k)dk = g(E)dE
 $$
+
 $$
 g(E)=2g(k(E))\frac{dk}{dE}
 $$
@@ -73,7 +73,7 @@ $$
 g(E) = \frac{2}{\Gamma(\frac{n}{2})}\left(\frac{L}{\hbar}\sqrt{\frac{m_e}{2\pi}}\right)^n (E)^{\frac{n}{2}-1}
 $$
 
-6.
+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
@@ -83,27 +83,24 @@ Total energy: $E = \int_{0}^{\infty}g(E)f(E)EdE $
 
 ### Exercise 3: a hypothetical material
 
-1.
-
+1. 
 $$
 E = \int_{0}^{\infty}\epsilon g(\epsilon)f(\epsilon)d\epsilon = 2.10^{10}\int_{0}^{\infty}\frac{\epsilon^{\frac{3}{2}}}{e^\frac{\epsilon-5.2}{k_BT}+1}d\epsilon
 $$
 
-2.
-
+2. 
 Substitute T=0 in the integral expression for total energy to find the ground state energy.
 
-3.
-
+3. 
 $$
 \Delta E = \frac{\pi^2}{6}(k_B T)^2\frac{\partial}{\partial \epsilon}\left(\epsilon g(\epsilon)\right)\bigg|_{\epsilon=\epsilon _F}
 $$
 
-5.
+5. 
 
 $C_v = 1.6713.10^6 eV/K$
 
-4, 6.
+4, 6. 
 
 ```python
 mu = 5.2
@@ -143,7 +140,7 @@ Check the source code written in python for solving integral using midpoint rule
 
 ### Exercise 4: graphene
 
-1.
+1. 
 
 ```python
 import numpy as np
@@ -169,35 +166,33 @@ ax.yaxis.set_label_coords(0.5,1)
 ax.xaxis.set_label_coords(1.0, 0.49)
 ```
 
-2.
-
-In two dimensions: $g(k) dk = (2+2)\frac{Ak}{2\pi}dk$
+2. The DOS for the positive energies is given by
 
 $$
-g(\epsilon)d\epsilon = \frac{2Ak}{\pi}dk
+g(\epsilon) = (2_s + 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{\epsilon}{c^2},
 $$
 
-Substituting $\epsilon(k) = \pm c\mid k\mid$,
-
+where $2_s$ is the spin degeneracy and $2_v$ is the valley degeneracy.
+If we account for the negative energies as well, we obtain
 $$
-g(\epsilon) = \frac{2A\mid \epsilon \mid}{\pi c^2}
+g(\epsilon) = (2_s + 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{|\epsilon|}{c^2}.
 $$
 
-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.
+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:
 
 $$
-n_{ex} = \frac{Ak_B^2T^2}{\pi c^2}
+\begin{align}
+n_{ex} &= \frac{1}{2} g(-k_B T) k_B T
+&= \frac{L^2 k_B^2T^2}{\pi c^2}.
+\end{align}
 $$
-
+From this it follows that the energy difference is given by
 $$
-E(T) - E_0 = \frac{Ak_B^3T^3}{\pi c^2}
+E(T) - E_0 = \frac{L^2 k_B^3T^3}{\pi c^2}.
 $$
 
-4.
-
+4. 
 $$
-C_v(T) = \frac{\partial E}{\partial T} = \frac{3Ak_B^3T^2}{\pi c^2}
+C_v(T) = \frac{\partial E}{\partial T} = \frac{3L^2k_B^3T^2}{\pi c^2}
 $$