Sommerfeld model

    from matplotlib import pyplot

import numpy as np

from common import draw_classic_axes, configure_plotting

configure_plotting()

(based on chapter 4 of the book)

Expected prerequisites

Before the start of this lecture, you should be able to:

Given an elementary (e.g. linear) dispersion relation, compute the density of states.
Write down the Fermi-Dirac distribution.
Write down the Schrödinger equation and solve it in free space to obtain the parabolic dispersion relation of free electrons.
Describe the relation between the wavevector, wavenumber, and wavelength of a plane wave.

Learning goals

After this lecture you will be able to:

Calculate the electron density of states in 1D, 2D, and 3D for the parabolic dispersion of free electrons.
Use the density of states to express the number and energy of electrons in a system as an integral over energy for $T = 0$.
Use the Fermi-Dirac distribution to extend the previous learning goal to $T > 0$.
Calculate the electron contribution to the specific heat of a solid.
Describe what is the Fermi energy, the Fermi temperature, the Fermi surface, the Fermi wavevector, and the Fermi wavelength.

The free electron dispersion¶

Two electrons are sitting on a bench. Another one approaches and asks: "May I join you guys?" The first two immediately reply: "Who do you think we are? Bosons?"

In the Debye model, we studied the properties and physical behavior of phonons.
The Sommerfeld model applies the same conceptual approach to electrons in metals. Sommerfeld considered the electrons as free particles that are not interacting with atomic nuclei, which is why the model is also called the free electron model. Similar to the Debye model, we consider a cubic box of size $L \times L \times L$ with periodic boundary conditions. The solutions to the Schrödinger equation of a free particle are plane waves: $$ \psi \propto \exp(i\mathbf{k} \cdot \mathbf{r}), $$ with $\mathbf{k}$ is the electron wave vector. Because of the periodic boundary conditions, the allowed wavevectors $\mathbf{k}$ become discretized into $(k_x, k_y, k_z) = \frac{2\pi}{L} (n_x, n_y, n_z)$. The plane waves, characterized by their wavevector, have eigenenergies given by the dispersion relation

\[ \varepsilon(\mathbf{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}, \]

with $m$ being the mass of the electron. Let us plot the free-electron dispersion $\varepsilon(k)$ as a function of $k$ for a 1D system:

kf = 3;
extrapol = 1.1;
ks = np.arange(-kf, kf+1);
kcont = np.linspace(-extrapol*kf, extrapol*kf, 200);

Edis = ks**2;
Econt = kcont**2;

fig = pyplot.figure();
ax = fig.add_subplot(111);
ax.plot(kcont, Econt);
ax.plot(ks, Edis, 'k.', markersize=10);
for i in range(2*kf + 1):
    ax.plot([ks[i], ks[i]], [0.0, Edis[i]], 'k:');
ax.set_xlim(-3.75, 3.75);
ax.set_ylim(0.0, 11);

ax.set_xlabel(r"$k \: \left[ \frac{2 \pi}{L} \right]$");
ax.set_ylabel(r"$\varepsilon$");

ax.set_xticklabels([""] + ks.tolist() + [""]);
ax.set_yticks([]);

draw_classic_axes(ax, xlabeloffset = .6);

svg

Here, each black dot is a possible electron state.

Comparing the analyses of electrons and phonons¶

In contrast with the linear dispersion of the phonon modes in the Debye model, the dispersion of free electrons in the Sommerfeld model is quadratic. In addition, the electrons obey fermionic instead of bosonic statistics. As a result, the occupation of the electron states is described by the Fermi-Dirac distribution

\[ n_{F}(\beta(\varepsilon-\mu)) = \frac{1}{e^{\beta(\varepsilon-\mu)}+1}. \]

Here $\beta = 1/k_{B}T$, $\varepsilon$ is the energy, and $\mu$ is the chemical potential. The chemical potential is the characteristic energy above which $n_{F}$ goes to zero. We obtain the number of electrons in the system by summing over the occupied states, as determined the by Fermi-Dirac distribution:

\[\begin{align*} N &= 2 \sum_{\mathbf{k}} n_{F}(\beta(\varepsilon-\mu))\\ &= 2 \left( \frac{L}{2 \pi} \right)^3 \int n_{F}(\beta(\varepsilon(\mathbf{k})-\mu))\mathrm{d} \mathbf{k}. \end{align*}\]

Just like with phonons, we replaced the discrete sum over $\mathbf{k}$ with a volume integral. The factor 2 here accounts for the number of distinct electron states per $\mathbf{k}$ value.

Check: why is there a degeneracy of 2? Weren't there 3 polarizations?

The factor $2$ accounts for the spin degeneracy. To keep track of the origin of this term we will denote the spin degeneracy as $2_s$.

Similar to the total energy of phonons in the Debye model, the expression for the total energy of electrons is

\[ E = 2_s \left( \frac{L}{2 \pi} \right)^3 \int \varepsilon(\mathbf{k}) n_{F}(\beta(\varepsilon(\mathbf{k})-\mu)) \mathrm{d} \mathbf{k}. \]

There are only two differences: the distribution is $n_{F}$ instead of $n_B$ and the degeneracy is 2 instead of 3. In the table below we compare the properties of both phonons and electrons.

	Phonons	Electrons
Dispersion relation	$\omega = v_s \lvert\mathbf{k}\rvert$	$\varepsilon = \frac{\hbar^2\mathbf{k}^2}{2m}$
Statistics	Bose-Einstein	Fermi-Dirac
$n(\varepsilon) =$	$1/(e^{\beta \varepsilon} - 1)$	$1/(e^{\beta(\varepsilon - \mu)} + 1)$
Degeneracy per $\mathbf{k}$	3 (polarization)	2 (spin)
Total particle number	temperature-dependent	constant

It is important to note the last difference in this table: warming a material up creates more thermally excited phonons. The number of electrons, on the other hand, stays the same: the electrons may not appear out of nowhere.¹

About $\hbar$

Within quantum mechanics energy and frequency are related by Planck's constant: $\varepsilon = \hbar\omega$. Similarly, $p = \hbar k$ relates a particle's momentum with its wave vector. This relation is so unambiguous that you may encounter these terms used synonymously in scientific literature.

The Fermi sea, energy, surface, wavevector, and wavelength¶

To determine the chemical potential $\mu$ let us consider a 2D system with zero temperature and a finite number of electrons. At $T=0$, the Fermi-Dirac distribution becomes a step function

\[ n_{F}(\beta(\varepsilon-\mu)) = \Theta(-(\varepsilon-\varepsilon_F)). \]

The chemical potential at $T=0$ is called the Fermi energy $\varepsilon_F$. All electron states with lower energies are occupied, and all the states with higher energies are empty. In the reciprocal space, the occupied $\mathbf{k}$-states form a circle (in 1D it is a line and in 3D a sphere).

# creating grid
N = 10
x = np.linspace(-N//2, N//2, N+1)
xx, yy = np.meshgrid(x,x)

# Initialzing figure
fig = pyplot.figure(figsize = (10,10));
ax = fig.add_subplot(111);

# Creating figure
bound = N//3
ax.scatter(xx[np.sqrt(xx**2+yy**2)<=bound],yy[np.sqrt(xx**2+yy**2)<=bound], color = 'k')
ax.scatter(xx[np.sqrt(xx**2+yy**2)>bound],yy[np.sqrt(xx**2+yy**2)>bound], facecolors='none', edgecolors='k')
ax.add_patch(pyplot.Circle((0, 0), bound+0.05, color='k', fill=False))
ax.set_xlim([-N//2, N//2])
ax.set_ylim([-N//2, N//2])
ax.set_xticks([1,2,N//2-0.5]);
ax.set_yticks([1,2,N//2-0.5]);
ax.set_xticklabels([r'$\frac{2 \pi}{L}$',r'$\frac{4 \pi}{L}$',r"$k_x$"])
ax.set_yticklabels([r'$\frac{2 \pi}{L}$',r'$\frac{4 \pi}{L}$',r"$k_y$"])
draw_classic_axes(ax, xlabeloffset = .8, ylabeloffset = 0.2);

svg

A good metaphor for describing this state of many electrons is a sea: electrons occupy a finite area in reciprocal space, starting from the "deepest" points with the lowest energy all the way up to the chemical potential—also called Fermi level. The border of the Fermi sea is called the Fermi surface (you should notice a pattern here), and in the free electron model it is a sphere with the radius equal to the magnitude of the Fermi wavevector (i.e., the Fermi wavenumber). To clarify the relation between these concepts let us take a look at the dispersion relation in 1D:

kf = 3.0;
extrapol = 4.0/3.0;
kfilled = np.linspace(-extrapol*kf, extrapol*kf, 100);
kstates = np.linspace(-extrapol*kf, extrapol*kf, 500);

Efilled = kfilled**2;
Estates = kstates**2;

fig = pyplot.figure();
ax = fig.add_subplot(111);

# Creating plot
trans = 1
ax.plot([kf, kf], [0.0, kf*kf], 'k:');
ax.plot(kstates, Estates, color = 'lightblue', linestyle = '-',alpha = trans);
ax.scatter(kfilled[np.abs(kfilled)<=kf], Efilled[np.abs(kfilled)<=kf], color = 'k', s = 3.3**2, zorder = 10);
ax.scatter(kfilled, Efilled, facecolors='none', edgecolors='k', s = 3.3**2, zorder = 10);
ax.axhline(kf*kf, linestyle = "dotted", color='k');

ax.set_xticks([kf]);
ax.set_yticks([kf*kf + 0.4]);
ax.set_xticklabels([r"$k_F$"]);
ax.set_yticklabels([r"$ε_F$"]);

ax.set_xlabel(r"$k$");
ax.set_ylabel(r"$ε$");

ax.set_xlim(-kf*extrapol, kf*extrapol)
ax.set_ylim(0.0, kf*kf*extrapol);
draw_classic_axes(ax, xlabeloffset=.6);

svg

By using the dispersion relation, we obtain the relation between the Fermi energy and the Fermi wavevector

\[ \varepsilon_F = \frac{\hbar^2 |\mathbf{k}_F|^2}{2m}. \]

The Fermi wavevector $\mathbf{k}_F$ also defines the Fermi momentum $\mathbf{p}_F = \hbar \mathbf{k}_F$ and the Fermi velocity:

\[ \mathbf{v}_F = \frac{\mathbf{p}_F}{m} = \frac{\hbar \mathbf{k}_F}{m}. \]

The Fermi energy of copper is ~7 eV. What is the corresponding Fermi velocity?

The Fermi velocity $v_F\approx$ 1700 km/s or 0.3% of the speed of light, and much faster than the Drude theory drift velocity!

Deriving the density of states for the parabolic dispersion of the free electron model¶

Like before, to compute the energy and heat capacity, we need to find the density of states—the number of states per energy interval. Once again, we compute the density of states at the energy $\varepsilon$ as a derivative of the total number of states below that energy.

\[ g(\varepsilon) = \frac{dN_\text{states}}{d\varepsilon}. \]

Let us calculate the density of states for a 3D system. Because the free-electron dispersion is isotropic², we use spherical coordinates. The total number of states below energy $\varepsilon$ is then

\[\begin{align*} N_\text{states} &\overset{\mathrm{3D}}{=}2_s \left(\frac{L}{2\pi}\right)^3\int_{\varepsilon(\mathbf{k}) < \varepsilon}\mathrm{d}{\mathbf{k}}\\ &=2_s \left(\frac{L}{2\pi}\right)^3 4\pi\int k^2\mathrm{d}k\\ &=\frac{V}{\pi^2}\int k^2\mathrm{d}k, \end{align*}\]

We rewrite this expression into an integral over energy using the dispersion relation, substituting $k=\hbar^{-1}\sqrt{2m\varepsilon}$ and $\mathrm{d}k=\hbar^{-1}\sqrt{m/2\varepsilon} d\varepsilon$:

\[\begin{align*} N_\text{states} &=\frac{V}{\pi^2}\int\frac{2m \varepsilon}{\hbar^3}\sqrt{\frac{m}{2\varepsilon}}\mathrm{d}\varepsilon\\ &=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\varepsilon}\ \mathrm{d}\varepsilon \\ & = \int g(\varepsilon) \mathrm{d}\varepsilon \\ \end{align*}\]

We thus find the density of states:

\[\begin{align*} g(\varepsilon) &= \frac{ \mathrm{d}N_\text{states}}{ \mathrm{d}\varepsilon}\\ & =\frac{Vm^{3/2}\sqrt{2\varepsilon}}{\pi^2\hbar^3} \propto\sqrt{\varepsilon} \end{align*}\]

We observe that the density of states for the 3D parabolic free-electron dispersion is proportional to the square root of energy:

\[ g(\varepsilon) \propto\sqrt{\varepsilon} \]

Repeating similar derivations, we find the density of states of 1D and 2D systems:

1D: $g(\varepsilon) = \frac{2 L}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}\varepsilon} \propto 1/\sqrt{\varepsilon}$
2D: $g(\varepsilon) = \frac{k L^2}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}\varepsilon} \propto \text{constant}$

We plot these three behaviors of $g(\varepsilon)$ below for a comparison.

E = np.linspace(0.001, 2, 500)
fig, ax = pyplot.subplots()

# Plotting the figure
ax.plot(E, 1/np.sqrt(E), label = '1D')
ax.plot(E, 9*np.ones(len(E)), label = '2D')
ax.plot(E, 15*np.sqrt(E), label = '3D')

ax.set_ylabel(r"$g(\varepsilon)$")
ax.set_xlabel(r"$\varepsilon$")
ax.legend()
ax.set_xticks([])
ax.set_yticks([])
draw_classic_axes(ax, xlabeloffset=.2)

svg

Relation between the Fermi energy and the number of electrons¶

The Fermi energy sets the number of electrons in the system $N$. We can see this by expressing $N$ at $T = 0$ as an integral over the density of states:

\[\begin{align*} N &= \int \limits_0^{\infty} n_{F}(\beta(\varepsilon-\mu)) g(\varepsilon) \mathrm{d}\varepsilon\\ &\overset{\mathrm{T = 0}}{=}\int \limits_0^{\varepsilon_F}g(\varepsilon)\mathrm{d}\varepsilon \\ &\overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2m\varepsilon_F)^{3/2}. \end{align*}\]

Solving this equation for the Fermi energy yields:

\[ \varepsilon_{F} = \frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{\frac{2}{3}}. \]

Having found the Fermi energy, we can use the dispersion relation ($\varepsilon(\mathrm{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}$) to find the length of the Fermi wavevector (i.e., the Fermi wavenumber) $k_F = |\mathbf{k_{F}}|$:

\[ k_F = \left( 3\pi^2\frac{N}{V} \right)^{\frac{1}{3}}. \]

Using the Fermi wavenumber, we calculate the Fermi wavelength $\lambda_F\equiv 2\pi/k_F$. Plugging in numbers (see exercise) shows that the Fermi wavelength is on the order of the atomic spacing for typical free electron densities in metals³.

The chemical potential and the Fermi energy¶

We now extend our discussion to $T > 0$ by taking a closer look at the Fermi-Dirac distribution

\[ n_{F}(\beta(\varepsilon-\mu)) = \frac{1}{e^{\beta(\varepsilon-\mu)}+1}. \]

The Fermi-Dirac distribution $n_{F}(\beta(\varepsilon-\mu))$ for $T = 0$ and $T > 0$ is shown below. For both cases we have the same chemical potential $\mu = \varepsilon_F$.

fig = pyplot.figure()
ax = fig.add_subplot(1,1,1)
xvals = np.linspace(0, 2, 200)
mu = .75
beta = 20
ax.plot(xvals, xvals < mu, ls='dashed', label='$T=0$')
ax.plot(xvals, 1/(np.exp(beta * (xvals-mu)) + 1),
        ls='solid', label='$T>0$')
ax.set_xlabel(r'$\varepsilon$')
ax.set_ylabel(r'$n_{F}(\varepsilon, T)$')
ax.set_yticks([0, 1])
ax.set_yticklabels(['$0$', '$1$'])
ax.set_xticks([mu])
ax.set_xticklabels([r'$\mu = \varepsilon_{F}$'])
ax.set_ylim(-.1, 1.1)
ax.legend()
draw_classic_axes(ax)
pyplot.tight_layout()

svg

At a finite temperature $T>0$, thermal excitations smear out the sharp change in the number of occupied electrons near $\varepsilon_F$. The Fermi energy is typically in the range of electronvolts, which corresponds to a Fermi temperature of $T_F = \varepsilon_F/k_B \approx 10 000$K. This would therefore be the temperature required for thermal excitations to give an electron a similar amount of energy as the Fermi energy itself! Therefore, at room temperature ($T = 300$K), the electron distribution over energies is very similar to that at $T=0$, and we can safely approximate the chemical potential at all temperatures by the Fermi energy: $\mu = \varepsilon_F$.

The electron heat capacity¶

To calculate the electron heat capacity, we compare the occupied electron states $g(\varepsilon) n_{F}(\beta(\varepsilon-\mu))$ at $T = 0$ (blue shaded area) to the occupied states at $T > 0$ (orange shaded area).

E = np.linspace(0, 2, 500)
fig, ax = pyplot.subplots()
ax.plot(E, np.sqrt(E), linestyle='dashed')
ax.text(1.7, 1.4, r'$g(ε)\propto \sqrt{ε}$', ha='center')
ax.fill_between(E, np.sqrt(E) * (E < 1), alpha=.3)

n = np.sqrt(E) / (1 + np.exp(20*(E-1)))
ax.plot(E, n)
ax.fill_between(E, n, alpha=.5)
w = .17
ax.annotate('', xy=(1, 1), xytext=(1-w, 1),
            arrowprops=dict(arrowstyle='<->', shrinkA=0, shrinkB=0))
ax.text(1-w/2, 1.1, r'$\sim k_BT$', ha='center')
ax.plot([1-w, 1+w], [1, 0], c='k', linestyle='dashed')
ax.annotate('', xy=(1, 0), xytext=(1, 1),
            arrowprops=dict(arrowstyle='<->', shrinkA=0, shrinkB=0))
ax.text(1.2, .7, r'$g(ε_F)$', ha='center')
ax.set_xticks([1])
ax.set_xticklabels([r'$ε_F$'])
ax.set_yticks([])

ax.set_ylabel(r"$g(ε)$")
ax.set_xlabel(r"$ε$")
draw_classic_axes(ax, xlabeloffset=.2)

svg

To estimate the increase in the electron energy caused by the increase in temperature, we approximate the difference between the blue and orange areas by triangles, as shown in the figure. This approximation is appropriate because the thermal smearing happens over an energy range $\Delta E \approx k_B T$ that is much smaller than the Fermi energy $\varepsilon_{F}$.

Sommerfeld expansion

A more rigorous way to estimate the energy of electrons at a finite temperature is to apply the Sommerfeld expansion. It still uses on the smallness of $k_B T$ compared to $\varepsilon_F$, but computes the electron energy without approximating the resulting integrals.

At a finite temperature, the electrons occupying the top triangle (blue) become thermally excited to occupy the bottom triangle (orange). Since the base of the triangle is proportional to $k_BT$ and the height is $\sim g(\varepsilon_F)$, the number of excited electrons is $N_\mathrm{exc} \approx g(\varepsilon_F)k_BT$ (neglecting constants not depending on $\varepsilon_{F}$).

These electrons have gained $k_BT$ of thermal energy, such that the total extra energy is

\[\begin{align*} E(T) &= E(T = 0) + N_\mathrm{exc}k_BT\\ &\approx E(T = 0) + g(\varepsilon_F)k_B^2T^2. \end{align*}\]

Therefore, the electron heat capacity $C_e$ is

\[\begin{align*} C_e &= \frac{ \mathrm{d}E}{ \mathrm{d}T}\\ &\approx 2 g(ε_F)k_B^2T\\ &\overset{\mathrm{3D}}{=} 3 Nk_B\frac{T}{T_F}\\ &\propto T, \end{align*}\]

where we used $N=\frac{2}{3}\varepsilon_Fg(\varepsilon_F)$ and the Fermi temperature $T_F \equiv \varepsilon_F/k_B$.

How does the electron heat capacity $C_e$ compare to the phonon heat capacity $C_p$?

At room temperature $C_\mathrm{p}\approx 3Nk_B\gg C_e \propto k_B T / T_F$, because $T \ll T_F$.
Near $T=0$, the phonon heat capacity $C_p\propto k_B (T/T_D)^3$, and it becomes smaller than the electron heat capacity at $T \lesssim \sqrt{T_D^3/T_F}$

Useful trick: scaling of $C_e$¶

Similar to how we understood the low temperature heat capacity of the Debye model, the behavior of $C_e$ can be quickly memorized and understood using the following argument:

Particles within an energy range of $\sim k_{B}T$ to the Fermi energy $\varepsilon_F$ become thermally excited, and each carries an extra energy $k_{B}T$.

Example 1: 3D free electrons¶

In 3D, $g(\varepsilon_F)$ is roughly constant. Thus the total energy obtained through thermal excitation is proportional to $T \times \left( T\times g(\varepsilon_F) \right)$, from which it follows that $C_e \propto T$.

Example 2: graphene¶

In exercise 2, we will analyze graphene. Unlike in metals, electrons in graphene cannot be treated as 'free'. As a result, graphene has a density of states $g(\varepsilon) \propto \varepsilon$, and Fermi energy $\varepsilon_F = 0$. Therefore, within the energy range of $k_BT$, $g(\varepsilon) \propto k_BT$. Thus the total energy is proportional to $T \times T^2$ and the heat capacity $C_e \propto T^2$.

Conclusions¶

The Sommerfeld free electron model treats electrons as free particles with energy dispersion $\varepsilon = \frac{\hbar^2|\mathbf{k}|^2}{2m}$.
The density of states $g(\varepsilon)$ follows from the dispersion relation by using a general procedure that is analogous to that for phonons.
The Fermi-Dirac distribution gives the probability of an electron state at energy $\varepsilon$ to be occupied.
The electron contribution to the heat capacity is proportional to $T$. It is much lower than phonon heat capacity at high temperatures, and much higher at low temperatures.
The scaling of heat capacity with $T$ can be quickly estimated by estimating the number of particles in an energy range $k_BT$ from the Fermi energy.

Exercises¶

Warm-up questions*¶

List the differences between electrons and phonons from your memory.
Write down the dispersion of free electrons.
Write down an integral expression for the total energy of particles with density of states $g(\varepsilon)$ and occupation number $n_{F}(\beta(\varepsilon - \mu))$.
Describe what is the Fermi surface. For the free-electron model, what does the Fermi surface look like in 1D, 2D, and 3D?
Argue if the heat capacity of a solid at temperatures near $T=0$ is dominated by electrons or phonons.

Exercise 1*: Deriving the density of states for a parabolic dispersion relation.¶

In this lecture, we found that the parabolic free-electron dispersion yields a density of states $g(\varepsilon)$ that is proportional to $1/\sqrt{\varepsilon}$ in 1D, constant in 2D, and proportional to $\sqrt{\varepsilon}$ in 3D. In this exercise, we will practice this important derivation again.

Write down the free-electron dispersion
What is the distance between nearest-neighbour points in $\mathbf{k}$-space? Assume periodic boundary conditions. What is the density of $\mathbf{k}$-points in 1, 2, and 3 dimensions?
Express the number of states between energies $0<\varepsilon<\varepsilon_0$ as an integral over k-space. Do so for 1D, 2D and 3D. Do not forget spin degeneracy.
Transform these integrals into integrals over energy for 1D, 2D and 3D. What relation do you need to do so? Indicate the integral boundaries. Extract the density of states. Are the integral boundaries important for the result?

Exercise 2*: Applying the free electron model to 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.).

Check the Fermi surface database. Explain why potassium and (most) other alkali metals can be described well with the Sommerfeld model.
Calculate the Fermi temperature, Fermi wave vector and Fermi velocity for potassium. Sketch the Fermi distribution at room temperature and indicate the role of the Fermi temperature.
Calculate the free electron density $n$ in potassium.
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?

Exercise 3: The electron dispersion, density of states, and heat capacity of graphene¶

One of the most famous recently discovered materials is graphene. It consists of carbon atoms arranged in a 2D honeycomb structure. Unlike in metals, electrons in graphene cannot be treated as 'free'. Instead, 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 $\varepsilon_F = 0$.

Make a sketch of the dispersion relation. Include both positive and negative energies. What other well-known particles have a linear dispersion relation?
Using the dispersion relation and assuming periodic boundary conditions, calculate the density of states $g(\varepsilon)$ 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 realizing it should have the same density of states (why?), but now for negative energies.
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 $\Delta E(T) = E(T)-E_0$. To do so, show first that the number of electrons that will get excited is (approximately) given by

\[ n_{ex} = \frac{1}{2} g(-k_B T) k_B T. \]
Calculate the heat capacity $C_e$ as a function of the temperature $T$.

Exercise 4: Two energy bands¶

An 'energy band' is a range of energies within which there are states available for the particles in the system. An energy band is therefore closely related to the dispersion relation. For instance, for the free electron dispersion $\varepsilon = \hbar^2k^2/2m$, the available states (=the energy band) lie between zero and infinite energy. As we will encounter more often in our course, many materials have multiple, possibly overlapping energy bands, which are each described by their own dispersion. Here we analyze a system with two, partially overlapping, energy bands. These bands are each described by a free-electron dispersion, but they are offset with repect to each other in energy. The goal is to calculate the density of states and electron occupation of the system.

The dispersion of energy band 1 is $\varepsilon_1(\mathbf{k}) = \tfrac{\hbar^2 \mathbf{k}^2}{2m}$ and that of band 2 is $\varepsilon_2(\mathbf{k}) = \tfrac{\hbar^2 \mathbf{k}^2}{2m} + \varepsilon_0$. The Fermi energy is $E_F$, and we assume $E_F \gg \varepsilon_0$. We consider a two-dimensional system.

Sketch the two dispersions in one plot. Indicate the Fermi energy.
Calculate the density of states and sketch it as a function of energy. Hint: the total density of states is obtained by adding the density of states associated with the individual bands.
Express the number of electrons in the system in terms of the Fermi energy $E_F$.
Express the number of electrons in the energy range $\varepsilon_a<\varepsilon<\infty$, for some energy $\varepsilon_a \geq \varepsilon_0$, as an integral over energy, for $T>0$.
Assuming $\varepsilon_a - E_F \gg k_B T$, explicitly calculate the integral of the previous subquestion.

This is not completely true, as we will see when learning about semiconductors ↩
An isotropic material means that the material is the same in all directions. ↩
The mean inter-particle distance is related to the electron density $n = N/V$ as $\langle r \rangle \propto n^{-1/3}$. The exact proportionality constant depends on the properties of the system. The Fermi wavelength sets the scale at which quantum interference effects of the electronic waves become important. In some materials (e.g. graphene) it can be on the 100 nm scale - accessible to nanofabrication techniques. Striking images of electron interference at the atomic scale are visible with a scanning tunneling microscope. ↩

	Phonons	Electrons
Dispersion relation	\(\omega = v_s \lvert\mathbf{k}\rvert\)	\(\varepsilon = \frac{\hbar^2\mathbf{k}^2}{2m}\)
Statistics	Bose-Einstein	Fermi-Dirac
\(n(\varepsilon) =\)	\(1/(e^{\beta \varepsilon} - 1)\)	\(1/(e^{\beta(\varepsilon - \mu)} + 1)\)
Degeneracy per \(\mathbf{k}\)	3 (polarization)	2 (spin)
Total particle number	temperature-dependent	constant