The Fermi Surface is the boundary of the fermi sea below which all states are occupied at zero temperature, it's respecitvely a line, circle or a sphere in the 1,2 or the 3rd dimension.
#### Question 5.
If the electrons in the solid can be described by the free electron model, the heat capacity is dominated by the electrons, since $T^3$ decreases faster than $T$ as $T->0$.
4. The Fermi Surface is the collection of occupied states at the Fermi energy. In the free electron model, the Fermi surface is two points in 1D, a circle in 2D, and the surface of a sphere in 3D.
5. If the electrons in the solid can be described by the free electron model, the heat capacity is dominated by the electrons, since $T^3$ decreases faster than $T$ as $T\rightarrow 0$.
### Exercise 1*: Deriving the density of states for a parabolic dispersion relation.
#### Question 1.
$$
\varepsilon(k) = \frac(\hbar^2 k^2}{2m}
$$
#### Question 2.
$ \Delta k = \frac{2 \pi}{L}, density: $(\frac{L}{2\pi})^d$, where d = 1,2 or 3 for 1D, 2D and 3D respecitvely.
Alkali metals mostly have a spherical Fermi surface. Their energy depends only on the magnitude of the Fermi wavevector.
#### Question 2.
Refer to the lecture notes for the plot. $k_F = 7.5 \cdot 10^9 m^{-1}$, $v_F = 8.6 \cdot 10^5 m/s$ and $T_F = 24600 K$.
2. The distance between points in k-space is $ \Delta k = \frac{2 \pi}{L}$. The density of k-points is: $(\frac{L}{2\pi})^d$, where d = 1, 2, or 3 for 1D, 2D and 3D respectively.
#### Question 3.
$$
n = \frac{N}{V} = \frac{1}{3 \pi^{2} \hbar^{3}}\left(2 m \varepsilon_{F}\right)^{3 / 2} = 1.4 \cdot 10^28 m^{-3}
4. To transform the integral from one over k-space to one over energy, we need the dispersion relation. The integral boundaries are $0<\varepsilon<\varepsilon_0$. Using $k=|\mathbf{k}| = \sqrt{2m\varepsilon/\hbar^2}$ and $dk = \sqrt{m/(2\varepsilon\hbar^2)}d\varepsilon $, we get:
where $\rho$ is the mass density, $N_A$ is the Avogadro's constant, $M$ is molar mass and $Z$ is the valence of potassium atom.
We see that the free-electron density is much lower than the total electron density. We conclude that only a small fraction of the electrons are available for conduction - roughly 1 free electron per potassium atom.
### Exercise 3: The electron dispersion, density of states, and heat capacity of graphene
1. The electron dispersion is plotted here as a function of $k=|\mathbf{k}|=\sqrt{k_x^2+k_y^2}$
2\. We recall from the Debye lecture that the DOS of a 2D linear dispersion is linear in energy. A similar calculation for graphene shows that the DOS, for positive energies is given by
$g(\varepsilon)$ vs $\varepsilon$ is a linear plot. Here, the region marked by $-k_B T$ is a triangle whose area gives the number of electrons that can be excited:
3\. Since we have $g(\varepsilon)\propto \varepsilon$ and $E_F=0$, the number of electrons that get thermally excited is given by the area of a triangle with base $k_BT$ and height $g(-k_BT)$:
We compute the density of states separately for each band and add the result. Note that for the first band there will be no states below $\varepsilon <0$ and for the second band no states below $\varepsilon<\varepsilon_0$.
\begin{align}
N_{1,states} &= 2_s (\frac{L}{2\pi})^2 2\pi \int k \textrm{d}k \\
N = \int_\varepsilon_a^\infty g(\varepsilon) n_F(\beta(\varepsilon-E_F)) \textrm{d}\varepsilon
$$
#### 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
#### Question 1.
Distance between nearest k-points is $\frac{2\pi}{L}$ and their density across n-dimensions is $(\frac{L}{2\pi})^n$.
1. We have sketched the two bands in the plot above. The Fermi energy is indicated by the red line.
#### Question 2.
$$
g_{1D}(k)\textrm{d} k = \left(\frac{L}{2\pi}\right) 2 \mathrm{d} k
$$
The factor 2 is due to positive and negative $k$-values having equal enery
\begin{align}
g_{2D}(k)\textrm{d} k &= \left(\frac{L}{2\pi}\right)^2 2\pi k \mathrm{d} k,\\
g_{3D}(k)\textrm{d} k &= \left(\frac{L}{2\pi}\right)^3 4\pi k^2 \mathrm{d} k
\end{align}
#### Question 3.
\begin{align}
g(k)\textrm{d} k &= \left(\frac{L}{2\pi}\right)^n S_{n-1}(k)\textrm{d} k \\
&= \left(\frac{L}{2\pi}\right)^n \frac{2\pi^{\frac{n}{2}}k^{n-1}}{\Gamma(\frac{n}{2})}\textrm{d} k
\end{align}
2. We compute the density of states separately for each band and add these partial density of states to obtain the total density of states. Note that for the first band there will be no states below $\varepsilon <0$ and for the second band no states below $\varepsilon<\varepsilon_0$.
#### Question 4.
See hint of the question
Therefore, we have $g(\varepsilon)=A$ for $0<\varepsilon<\varepsilon_0$ and $g(\varepsilon)=2A$ for $\varepsilon>\varepsilon_0$, where $A=2_s \frac{L^2}{2\pi}\frac{m}{\hbar^2}$ is the constant DOS we found for a 2D parabolic dispersion in exercise 1.4.
#### Question 5.
3. At $T = 0$, the Fermi-Dirac distribution is a step function. The number of electrons is
$$
N = \int_0^{E_F}g(\varepsilon)d\varepsilon = A\varepsilon_0 + 2A(E_F-\varepsilon_0)
Check the source code written in python for solving integral using midpoint rule.
$$
N = \int_{\varepsilon_a}^\infty g(\varepsilon) \frac{1}{e^{\beta (\varepsilon-E_F)}+1} \textrm{d}\varepsilon
$$
5. Because $\varepsilon_a-E_F \gg k_B T$, the Fermi-Dirac can be approximated by the Boltzmann distribution: $n_F \approx e^{-\beta (\varepsilon-E_F)}$. Working out the integral we obtain
*[DOS]: Density of states
\ No newline at end of file
$$
N = \int_{\varepsilon_a}^\infty 2A e^{-\beta (\varepsilon-E_F)} \textrm{d}\varepsilon = 2A k_B T e^{-\beta (\varepsilon_a-E_F)}
