Skip to content
Snippets Groups Projects

lecture_4

Merged lecture_4
lecture_4 into master
Merged Bowy La Riviere requested to merge lecture_4 into master
Compare
and
1 file
+ 286
− 130
Compare changes
  • Side-by-side
  • Inline
src/4_sommerfeld_model.md
+ 286
− 130
@@ -10,7 +10,7 @@ configure_plotting()
_(based on chapter 4 of the book)_
!!! success "Expected prior knowledge"
!!! success "Expected prerequisites"
Before the start of this lecture, you should be able to:
@@ -23,43 +23,36 @@ _(based on chapter 4 of the book)_
After this lecture you will be able to:
- calculate the electron density of states in 1D, 2D, and 3D using the Sommerfeld free-electron model.
- express the number and energy of electrons in a system in terms of integrals over k-space.
- use the Fermi distribution to extend the previous learning goal to finite T.
- calculate the electron contribution to the specific heat of a solid.
- describe central terms such as the Fermi energy, Fermi temperature, and Fermi wavevector.
- Calculate the electron density of states in 1D, 2D, and 3D using the Sommerfeld free-electron model.
- Express the number and energy of electrons in a system in terms of integrals over k-space 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 terms such as the Fermi energy and the Fermi wavevector.
## Electrons vs phonons
> 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?"
Having learned the statistical properties of phonons and the [Debye model](2_debye_model.md), let us use these as a starting point for comparing with the electrons.
Here is a table comparing the most important properties of electrons and phonons:
| | Phonons | Electrons |
| - | - | - |
| Governed by | Wave equation | Schrödinger equation |
| | $ d² δ\mathbf{r} / dt² = v²∇²δ\mathbf{r}$ | $ i ħdψ/dt = -ħ²∇²ψ/2m$ |
| Dispersion relation | $ω=v \|\mathbf{k}\|$ | $ε = ħ²k²/2m$ |
| Statistics | Bose-Einstein | Fermi-Dirac |
| $n =$ | $1/[\exp(βE) - 1]$ | $1/[\exp(β[E - μ]) + 1]$ |
| Number per $\mathbf{k}$ | 3 (polarization) | 2 (spin) |
| Total number | temperature-dependent | as many as there are |
On the one hand, there are important differences, but on the other hand there are a lot of similarities.
* The solutions of the equation of motion are still plane waves $ψ ∝ \exp(i\mathbf{k}\mathbf{r})$.
* The periodic boundary conditions work exactly the same: $k_{x,y,z}=…, \frac{-4\pi}{L}, \frac{-2\pi}{L}, 0, \frac{2\pi}{L}, \frac{4\pi}{L}, …$
* The density of $k$-states per unit volume in the reciprocal space is the same: $(L/2π)³$.
* The expression for the total energy of the system is *very similar*:
$E_\textrm{total} = 2_s (L/2π)³∫ ε(\mathbf{k}) n_F[ε(\mathbf{k})]d³\mathbf{k}.$
With this comparison we have everything to analyze the electron behavior.
In the previous lectures, we learned the properties and physical behavior of phonons in the [Debye model](2_debye_model.md)
We now do the same for electrons in a so-called Sommerfeld model.
Sommerfeld considered the electrons as _free particles_ in a cubic box of size $L^3$ with periodic boundary conditions (hence the name _Sommerfeld free electron model_ or simply _Sommerfeld model_).
The solutions to the particle in a box Schrödinger equation are plane waves
$$
\psi ∝ \exp(i\mathbf{k} \cdot \mathbf{r})
$$
where $\mathbf{k}$ is the electron's wavevector.
As a result of the periodic boundary conditions, $\mathbf{k}$ must take values $\frac{2\pi}{L} (n_x, n_y, n_z)$.
The plane wave's eigenenergies are given by the following dispersion relation
## Fermi sea
$$
\epsilon(\mathbf{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}.
$$
For a warm-up let us imagine what happens in 1D, when we only have a few electron states available.
Here $m$ is the mass of the electron.
Despite looking continuous, $\mathbf{k}$ can only take discrete values, and thus $\epsilon$ must also be discrete.
To visualize this, we have plotted $\epsilon$ as a function of $k$ for a 1D system in the figure below.
Each black dot corresponds to an occupiable $k$ state.
```python
kf = 3;
@@ -71,7 +64,6 @@ Edis = ks**2;
Econt = kcont**2;
fig = pyplot.figure();
#fig.set_size_inches(2, 2)
ax = fig.add_subplot(111);
ax.plot(kcont, Econt);
ax.plot(ks, Edis, 'k.', markersize=10);
@@ -80,91 +72,118 @@ for i in range(2*kf + 1):
ax.set_xlim(-3.75, 3.75);
ax.set_ylim(0.0, 11);
ax.set_xlabel(r"$k \enspace \left[ \frac{2 \pi}{L} \right]$");
ax.set_ylabel(r"$ε$");
ax.set_xlabel(r"$k \: \left[ \frac{2 \pi}{L} \right]$");
ax.set_ylabel(r"$\epsilon$");
ax.set_xticklabels([""] + ks.tolist() + [""]);
ax.set_yticks([]);
draw_classic_axes(ax, xlabeloffset=.6);
draw_classic_axes(ax, xlabeloffset = .6);
```
At $T=0$ the states with zero energy get completely filled up because the electrons cannot disappear.
![](figures/fermi_circle_periodic.svg)
## Density of states
Our first goal is to compute the density of states, which we will do now using a more explicit algorithm.
By definition density of states is the number of states per energy interval.
Therefore if we compute the *total* number of states $N(ε)$ with energy lower than $ε$, then $g(ε) = dN(ε)/dε$.
There is another difference between the statistical properties of phonons and electrons.
Dispersion is not the only difference between electrons and phonons.
Phonons are bosons whereas electrons are fermions.
As a result, statistical properties of electrons are governed by the Fermi-Dirac distribution
Let us apply this to the 3D case for a start.
Assuming three dimensions and spherical symmetry (the dispersion in the free electron model is isotropic), we find
$$
N=2\left(\frac{L}{2\pi}\right)^3\int\mathrm{d}{\bf k}=2 \left(\frac{L}{2\pi}\right)^34\pi\int k^2\mathrm{d}k=\frac{V}{\pi^2}\int k^2\mathrm{d}k,
n_{FD}(\beta(\epsilon-\mu)) = \frac{1}{e^{\beta(\epsilon-\mu)}+1}.
$$
where the factor 2 is due to spin, and $\left(\frac{L}{2\pi}\right)^3$ is once again the density of points in k-space.
Using $k=\frac{\sqrt{2mε}}{\hbar}$ and $\mathrm{d}k=\frac{1}{\hbar}\sqrt{\frac{m}{2ε}}\mathrm{d}ε$ we can rewrite this as:
We denote $\beta = \frac{1}{k_{B}T}$, $\epsilon$ the energy of an electron and $\mu$ the _chemical potential_ of an electron.
The Fermi-Dirac distribution defines the number of electrons in the system:
$$
N=\frac{V}{\pi^2}\int\frac{2mε}{\hbar^3}\sqrt{\frac{m}{2ε}}\mathrm{d}ε=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2ε}\ \mathrm{d}ε
\begin{align}
N &= 2 \sum_{\mathbf{k}} n_{FD}(\beta(\epsilon-\mu))\\
&\approx 2 \left( \frac{L}{2 \pi} \right)^3 \int n_{FD}(\beta(\epsilon-\mu))\mathrm{d} \mathbf{k}.
\end{align}
$$
So we find for the density of states:
Here the factor 2 accounts for the degeneracy per $\mathbf{k}$ value.
??? question "Why is there a degeneracy of 2? Wasn't the phonon degeneracy 3?"
The factor $2$ accounts for the spin degeneracy. From now on we will denote the spin degeneracy as $2_s$.
In a similar fashion, the expression for the total energy is given by
$$
g(ε)=\frac{ \mathrm{d}N}{ \mathrm{d}ε}=\frac{Vm^{3/2}\sqrt{2ε}}{\pi^2\hbar^3}\propto\sqrt{ε}
E = 2_s \left( \frac{L}{2 \pi} \right)^3 \int \epsilon(\mathbf{k}) n_{FD}(\beta(\epsilon-\mu)) \mathrm{d} \mathbf{k}.
$$
```python
E = np.linspace(0, 2, 500)
fig, ax = pyplot.subplots()
The above expression is extremely similar to the total energy of phonons in the Debye model.
There are only two differences: the distribution is $n_{FD}$ instead of $n_{BE}$ and the degeneracy is 2 instead of 3.
In the table below we summarize the properties of both phonons and electrons.
ax.plot(E, np.sqrt(E))
ax.set_ylabel(r"$g(ε)$")
ax.set_xlabel(r"$ε$")
draw_classic_axes(ax, xlabeloffset=.2)
```
| | Phonons | Electrons |
| - | - | - |
| Dispersion relation | $\omega = v_s \lvert\mathbf{k}\rvert$ | $\epsilon = \frac{\hbar^2\mathbf{k}^2}{2m}$ |
| Statistics | Bose-Einstein | Fermi-Dirac |
| $n(\epsilon) =$ | $\frac{1}{e^{\beta \epsilon} - 1}$ | $\frac{1}{e^{\beta(\epsilon - \mu)} + 1}$ |
| Degeneracy per $\mathbf{k}$ | 3 (polarization) | 2 (spin) |
| Total number | temperature-dependent | as many as there are |
Similarly,
The comparison gives us everything we need to analyze the electron's behavior in a solid.
- For 1D: $g(ε) = \frac{2 L}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}ε} \propto 1/\sqrt{ε}$
- For 2D: $g(ε) = \frac{k L^2}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}ε} \propto \text{constant}$
## Fermi sea
Let us imagine what happens when $T = 0$ in a 2D system.
At $T = 0$, the system occupies the ground state.
Electrons are fermions and thus they obey the Pauli exclusion principle.
As a consequence, each $\mathbf{k}$-state can only be occupied by two electrons with oposite spin.
Accordingly, the electrons fill up all the lowest possible $\mathbf{k}$-states.
In the reciprocal space, the occupied $\mathbf{k}$-states form a circle (in 1D it is a line and in 3D a sphere).
```python
# creating grid
N = 10
x = np.linspace(-N//2, N//2, N+1)
xx, yy = np.meshgrid(x,x)
Given the number of electrons in a system, we can now fill up these states starting from the lowest energy until we run out of electrons, at which point we reach the _Fermi energy_.
# Initialzing figure
fig = pyplot.figure(figsize = (8,8));
ax = fig.add_subplot(111);
## Fermi energy, Fermi wavevector, Fermi wavelength
# 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);
At $T=0$, the total number of electrons is given by the integral over the density of states up to the _Fermi energy_ $ε_\mathrm{F}$:
```
The circle is defined by the Fermi-Dirac distribution $n_{FD}(\beta(\epsilon-\mu))$.
At T = 0, $n_{FD}(\beta(\epsilon-\mu))$ becomes a step function:
$$
N=\int_0^{ε_\mathrm{F}}g(ε)\mathrm{d}ε \overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2mε_\mathrm{F})^{3/2}.
n_{FD}(\beta(\epsilon-\mu)) = \Theta(-(\epsilon-\epsilon_F)).
$$
Note that $N$ now denotes the total number of electrons in the system.
Alternatively, we can express $N$ as an integral over k-space up to the _Fermi wavenumber_, which is the wavenumber associated with the Fermi energy $k_\mathrm{F}=\sqrt{2mε_\mathrm{F}}/\hbar$
Here we have introduced a very important quantity - the _Fermi energy_ $\epsilon_F$.
The Fermi energy is the chemical potential $\mu$ of an electron at $T = 0$ (for $T > 0$ the chemical potential is called the Fermi level):
$$
N \overset{\mathrm{3D}}{=} 2\frac{L^3}{(2\pi)^3}\int_0^{k_\mathrm{F}} 4\pi k^2\mathrm{d}k= 2\frac{V}{(2\pi)^3} \frac{4}{3}\pi k_\mathrm{F}^3
\epsilon_{\mathrm{F}} = \mu, \quad T = 0.
$$
All states occupied up until the Fermi energy are a part of a _Fermi sea_.
Another way to describe $\epsilon_F$ is as the energy of the highest occupied state at $T = 0$.
These equations allow us to relate $ε_\mathrm{F}$ and $k_\mathrm{F}$ to the electron density $N/V$:
In a similar fashion, we define the _Fermi wavevector_ $\mathbf{k}_F$.
It is a wavevector that corresponds to the $\epsilon_F$ through the dispersion relation:
$$
ε_\mathrm{F}=\frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{2/3}, \mathrm{and} \quad k_\mathrm{F}=\left( 3\pi^2\frac{N}{V} \right)^{1/3}.
\epsilon_F = \frac{\hbar^2 \mathbf{k}_F^2}{2m}
$$
From the last equation it follows that the _Fermi wavelength_ $\lambda_\mathrm{F}\equiv 2\pi/k_\mathrm{F}$ is on the order of the atomic spacing for typical free electron densities in metals.
Example: For copper, the Fermi energy is ~7 eV. It would take a temperature of $\sim 70 000$K for electrons to gain such energy through a thermal excitation! The _Fermi velocity_ $v_\mathrm{F}=\frac{\hbar k_\mathrm{F}}{m}\approx$ 1750 km/s $\rightarrow$ electrons run with a significant fraction of the speed of light, only because lower energy states are already filled by other electrons.
To get a better understanding of these concepts, we have plotted the energy as a function of $k$ for a 1D system at $T=0$.
We see that all states up until $k_F$ are occupied and all states beyond that are empty.
```python
kf = 3.0;
extrapol = 4.0/3.0;
kfilled = np.linspace(-kf, kf, 500);
kfilled = np.linspace(-kf, kf, 40);
kstates = np.linspace(-extrapol*kf, extrapol*kf, 500);
Efilled = kfilled**2;
@@ -172,11 +191,14 @@ Estates = kstates**2;
fig = pyplot.figure();
ax = fig.add_subplot(111);
ax.fill_between(kfilled, Efilled, kf*kf, alpha=0.5);
# Creating plot
trans = 1
ax.plot([kf, kf], [0.0, kf*kf], 'k:');
ax.plot(kstates, Estates, 'k--');
ax.plot(kfilled, Efilled, linewidth=4);
ax.axhline(kf*kf, linestyle="dotted", color='k');
ax.plot(kstates[np.abs(kstates)<=kf], Estates[np.abs(kstates)<=kf], 'lightblue',alpha = trans);
ax.plot(kstates, Estates, color = 'lightblue', linestyle = '--',alpha = trans);
ax.scatter(kfilled, Efilled, color = '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]);
@@ -191,26 +213,111 @@ ax.set_ylim(0.0, kf*kf*extrapol);
draw_classic_axes(ax, xlabeloffset=.6);
```
The bold line represents all filled states at $T=0$. This is called the _Fermi sea_.
We mentioned earlier that the occupied states at $T = 0$ form a circle in the reciprocal space (or sphere or line depending on the dimensionality of the system).
This circle is called the _Fermi surface_.
The shape of the Fermi surface is determined by the dispersion relation.
The Fermi wavevector $\mathbf{k}_F$ also defines the _Fermi momentum_ $\mathbf{p}_F = \hbar \mathbf{k}_F$ and the _Fermi velocity_
$$
\mathbf{k}_F = \frac{\mathbf{p}_F}{m} = \frac{\hbar $\mathbf{k}_F$}{m}.
$$
New concept: _Fermi surface_ = all points in k-space with $ε=ε_\mathrm{F}$. For free electrons in 3D, the Fermi surface is the surface of a sphere.
![](figures/transport.svg)
## Density of states
Our aim is to compute the density of states (often written as DOS) using the algorithm we outline below.
The density of states is the number of states per energy interval.
Therefore we compute the *total* number of states $N(\epsilon)$ with energy lower than $\epsilon$:
$$
g(\epsilon) = \frac{dN(\epsilon)}{d\epsilon}.
$$
Let us apply this to the 3D case.
Assuming spherical symmetry (the dispersion in the free electron model is isotropic[^1]), we find:
$$
\begin{align}
N &=2_s \left(\frac{L}{2\pi}\right)^3\int\mathrm{d}{\bf k}\\
&=2_s \left(\frac{L}{2\pi}\right)^3 4\pi\int k^2\mathrm{d}k\\
&=\frac{V}{\pi^2}\int k^2\mathrm{d}k,
\end{align}
$$
where the factor $2_s$ is due to spin, and $\left(\frac{L}{2\pi}\right)^3$ is once again the density of points in k-space.
The orange circle represents the Fermi surface at finite current $\rightarrow$ this circle will shift only slightly before the electrons reach terminal velocity $\rightarrow$ all transport takes place near the Fermi surface.
We rewrite the expression above by substituting $k=\frac{\sqrt{2m\epsilon}}{\hbar}$ and $\mathrm{d}k=\frac{1}{\hbar}\sqrt{\frac{m}$:
$$
\begin{align}
N &=\frac{V}{\pi^2}\int\frac{2m \epsilon}{\hbar^3}\sqrt{\frac{m}{2\epsilon}}\mathrm{d}\epsilon\\
&=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\epsilon}\ \mathrm{d}\epsilon
\end{align}
$$
So we find for the density of states:
$$
\begin{align}
g(\epsilon) &= \frac{ \mathrm{d}N}{ \mathrm{d}\epsilon}\\
& =\frac{Vm^{3/2}\sqrt{2\epsilon}}{\pi^2\hbar^3} \propto\sqrt{\epsilon}
\end{align}
$$
## Finite temperature, heat capacity
We now extend our discussion to $T>0$ by including a temperature dependent occupation function $n_F(ε,T)$ into our expression for the total number of electrons:
Hence, we observe that the density of states of a 3D solid scales with a square root of energy:
$$
g(\epsilon) \propto\sqrt{\epsilon}
$$
Similarly, we find:
- For 1D: $g(\epsilon) = \frac{2 L}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}\epsilon} \propto 1/\sqrt{\epsilon}$
- For 2D: $g(\epsilon) = \frac{k L^2}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}\epsilon} \propto \text{constant}$
The plot below shows the electronic density of states for a 1D, 2D and 3D solid.
```python
E = np.linspace(0, 2, 500)
fig, ax = pyplot.subplots()
ax.plot(E, np.sqrt(E))
ax.set_ylabel(r"$g(\epsilon)$")
ax.set_xlabel(r"$\epsilon$")
draw_classic_axes(ax, xlabeloffset=.2)
```
## Relationship between the Fermi energy/wavevector to system parameters
We'd like to find the relationship between the Fermi energy $\epsilon_{\mathrm{F}}$ and the system parameters $N$ and $V$.
In order to do so, we calculate the number of electrons in the system at $T = 0$:
$$
\begin{align}
N &= \int \limits_0^{\infty} n_{FD}(\beta(\epsilon-\mu)) g(\epsilon) \mathrm{d}\epsilon
&\overset{\mathrm{T = 0}}{=}\int \limits_0^{\epsilon_\mathrm{F}}g(\epsilon)\mathrm{d}\epsilon \\
&\overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2m\epsilon_\mathrm{F})^{3/2}.
\end{align}
$$
We invert the above equation:
$$
N=\int_0^\infty n_F(ε,T)g(ε)\mathrm{d}ε,
\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{\frac{2}{3}}.
$$
where the probability for a certain electron state to be occupied is given by the Fermi-Dirac distribution
We use the dispersion relation ($\epsilon = \frac{\hbar^2 \mathbf{k}^2}{2m}$) to express the Fermi wavevector $k_{\mathrm{F}}$:
$$
n_F(ε,T)=\frac{1}{ \mathrm{e}^{(ε-\mu)/k_\mathrm{B}T}+1}
k_\mathrm{F} = \left( 3\pi^2\frac{N}{V} \right)^{\frac{1}{3}}.
$$
We calculate _Fermi wavelength_ $\lambda_\mathrm{F}\equiv 2\pi/k_\mathrm{F}$ and observe that it is on the order of the atomic spacing for typical free electron densities in metals[^2].
Example: For copper, the Fermi energy is ~7 eV. It would take a temperature of $\sim 70 000$K for electrons to gain such energy through a thermal excitation! The _Fermi velocity_ $v_\mathrm{F}=\frac{\hbar k_\mathrm{F}}{m}\approx$ 1750 km/s $\rightarrow$ electrons run with a significant fraction of the speed of light, only because lower energy states are already filled by other electrons.
<!--
![](figures/transport.svg)
We think that this image is very unclear and somewhat misleading.
We propose to explain with the Fermi-Dirac distribution why transport only takes place near the Fermi surface.
-->
## Finite temperature, heat capacity
We now extend our discussion to $T > 0$ by taking a close look at the Fermi-Dirac distribution $n_{FD}(\beta(\epsilon-\mu))$
$$
n_{FD}(\beta(\epsilon-\mu)) = \frac{1}{e^{\beta(\epsilon-\mu)}+1}.
$$
The Fermi-Dirac distribution $n_{FD}(\beta(\epsilon-\mu))$ for $T = 0$ and $T > 0$ is shown below.
Both cases have the same chemical potential $\mu$.
```python
fig = pyplot.figure()
ax = fig.add_subplot(1,1,1)
@@ -220,20 +327,21 @@ 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'$ε$')
ax.set_ylabel(r'$n_F(ε, T)$')
ax.set_xlabel(r'$\epsilon$')
ax.set_ylabel(r'$n_{FD}(\epsilon, T)$')
ax.set_yticks([0, 1])
ax.set_yticklabels(['$0$', '$1$'])
ax.set_xticks([mu])
ax.set_xticklabels([r'$\mu$'])
ax.set_xticklabels([r'$\mu = \epsilon_{\mathrm{F}}$'])
ax.set_ylim(-.1, 1.1)
ax.legend()
draw_classic_axes(ax)
pyplot.tight_layout()
```
where the chemical potential $\mu=ε_\mathrm{F}$ if $T=0$. Typically $ε_\mathrm{F}/k_\mathrm{B}$~70 000 K (~7 eV), whereas room temperature is only 300 K (~30 meV). Therefore, thermal smearing occurs only very close to the Fermi energy.
Having included the temperature dependence, we can now calculate the electronic contribution to the heat capacity $C_\mathrm{V,e}$.
Where $\epsilon_\mathrm{F} \equiv \mu$ if $T = 0$.
We observe that at finite temperature $T>0$, thermal excitations _smear out_ n_{FD}(\beta(\epsilon-\mu)).
We plot $g(\epsilon)\dot n_{FD}(\beta(\epsilon-\mu))$ at $T = 0$ and $T > 0$ alongside the 3D solid density of states to further illustrate the smearing effect.
```python
E = np.linspace(0, 2, 500)
@@ -260,76 +368,106 @@ ax.set_ylabel(r"$g(ε)$")
ax.set_xlabel(r"$ε$")
draw_classic_axes(ax, xlabeloffset=.2)
```
In order to estimate the fraction of thermally excited electrons, we approximate the blue and orange areas by triangles.
The validity of the approximation depends on the ratio between the thermal energy $E_{\mathrm{T}} = k_{B}T$ and the Fermi energy $\epsilon_{\mathrm{F}}$.
If the ratio is large, there is a lot of thermal smearing and the approximation breaks down because the area cannot be accurately approximated by triangles.
However, if the ratio is small, there is barely any thermal smearing, and $n_{FD}(\beta(\epsilon-\mu))$ is approximately a step function.
In such a case, the approximation holds.
The approximation was first used by Sommerfeld, hence the name _Sommerfeld expansion_[^4]
**Example**: Copper's Fermi energy is $\epsilon_{\mathrm{F}} ~ 7$ eV.
The electron temperature would need to be $\sim 70 000$K in order to gain similar energy through thermal excitations!
Therefore at room temperature ($T = 300 \: \rm{K}$) there is barely any thermal smearing.
Hence, the Sommerfeld expansion is very accurate in this case.
??? question "Suppose we have a material with a Fermi energy of 5 m$\mathbf{eV}$, can we still apply the Sommerfeld expansion at room temperarure?"
No, the thermal energy at room temperature is approximately 25 m$eV$. Therefore, thermal smearing is significant and the Sommerfeld expansion is not accurate.
We will estimate $C_\mathrm{V,e}$ and derive its scaling with temperature using the triangle method depicted in the figure. A finite temperature causes electrons in the top triangle to be excited to the bottom triangle. Because the base of this triangle scales with $k_\mathrm{B}T$ and its height with $ g(ε_\mathrm{F})$, it follows that the number of excited electrons $N_\mathrm{exc} \approx g(ε_\mathrm{F})k_\mathrm{B}T$ (neglecting pre-factors of order 1).
We now use the Sommerfeld expansion to approximate the electronic contribution to the heat capacity $C_\mathrm{e}$.
At finite temperature, the electrons from the top triangle (blue) excite to the bottom triangle (orange).
The base of the triangle scales with $k_\mathrm{B}T$ and the height with $ g(\epsilon_\mathrm{F})$.
Hence the number of excited electrons is $N_\mathrm{exc} \approx g(\epsilon_\mathrm{F})k_\mathrm{B}T$ (neglecting constants not depending on $\epsilon_{\mathrm{F}}$).
These electrons gain $k_\mathrm{B}T$ of energy, so the total extra energy is
$$
E(T)-E(0)=N_\mathrm{exc}k_\mathrm{B}T\approx g(ε_\mathrm{F})k_\mathrm{B}^2T^2.
\begin{align}
E(T) &= E(T = 0) + N_\mathrm{exc}k_\mathrm{B}T\\
&\approx E(T = 0) + g(\epsilon_\mathrm{F})k_\mathrm{B}^2T^2.
\end{align}
$$
Therefore, the heat capacity is given by
$$
C_\mathrm{V,e}=\frac{ \mathrm{d}E}{ \mathrm{d}T} \approx 2 g(ε_\mathrm{F})k_\mathrm{B}^2T=\ ...\ =3 Nk_\mathrm{B}\frac{T}{T_\mathrm{F}}\propto T,
\begin{align}
C_\mathrm{e} &= \frac{ \mathrm{d}E}{ \mathrm{d}T}\\
&\approx 2 g(ε_\mathrm{F})k_\mathrm{B}^2T\\
&\overset{\mathrm{3D}}{=} 3 Nk_\mathrm{B}\frac{T}{T_\mathrm{F}}\\
\propto T,
\end{align}
$$
where we used $N=\frac{2}{3}ε_\mathrm{F}g(ε_\mathrm{F})$ and we defined the _Fermi temperature_ $T_\mathrm{F}=\frac{ε_\mathrm{F}}{k_\mathrm{B}}$.
where we used $N=\frac{2}{3}ε_\mathrm{F}g(ε_\mathrm{F})$ and defined the _Fermi temperature_ $T_\mathrm{F}=\frac{\epsilon_\mathrm{F}}{k_\mathrm{B}}$.
How does $C_\mathrm{V,e}$ relate to the phonon contribution $C_\mathrm{V,p}$?
How does $C_\mathrm{e}$ relate to the phonon contribution $C_\mathrm{p}$?
- At room temperature, $C_\mathrm{V,p}=3Nk_\mathrm{B}\gg C_\mathrm{V,e}$
- Near $T=0$, $C_\mathrm{V,p}\propto T^3$ and $C_\mathrm{V,e}\propto T$ $\rightarrow$ competition.
- At room temperature ($T \gg T_{\mathrm{F}}$, else the Sommerfeld expansion is invalid), $C_\mathrm{p}=3Nk_\mathrm{B}\gg C_\mathrm{e}$
- Near $T=0$, $C_\mathrm{p}\propto T^3$ and $C_\mathrm{e}\propto T$ $\rightarrow$ competition.
## Useful trick: scaling of $C_V$
## Useful trick: scaling of $C$
<!--
We are not sure if this part is that useful.
It seems really confusing for students as to why we can also use the Sommerfeld approximation for bosonic modes.
-->
The behavior of $C$ can be quickly memorized and understood using the following mnemonic rule:
Behavior of $C_\mathrm{V}$ can be very quickly memorized or understood using the following mnemonic rule
> Particles in within an energy range ~$kT$ become thermally excited, and each carries extra energy $kT$.
> Particles within an energy range ~$k_{B}T$ become thermally excited, and each carries extra energy $k_{B}T$.
#### Example 1: electrons
$g(ε_\mathrm{F})$ roughly constant ⇒ total energy in the thermal state is $T \times [T\times g(ε_\mathrm{F})]$ ⇒ $C_\mathrm{V} \propto T$.
$g(\epsilon_\mathrm{F})$ roughly constant ⇒ total energy in the thermal state is proportional to $T \times [T\times g(\epsilon_\mathrm{F})]$ ⇒ $C_\mathrm{V} \propto T$.
#### Example 2: graphene with $E_F=0$ (midterm 2018)
#### Example 2: graphene with $E_{\mathrm{F}}=0$
$g(ε) \propto ε$ ⇒ total energy is $T \times T^2$ ⇒ $C_\mathrm{V} \propto T^2$.
$g(\epsilon) \propto \epsilon$ ⇒ total energy is $T \times T^2$ ⇒ $C \propto T^2$.
#### Example 3: phonons in 3D at low temperatures.
$g(ε) \propto ε^2$ ⇒ total energy is $T \times T^3$ ⇒ $C_\mathrm{V} \propto T^3$.
$g(\epsilon) \propto \epsilon^2$ ⇒ total energy is $T \times T^3$ ⇒ $C_\mathrm{V} \propto T^3$.
## Conclusions
1. The Sommerfeld free electron model treats electrons as waves with dispersion $ε=\frac{\hbar^2k^2}{2m}$.
1. The Sommerfeld free electron model treats electrons as plane waves with energy dispersion $\epsilon = \frac{\hbar^2k^2}{2m}$.
2. The density of states (DOS) can be derived from the dispersion relation. This procedure is general, and analogous to e.g. that for phonons (see lecture 2 - Debye model).
3. The Fermi-Dirac distribution describes the probability an electron state is occupied.
4. The free-electron heat capacity is linear with $T$.
5. The scaling of heat capacity with $T$ can be derived quickly by estimating the nr of particles in an energy range $k_\mathrm{B}T$, using the DOS.
4. The electronic contribution to the heat capacity is linear with $T$.
5. The scaling of heat capacity with $T$ can be quickly derived by estimating the number of particles in an energy range $k_\mathrm{B}T$ using the DOS.
## Exercises
### Warm-up questions
1. List the differences between electrons and phonons from your memory.
2. Write down the expression for the total energy of particles with the density of states $g(E)$ and the occupation number $n(E, T)$.
3. Explain what happens if a material is heated up to its Fermi temperature (assuming that materials where it is possible exist).
2. Write down the expression for the total energy of particles with the density of states $g(E)$ and the occupation number $n_{FD}(\beta(E - \mu))$.
3. Explain what happens if a material is heated up to its Fermi temperature (assuming that material where this is possible exists).
### Exercise 1: potassium
The Sommerfeld model provides a good description of free electrons in alkali metals such as potassium, which has a Fermi energy of 2.12 eV (data from Ashcroft, N. W. and Mermin, N. D., Solid State Physics, Saunders, 1976.).
The Sommerfeld model provides a good description of free electrons in alkali metals such as potassium (element K), which has a Fermi energy of **$\epsilon_{\mathrm{F}} = 2.12$ eV** (data from Ashcroft, N. W. and Mermin, N. D., Solid State Physics, Saunders, 1976.).
1. Check the [Fermi surface database](http://www.phys.ufl.edu/fermisurface/). Explain why potassium and (most) other alkali metals can be described well with the Sommerfeld model.
2. Calculate the Fermi temperature, Fermi wave vector and Fermi velocity for potassium.
3. Why is the Fermi temperature much higher than room temperature?
4. Calculate the free electron density in potassium.
4. Calculate the free electron density $n$ in potassium.
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?
### Exercise 2: the $n$-dimensional free electron model.
In the lecture, it has been explained that the density of states (DOS) of the free electron model is proportional to $1/\sqrt{\epsilon}$ in 1D, constant in 2D and proportional to $\sqrt{\epsilon}$ in 3D. In this exercise, we are going to derive the DOS of the free electron model for an arbitrary number of dimensions.
Suppose we have an $n$-dimensional hypercube with length $L$ for each side and contains free electrons.
Suppose we have a $n$-dimensional hypercube with length $L$ for each side that houses free electrons.
1. Assuming periodic boundary conditions, what is the distance between nearest-neighbour points in $\mathbf{k}$-space? What is the density of $\mathbf{k}$-points in n-dimensional $\mathbf{k}$-space?
2. The number of $\mathbf{k}$-points with a magnitude between $k$ and $k + dk$ is given by $g(k)dk$. Using the answer for 1, find $g(k)$ for 1D, 2D and 3D.
1. 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 n-dimensional $\mathbf{k}$-space?
2. The number of $\mathbf{k}$-points between $k$ and $k + dk$ is given by $g(k)dk$.
Using the answer for 1, find $g(k)$ for 1D, 2D and 3D.
3. Now show that $g(k)$ for $n$ dimensions is given by
$$g(k) = \frac{1}{\Gamma(n/2)} \left( \frac{L }{ \sqrt{\pi}} \right)^n \left( \frac{k}{2} \right)^{n-1},$$ where $\Gamma(z)$ is the [gamma function](https://en.wikipedia.org/wiki/Gamma_function).
@@ -350,7 +488,8 @@ A hypothetical metal has a Fermi energy $\epsilon_F = 5.2 \, \mathrm{eV}$ and a
1. Give an integral expression for the total energy of the electrons in this hypothetical material in terms of the DOS $g(\epsilon)$, the temperature $T$ and the chemical potential $\mu = \epsilon_F$.
2. Find the ground state energy at $T = 0$.
3. In order to obtain a good approximation of the integral for non-zero $T$, one can make use of the [Sommerfeld expansion](https://en.wikipedia.org/wiki/Sommerfeld_expansion). Using this expansion, find the difference between the total energy of the electrons for $T = 1000 \, \mathrm{K}$ with that of the ground state.
3. In order to obtain a good approximation of the integral for non-zero $T$, one can make use of the [Sommerfeld expansion](https://en.wikipedia.org/wiki/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.
4. Now, find this difference in energy by calculating the integral found in 1 numerically. Compare your result with 3.
??? hint
@@ -360,10 +499,27 @@ A hypothetical metal has a Fermi energy $\epsilon_F = 5.2 \, \mathrm{eV}$ and a
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.
### Exercise 4: graphene
One of the most famous recently discovered materials is [graphene](https://en.wikipedia.org/wiki/Graphene), which 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:
$$ \epsilon(\mathbf{k}) = \pm c|\mathbf{k}|.$$ Note that the $\pm$ here means that there are two energy levels at a specified $\mathbf{k}$. The Fermi level is set at $\epsilon_F = 0$.
One of the most famous recently discovered materials is [graphene](https://en.wikipedia.org/wiki/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:
$ \epsilon(\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 $\epsilon_F = 0$.
1. Make a sketch of the dispersion relation.
What other well-known particles have a linear dispersion relation?
2. Using the dispersion relation and assuming periodic boundary conditions, derive an expression for the DOS 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 $|\epsilon|$.
??? hint
It is convenient to first start by only considering the positive energy contributions $\epsilon(\mathbf{k}) = + c|\mathbf{k}|$ and calculate the DOS for it. Then account for the negative energy contributions $\epsilon(\mathbf{k}) = - c|\mathbf{k}|$ by adding it to the DOS for the positive energies. You can also make use of $\frac{\rm{d} |k|}{\rm{d}k} = \frac{k}{|k|}$.
1. Make a sketch of the dispersion relation. What other well-known particles have a linear dispersion relation?
2. Using the dispersion relation and assuming periodic boundary conditions, derive an expression for the DOS of graphene. Your result should be linear with $|\epsilon|$. Do not forget spin degeneracy, and take into account that graphene has an additional two-fold 'valley degeneracy'.
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 $$n_{ex} = \frac{1}{2} g(-k_B T) k_B T.$$
4. Calculate the heat capacity $C_V$ as a function of the temperature $T$.
[^1]: An [isotropic](https://en.wikipedia.org/wiki/Isotropic_solid) material means that the material is the same in all directions.
[^2]: The mean inter-particle distance is related to the electron density $n = \frac{N}{V}$ by
$$
\langle r \rangle \propto \frac{1}{n^{\frac{1}{3}}}.
$$
The exact proportionality constant depends on the properties of the system.