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@@ -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 +23,35 @@ _(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 [Debye model](2_debye_model.md), we learned 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, $\mathbf{k}$ must take discrete values $\frac{2\pi}{L} (n_x, n_y, n_z)$.
The plane waves have eigenenergies given by the dispersion relation
## Fermi sea
$$
\varepsilon(\mathbf{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m},
$$
For a warm-up let us imagine what happens in 1D, when we only have a few electron states available.
with $m$ being the mass of the electron.
Let us plot $\varepsilon(k)$ as a function of $k$ for a 1D system:
```python
kf = 3;
@@ -71,7 +63,6 @@ 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 +71,111 @@ 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"$\varepsilon$");
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)
Here each black dot is a possible electron state.
## 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 $g(ε) = dN(ε)/dε$.
In addition to having a quadratic dispersion and not linear dispersion, electrons also obey fermionic, and not bosonic statistics.
As a result, the occupation of electron states is described 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
$$
N=2\left(\frac{L}{2\pi}\right)^3\int\mathrm{d}{\bf k}=2 \left(\frac{L}{2\pi}\right)^34\pi\int k^2\mathrm{d}k=\frac{V}{\pi^2}\int k^2\mathrm{d}k,
n_{F}(\beta(\varepsilon-\mu)) = \frac{1}{e^{\beta(\varepsilon-\mu)}+1}.
$$
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:
$$
N=\frac{V}{\pi^2}\int\frac{2mε}{\hbar^3}\sqrt{\frac{m}{2ε}}\mathrm{d}ε=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2ε}\ \mathrm{d}ε
$$
Here $\beta = 1/k_{B}T$, $\varepsilon$ is the energy, and $\mu$ the _chemical potential_ of an electron.
The Fermi-Dirac distribution defines the number of electrons in the system:
So we find for the density of states:
$$
g(ε)=\frac{ \mathrm{d}N}{ \mathrm{d}ε}=\frac{Vm^{3/2}\sqrt{2ε}}{\pi^2\hbar^3}\propto\sqrt{ε}
\begin{align}
N &= 2 \sum_{\mathbf{k}} n_{F}(\beta(\varepsilon-\mu))\\
&= 2 \left( \frac{L}{2 \pi} \right)^3 \int n_{F}(\beta(\varepsilon-\mu))\mathrm{d} \mathbf{k}.
\end{align}
$$
```python
E = np.linspace(0, 2, 500)
fig, ax = pyplot.subplots()
Just like with phonons, we replaced a discrete sum over $k$ with a volume integral.
The factor 2 here accounts for the number of distinct electron states per $\mathbf{k}$ value.
ax.plot(E, np.sqrt(E))
??? question "Check: why is there a degeneracy of 2? Weren't there 3 polarizations?"
ax.set_ylabel(r"$g(ε)$")
ax.set_xlabel(r"$ε$")
draw_classic_axes(ax, xlabeloffset=.2)
```
The factor $2$ accounts for the spin degeneracy. To keep track of the origin of this term we will denote the spin degeneracy as $2_s$.
Similarly,
Similar to the tota energy of phonons in the Debye model, the expression for the total energy of electrons is
- For 1D: $g(ε) = \frac{2 L}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}ε} \propto 1/\sqrt{ε}$
- For 2D: $g(ε) = \frac{k L^2}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}ε} \propto \text{constant}$
$$
E = 2_s \left( \frac{L}{2 \pi} \right)^3 \int \varepsilon(\mathbf{k}) n_{F}(\beta(\varepsilon-\mu)) \mathrm{d} \mathbf{k}.
$$
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 summarize the properties of both phonons and electrons.
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_.
| | 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 |
## Fermi energy, Fermi wavevector, Fermi wavelength
!!! note "About $\hbar$"
At $T=0$, the total number of electrons is given by the integral over the density of states up to the _Fermi energy_ $ε_\mathrm{F}$:
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.
$$
N=\int_0^{ε_\mathrm{F}}g(ε)\mathrm{d}ε \overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2mε_\mathrm{F})^{3/2}.
$$
Note that $N$ now denotes the total number of electrons in the system.
The last difference is important: 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]
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 $k_\mathrm{F}=\sqrt{2mε_\mathrm{F}}/\hbar$
## The Fermi sea
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
$$
N \overset{\mathrm{3D}}{=} 2\frac{L^3}{(2\pi)^3}\int_0^{k_\mathrm{F}} 4\pi k^2\mathrm{d}k= 2\frac{V}{(2\pi)^3} \frac{4}{3}\pi k_\mathrm{F}^3
n_{F}(\beta(\varepsilon-\mu)) = \Theta(-(\varepsilon-\varepsilon_F)).
$$
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).
These equations allow us to relate $ε_\mathrm{F}$ and $k_\mathrm{F}$ to the electron density $N/V$:
```python
# creating grid
N = 10
x = np.linspace(-N//2, N//2, N+1)
xx, yy = np.meshgrid(x,x)
$$
ε_\mathrm{F}=\frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{2/3}, \mathrm{and} \quad k_\mathrm{F}=\left( 3\pi^2\frac{N}{V} \right)^{1/3}.
$$
# Initialzing figure
fig = pyplot.figure(figsize = (10,10));
ax = fig.add_subplot(111);
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.
# 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);
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.
```
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 *Fermi wave vector*.
To clarify the relation between these concepts let us take a look at the dispersion relation in 1D:
```python
kf = 3.0;
extrapol = 4.0/3.0;
kfilled = np.linspace(-kf, kf, 500);
kfilled = np.linspace(-extrapol*kf, extrapol*kf, 100);
kstates = np.linspace(-extrapol*kf, extrapol*kf, 500);
Efilled = kfilled**2;
@@ -172,11 +183,14 @@ 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, 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]);
@@ -186,30 +200,115 @@ 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_xlim(-kf*extrapol, kf*extrapol)
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_.
By using the dispersion relation, we arrive to the relation
$$
\varepsilon_F = \frac{\hbar^2 \mathbf{k}_F^2}{2m}.
$$
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.
The Fermi wavevector $\mathbf{k}_F$ also defines the _Fermi momentum_ $\mathbf{p}_F = \hbar \mathbf{k}_F$ and the _Fermi velocity_:
![](figures/transport.svg)
$$
\mathbf{v}_F = \frac{\mathbf{p}_F}{m} = \frac{\hbar \mathbf{k}_F}{m}.
$$
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.
??? question "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!
## 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:
## Density of states
Like before, in order to compute 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.
$$
N=\int_0^\infty n_F(ε,T)g(ε)\mathrm{d}ε,
g(\varepsilon) = \frac{dN}{d\varepsilon}.
$$
where the probability for a certain electron state to be occupied is given by the Fermi-Dirac distribution
Let us calculate the density of states for a 3D system.
Because the dispersion of the free electron is isotropic[^2], we utilize spherical coordinates.
The total number of states is then given as
$$
n_F(ε,T)=\frac{1}{ \mathrm{e}^{(ε-\mu)/k_\mathrm{B}T}+1}
\begin{align}
N &\overset{\mathrm{3D}}{=}2_s \left(\frac{L}{2\pi}\right)^3\int_{\varepislon(\mathbf{k} < \varepsilon)}\mathrm{d}{\mathbf{k}}\\
&=2_s \left(\frac{L}{2\pi}\right)^3 4\pi\int k^2\mathrm{d}k\\
&=\frac{V}{\pi^2}\int k^2\mathrm{d}k,
\end{align}
$$
We rewrite the expression above by substituting $k=\hbar^{-1}\sqrt{2m\varepsilon}}$ and $\mathrm{d}k=\hbar^{-1}\sqrt{2m/\varepsilon}}\.d\varepsilon$:
$$
\begin{align}
N &=\frac{V}{\pi^2}\int\frac{2m \varepsilon}{\hbar^3}\sqrt{\frac{m}{2\varepsilon}}\mathrm{d}\varepsilon\\
&=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\varepsilon}\ \mathrm{d}\varepsilon
\end{align}
$$
Therefore we find the density of states:
$$
\begin{align}
g(\varepsilon) &= \frac{ \mathrm{d}N}{ \mathrm{d}\varepsilon}\\
& =\frac{Vm^{3/2}\sqrt{2\varepsilon}}{\pi^2\hbar^3} \propto\sqrt{\varepsilon}
\end{align}
$$
We observe that the density of states of a 3D solid is proportional to a square root of energy:
$$
g(\varepsilon) \propto\sqrt{\varepsilon}
$$
Repeating the 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.
```python
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()
draw_classic_axes(ax, xlabeloffset=.2)
```
## Relationship between the Fermi energy and the system parameters
We would like to relate the Fermi energy $\varepsilon_{F}$ to the system parameters the number of electrons in the system ($N$) and the volume of the box $V = L^3$.
To do so, we calculate the number of electrons in the system at $T = 0$ using the previously found density of states:
$$
\begin{align}
N &= \int \limits_0^{\infty} n_{F}(\beta(\varepsilon-\mu)) g(\varepsilon) \mathrm{d}\varepsilon\\
&\overset{\mathrm{T = 0}}{=}\int \limits_0^{\varepsilon_F}g(\varepsilon)\mathrm{d}\varepsilon \\
&\overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2m\varepsilon_F)^{3/2}.
\end{align}
$$
Inverting the above equation yields:
$$
\varepsilon_{F} = \frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{\frac{2}{3}}.
$$
We use the dispersion relation ($\varepsilon(\mathrm{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}$) to express the Fermi wavevector $k_{F}$:
$$
k_F = \left( 3\pi^2\frac{N}{V} \right)^{\frac{1}{3}}.
$$
Using the Fermi wavevector, we calculate _Fermi wavelength_ $\lambda_F\equiv 2\pi/k_F$ and observe that it is on the order of the atomic spacing for typical free electron densities in metals[^3].
## Heat capacity
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 $T > 0$ are shown below.
For both cases we have the same chemical potential $\mu = \varepsilon_F$.
```python
fig = pyplot.figure()
@@ -220,20 +319,23 @@ 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'$\varepsilon$')
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'$\mu$'])
ax.set_xticklabels([r'$\mu = \varepsilon_{F}$'])
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 $C_\mathrm{V,e}$.
At a finite temperature $T>0$, thermal excitations _smear out_ the sharp change in the number of occupied electrons near $\varepsilon_F$.
Because the Fermi energy is typically in the range of electronvolts, the temperature of $\sim 10 000$K would be required in order for thermal excitations to give an electron a similar amount of energy!
Therefore at room temperature $T = 300$K the electron distribution over energies is very similar to that at $T=0$.
Below we compare the number of occupied electron states at each energy $g(\varepsilon) n_{F}(\beta(\varepsilon-\mu))$ at $T = 0$ (blue shaded area) with $T > 0$ (orange shaded area).
```python
E = np.linspace(0, 2, 500)
@@ -261,75 +363,96 @@ ax.set_xlabel(r"$ε$")
draw_classic_axes(ax, xlabeloffset=.2)
```
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).
In order to estimate the electron energy increase, we approximate difference between the blue and orange areas by triangles, as shown in the figure.
This approximation is appropriate because the thermal smearing happens at the energies $E ∼ k_B T$, and it is much smaller than the Fermi energy $\varepsilon_{F}$.
These electrons gain $k_\mathrm{B}T$ of energy, so the total extra energy is
!!! note "Sommerfeld expansion"
$$
E(T)-E(0)=N_\mathrm{exc}k_\mathrm{B}T\approx g(ε_\mathrm{F})k_\mathrm{B}^2T^2.
$$
A more rigorous way to estimate the energy of electrons at a finite temperature is to apply the [Sommerfeld expansion](https://en.wikipedia.org/wiki/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.
Therefore, the heat capacity is given by
At a finite temperature, the electrons occupying the top triangle (blue) are thermally excited to occupy the bottom triangle (orange).
The base of the triangle is proportional to $k_BT$ and the height is $\sim g(\varepsilon_F)$.
Hence the number of excited electrons is $N_\mathrm{exc} \approx g(\varepsilon_F)k_BT$ (neglecting constants not depending on $\varepsilon_{F}$).
These electrons gain $k_BT$ of thermal energy, such that the total extra energy is
$$
C_\mathrm{V,e}=\frac{ \mathrm{d}E}{ \mathrm{d}T} \approx 2 g(ε_\mathrm{F})k_\mathrm{B}^2T=\ ...\ =3 Nk_\mathrm{B}\frac{T}{T_\mathrm{F}}\propto T,
\begin{align}
E(T) &= E(T = 0) + N_\mathrm{exc}k_BT\\
&\approx E(T = 0) + g(\varepsilon_F)k_B^2T^2.
\end{align}
$$
where we used $N=\frac{2}{3}ε_\mathrm{F}g(ε_\mathrm{F})$ and we defined the _Fermi temperature_ $T_\mathrm{F}=\frac{ε_\mathrm{F}}{k_\mathrm{B}}$.
How does $C_\mathrm{V,e}$ relate to the phonon contribution $C_\mathrm{V,p}$?
- 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.
## Useful trick: scaling of $C_V$
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 defined the _Fermi temperature_ $T_F \equiv \varepsilon_F/k_B$.
Behavior of $C_\mathrm{V}$ can be very quickly memorized or understood using the following mnemonic rule
How does $C_e$ relate to the phonon contribution $C_p$?
> Particles in within an energy range ~$kT$ become thermally excited, and each carries extra energy $kT$.
- 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 $T \lesssim \sqrt{T_D^3/T_F}$
#### Example 1: electrons
## Useful trick: scaling of $C$
$g(ε_\mathrm{F})$ roughly constant ⇒ total energy in the thermal state is $T \times [T\times g(ε_\mathrm{F})]$ ⇒ $C_\mathrm{V} \propto T$.
The behavior of $C$ can be quickly memorized and understood using the following mnemonic rule:
#### Example 2: graphene with $E_F=0$ (midterm 2018)
> Particles within an energy range of $\sim k_{B}T$ to the Fermi energy $\varepsilon_F$ become thermally excited, and each carry an extra energy $k_{B}T$.
$g(ε) \propto ε$ ⇒ total energy is $T \times T^2$ ⇒ $C_\mathrm{V} \propto T^2$.
The contribution to the energy through thermal excitations can be calculated by triangulating the thermal smearing of the density of states around the Fermi energy $\varepsilon_F$.
The triangulated area is proportional to the number of excited electrons $N_{exc}$.
Let us illustrate this with two examples:
#### Example 3: phonons in 3D at low temperatures.
### Example 1: 3D free electrons
$g(ε) \propto ε^2$ ⇒ total energy is $T \times T^3$ ⇒ $C_\mathrm{V} \propto T^3$.
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
Graphene has a Fermi energy of $\varepsilon_F = 0$ and a density of states $g(\varepsilon) \propto \varepsilon$.
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
1. The Sommerfeld free electron model treats electrons as waves with dispersion $ε=\frac{\hbar^2k^2}{2m}$.
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.
1. The Sommerfeld free electron model treats electrons as free particles with energy dispersion $\varepsilon = \frac{\hbar^2k^2}{2m}$.
2. The density of states $g(\varepsilon)$ can be derived from the dispersion relation.
This procedure is general, and analogous to e.g. that for phonons.
3. The Fermi-Dirac distribution describes the probability of an electron state to be occupied.
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 $k_BT$ using the density of states.
## 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 $g(\varepsilon)$ and the occupation number $n_{F}(\beta(\varepsilon - \mu))$.
3. Explain what happens if a material is heated up to its Fermi temperature (assuming that material where this is possible exists).
4. Why can we not use the Sommerfeld expansion with a Fermi energy of the order of the thermal energy?
5. Is the heat capacity of a solid at temperatures near $T=0$ dominated by electrons or phonons?
### 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 $\varepsilon_{F} = 2.12$ eV (data from Ashcroft, N. W. and Mermin, N. D., Solid State Physics, Saunders, 1976.).
1. Check the [Fermi surface database](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.
In the lecture, it has been explained that the density of states of the free electron model is proportional to $1/\sqrt{\varepsilon}$ in 1D, constant in 2D and proportional to $\sqrt{\varepsilon}$ in 3D. In this exercise, we are going to derive the density of states of the free electron model for an arbitrary number of dimensions.
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).
@@ -341,16 +464,17 @@ Suppose we have an $n$-dimensional hypercube with length $L$ for each side and c
??? hint
Check [Wikipedia](https://en.wikipedia.org/wiki/Particular_values_of_the_gamma_function) to find out how to deal with half-integer values in the gamma function.
5. Using the expression in 3, calculate the DOS (do not forget the spin degeneracy).
6. Give an integral expression for the total number of electrons and for their total energy in terms of the DOS, the temperature $T$ and the chemical potential $\mu$ (_you do not have to work out these integrals_).
5. Using the expression in 3, calculate the density of states (do not forget the spin degeneracy).
6. Give an integral expression for the total number of electrons and for their total energy in terms of the density of states, the temperature $T$ and the chemical potential $\mu$ (_you do not have to work out these integrals_).
7. Work out these integrals for $T = 0$.
### Exercise 3: a hypothetical material
A hypothetical metal has a Fermi energy $\epsilon_F = 5.2 \, \mathrm{eV}$ and a DOS per unit volume $g(\epsilon) = 2 \times 10^{10} \, \mathrm{eV}^{-\frac{3}{2}} \sqrt{\epsilon}$.
### Exercise 3: a hypothetical material
A hypothetical metal has a Fermi energy $\varepsilon_F = 5.2 \, \mathrm{eV}$ and a density of states per unit volume $g(\varepsilon) = 2 \times 10^{10} \, \mathrm{eV}^{-\frac{3}{2}} \sqrt{\varepsilon}$.
1. Give an integral expression for the total energy of the electrons in this hypothetical material in terms of the DOS $g(\epsilon)$, the temperature $T$ and the chemical potential $\mu = \epsilon_F$.
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 $\mu = \varepsilon_F$.
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 +484,29 @@ 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:
$ \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?
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 $\frac{\rm{d} |k|}{\rm{d}k} = \frac{k}{|k|}$.
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 $T$.
4. Calculate the heat capacity $C_e$ as a function of the temperature $T$.
[^1]: This is not completely true, as we will see when learning about [semiconductors](13_semiconductors)
[^2]: An [isotropic](https://en.wikipedia.org/wiki/Isotropic_solid) material means that the material is the same in all directions.
[^3]: 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.