Electrons in metals II

Exercises

Preliminary provocations

1. Write down the expression for the total energy of particles with the density of states $g(\varepsilon)$ and the occupation number $n_{F}(\beta(\varepsilon - \mu))$.

   $$E = \int \limits_0^{\infty} g(\varepsilon) n_{F}(\beta (\varepsilon - \mu)) \varepsilon \mathrm{d} \varepsilon$$

2. Explain what happens if a material is heated up to its Fermi temperature (assuming that material where this is possible exists).

   The electronic heat capacity $C_e$ approaches $3N k_B$.

3. Why can we not use the Sommerfeld expansion with a Fermi energy of the order of the thermal energy?

   Thermal smearing is too significant and we can not accurately approximate the fraction of the excited electron with triangles anymore. Thus the Sommerfeld expansion breaks down.

4. Is the heat capacity of a solid at temperatures near $T=0$ dominated by electrons or vibrations?

   Electrons.

Exercise 1: potassium

The Sommerfeld model provides a good description of free electrons in alkali metals such as potassium (element K), which has a Fermi energy of $\varepsilon_{F} = 2.12$ eV (data from Ashcroft, N. W. and Mermin, N. D., Solid State Physics, Saunders, 1976.).

1. Check the Fermi surface database. Explain why potassium and (most) other alkali metals can be described well with the Sommerfeld model.

   Alkali metals mostly have a spherical Fermi surface. Their energy depends only on the magnitude of the Fermi wavevector.

2. Calculate the Fermi temperature, Fermi wave vector and Fermi velocity for potassium.

3. Why is the Fermi temperature much higher than room temperature?

   Electrons are fermions and obey the 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. Calculate the free electron density $n$ in potassium.

   $$n = \frac{N}{V} = \frac{1}{3 \pi^{2} \hbar^{3}}\left(2 m \varepsilon_{F}\right)^{3 / 2}$$

5. Compare this with the actual electron density of potassium, which can be calculated by using the density, atomic mass and atomic number of potassium. What can you conclude from this?

   $$n = \frac{\rho N_A Z}{M},$$ 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: a hypothetical material

A hypothetical metal has a Fermi energy $\varepsilon_F = 5.2 \, \mathrm{eV}$ and a density of states $g(\varepsilon) = 2 \times 10^{10} \, \mathrm{eV}^{-\frac{3}{2}} \sqrt{\varepsilon}$.

1. Give an integral expression for the total energy of the electrons in this hypothetical material in terms of the density of states $g(\varepsilon)$, the temperature $T$ and the chemical potential $\mu = \varepsilon_F$.

   $$E = \int_{0}^{\infty}\varepsilon g(\varepsilon) n_{F}(\beta (\varepsilon - \mu)) \textrm{d} \varepsilon = 2.10^{10}eV^{-\frac{3}{2}} \int_{0}^{\infty}\frac{\varepsilon^{\frac{3}{2}}}{e^\frac{\varepsilon-5.2}{k_BT}+1} \textrm{d} \varepsilon$$

2. Find the ground state energy at $T = 0$.

   $$E = \frac{4}{5} (5.2)^{\frac{5}{2}} 10^{10} eV$$

3. In order to obtain a good approximation of the integral for non-zero $T$, one can make use of the Sommerfeld expansion (the first equation is all you need and you can neglect the $O\left(\frac{1}{\beta \mu}\right)^{4}$ term). Using this expansion, find the difference between the total energy of the electrons for $T = 1000 \, \mathrm{K}$ with that of the ground state.

   $$\begin{align} E(T)-E(T=0) &= \frac{\pi^2}{6}(k_B T)^2\frac{\partial}{\partial \varepsilon}\left(\varepsilon g(\varepsilon)\right)\bigg|_{\varepsilon=\varepsilon _F}\\ &\approx 8.356 10^8 eV \end{align}$$

4. Now, find this difference in energy by calculating the integral found in 1 numerically. Compare your result with 3.

   Hint

   You can do numerical integration in python with scipy.integrate.quad(func, xmin, xmax)

5. Calculate the heat capacity for $T = 1000 \, \mathrm{K}$ in eV/K.

   $C_v = 1.6713.10^6 eV/K$

6. Numerically compute the heat capacity by approximating the derivative of energy difference found in 4 with respect to $T$. To this end, make use of the fact that $$\frac{dy}{dx}=\lim_{\Delta x \to 0} \frac{y(x + \Delta x) - y(x - \Delta x)}{2 \Delta x}.$$ Compare your result with 5.

mu = 5.2
kB = 8.617343e-5
T = 1000 #kelvin

import numpy as np
from scipy import integrate

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 numerically using scipy's integrate
dE = integrate.quad(integral, 0, 1e1, args=(T))[0] - 0.8e10 * 5.2**(5./2)

dT = 0.001
dEplus = integrate.quad(integral, 0, 1e1, args=(T+dT))[0] - 0.8e10 * 5.2**(5./2)
dEmin = integrate.quad(integral, 0, 1e1, args=(T-dT))[0] - 0.8e10 * 5.2**(5./2)

CV = (dEplus - dEmin) / (2*dT);

print(f'dE = {dE:.4e} eV')
print(f'Cv = {CV:.4e} eV/K')

Exercise 3: graphene

One of the most famous recently discovered materials is graphene. It consists of carbon atoms arranged in a 2D honeycomb structure. In this exercise, we will focus on the electrons in bulk graphene. Unlike in metals, electrons in graphene cannot be treated as 'free'. However, close to the Fermi level, the dispersion relation can be approximated by a linear relation: $\varepsilon(\mathbf{k}) = \pm c|\mathbf{k}|.$ Note that the $\pm$ here means that there are two energy levels at a specified $\mathbf{k}$. The Fermi level is set at $\varepsilon_F = 0$.

1. Make a sketch of the dispersion relation. What other well-known particles have a linear dispersion relation?

   import numpy as np
   import matplotlib.pyplot as plt
   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(r'$\varepsilon$', fontsize=18, rotation='horizontal')
   ax.yaxis.set_label_coords(0.5,1)
   ax.xaxis.set_label_coords(1.0, 0.49)
   

2. Using the dispersion relation and assuming periodic boundary conditions, derive an expression for the density of states of graphene. Do not forget spin degeneracy, and take into account that graphene has an additional two-fold 'valley degeneracy' (hence there is a total of a fourfold degeneracy instead of two). Your result should be linear with $|\varepsilon|$.

   Hint

   It is convenient to first start by only considering the positive energy contributions $\varepsilon(\mathbf{k}) = + c|\mathbf{k}|$ and calculate the density of states for it. Then account for the negative energy contributions $\varepsilon(\mathbf{k}) = - c|\mathbf{k}|$ by adding it to the density of states for the positive energies. You can also make use of $\frac{\rm{d} |k|}{\rm{d}k} = \frac{k}{|k|}$.

   $$g(\varepsilon) = 2_s 2_v 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{\varepsilon}{c^2},$$ 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 $$g(\varepsilon) = 2_s 2_v 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{|\varepsilon|}{c^2}.$$

3. At finite temperatures, assume that electrons close to the Fermi level (i.e. not more than $k_B T$ below the Fermi level) will get thermally excited, thereby increasing their energy by $k_B T$. Calculate the difference between the energy of the thermally excited state and that of the ground state $E(T)-E_0$. To do so, show first that the number of electrons that will get excited is given by $$n_{ex} = \frac{1}{2} g(-k_B T) k_B T.$$

   $g(\varepsilon)$ 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: $$\begin{align} n_{ex} &= \frac{1}{2} g(-k_B T) k_B T\\ &= \frac{L^2 k_B^2T^2}{\pi c^2}. \end{align}$$ From this it follows that the energy difference is given by $$E(T) - E_0 = \frac{L^2 k_B^3T^3}{\pi c^2}.$$

4. Calculate the heat capacity $C_e$ as a function of the temperature $T$.

   $$C_v(T) = \frac{\partial E}{\partial T} = \frac{3L^2k_B^3T^2}{\pi c^2}$$

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Last update: October 30, 2023

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