Update 4_sommerfeld_model.md - polish

src/4_sommerfeld_model.md

@@ -21,7 +21,7 @@ _(based on chapter 4 of the book)_
 
 ### Sommerfeld theory (free electron model)
 
-Atoms in a metal provide conduction electrons from their outer shells (often s-shells). These electrons can be described as waves in the crystal, analogous to phonons. The Hamiltonian of a free electron is:
+Atoms in a metal provide conduction electrons from their outer shells (often s-shells). These electrons can be described as waves, analogous to phonons. The Hamiltonian of a free electron is:
 
 $$
 \mathcal{H}=\frac{ {\bf p}^2}{2m}=-\frac{\hbar^2}{2m}\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \right)\ \Rightarrow\ \varepsilon=\frac{\hbar^2}{2m}\left( k_x^2+k_y^2+k_z^2 \right)
@@ -68,13 +68,15 @@ Given the number of electrons in a system, we can now fill up these states start
 
 ![](figures/fermi_circle_periodic.svg)
 
-To compute the density of states, we need to perform an integration of k-space. Assuming three dimensions and spherical symmetry (the dispersion in the free electron model is isotropic) we find for the total number of states:
+Our goal now is to compute the density of states. We start by expressing the total number of states $N$ as an integral over k-space. Assuming three dimensions and spherical symmetry (the dispersion in the free electron model is isotropic), we find 
 
 $$
 N=2\left(\frac{L}{2\pi}\right)^3\int{\rm d}{\bf k}=2 \left(\frac{L}{2\pi}\right)^34\pi\int k^2{\rm d}k=\frac{V}{\pi^2}\int k^2{\rm d}k ,
 $$
 
-where the factor 2 represents spin degeneracy. Using $k=\frac{\sqrt{2m\varepsilon}}{\hbar}$ and ${\rm d}k=\frac{1}{\hbar}\sqrt{\frac{m}{2\varepsilon}}{\rm d}\varepsilon$ we can rewrite this as:
+where the factor 2 represents spin degeneracy, and $\left(\frac{L}{2\pi}\right)^3$ is the density of points in k-space. 
+
+Using $k=\frac{\sqrt{2m\varepsilon}}{\hbar}$ and ${\rm d}k=\frac{1}{\hbar}\sqrt{\frac{m}{2\varepsilon}}{\rm d}\varepsilon$ we can rewrite this as:
 
 $$
 N=\frac{V}{\pi^2}\int\frac{2m\varepsilon}{\hbar^3}\sqrt{\frac{m}{2\varepsilon}}{\rm d}\varepsilon=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\varepsilon}\ {\rm d}\varepsilon
@@ -99,8 +101,8 @@ draw_classic_axes(ax, xlabeloffset=.2)
 
 Similarly,
 
-- For 1D: $g(\varepsilon) = \frac{2 V}{\pi} \frac{ {\rm d}k}{ {\rm d}\varepsilon} \propto 1/\sqrt{\varepsilon}$
-- For 2D: $g(\varepsilon) = \frac{k V}{\pi} \frac{ {\rm d}k}{ {\rm d}\varepsilon} \propto \text{constant}$
+- For 1D: $g(\varepsilon) = \frac{2 L}{\pi} \frac{ {\rm d}k}{ {\rm d}\varepsilon} \propto 1/\sqrt{\varepsilon}$
+- For 2D: $g(\varepsilon) = \frac{k L^2}{\pi} \frac{ {\rm d}k}{ {\rm d}\varepsilon} \propto \text{constant}$
 
 Total number of electrons: