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@@ -10,12 +10,12 @@ 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:
- Write down the Fermi-Dirac distribution function.
- Write down the Schrödinger equation and solve it in free space.
- Write down the solutions to the Schrödinger Equation for a free particle.
- Apply periodic boundary conditions, and compute the density of states
given a sufficiently simple dispersion relation.
@@ -23,44 +23,71 @@ _(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:
Having learned the statistical properties of [phonons in the Debye model](2_debye_model.md), let us use these as a starting point for comparing with the electrons.
We consider the electrons as _free particles_ in a cubic box of size $L^3$ with periodic boundary conditions.
In that case, the solutions to the Schrödinger equation will be a 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 corresponding eigenenergies are given by the following dispersion relation
$$
\epsilon(\mathbf{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}
$$
Here $m$ is the electron's mass.
There is another difference between the statistical properties of phonons and electrons.
Because electrons are fermions, their statistical properties are governed by the Fermi-Dirac distribution
$$
n_{FD}(\beta(\epsilon-\mu)) = \frac{1}{e^{\beta(\epsilon-\mu)}+1}
$$
where $\beta = \frac{1}{k_{\rm B}T}$, \epsilon is the energy of the electrons, described by the previously mentioned dispersion relations, and $\mu$ is the electron's _chemical potential_.
Using the Fermi-Dirac distribution, we can write the number of electrons in the system as
$$
\begin{align}
N &= 2 \sum_{\mathbf{k}} n_{FD}(\beta(\epsilon-\mu))\\
&\approx 2 \left( \frac{L}{2 \pi} \right)^3 \int \rm{d} \mathbf{k} n_{FD}(\beta(\epsilon-\mu))
\end{align}
$$
Here the factor 2 accounts for the degeneracy per $mathbf{k}$ value.
??? question "Why is there a degeneracy of 2 instead of 3?"
The factor $2$ accounts for the spin degeneracy. From now on we will denote the spin degeneracy as 2_s
In similar fashion the expression for the total energy is given by
$$
E = 2_s \left( \frac{L}{2 \pi} \right)^3 \int \rm{d} \mathbf{k} \epsilon(\mathbf{k}) n_{FD}(\beta(\epsilon-\mu))
$$
Note that this expression is extremely similar to the expression of the total energy of phonons in the Debye model.
The only differences are that we used $n_{FD}$ instead of $n_{BE}$ and a factor 2 instead of 3.
In the table below we summarize the properties of both phonons and electrons
| | 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 | $ω=v \|\mathbf{k}\|$ | $ε = ħ²k²/2m$ |
| Dispersion relation | $\omega = v_s \lvert\mathbf{k}\rvert$ | $\epsilon = \frac{\hbar^2\mathbf{k}^2}{2m}$ |
| Statistics | Bose-Einstein | Fermi-Dirac |
| $n =$ | $1/[\exp(βE) - 1]$ | $1/[\exp(β[E - μ]) + 1]$ |
| Number per $\mathbf{k}$ | 3 (polarization) | 2 (spin) |
| $n(\epsilon) =$ | $\frac{1}{e^{\beta \epsilon} - 1}$ | $\frac{1}{e^{\beta(\epsilon - \mu)} + 1}$ |
| Degeneracy per $\mathbf{k}$ | 3 (polarization) | 2 (spin) |
| Total number | temperature-dependent | as many as there are |
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.
With this comparison we have everything we need to analyze the electron's behaviour in a solid.
## Fermi sea
For a warm-up let us imagine what happens in 1D, when we only have a few electron states available.
<!---
We are not entirely sure if this graph helps out with understanding what a fermi sea/energy is.
It is a very usefull graph for understanding how states are occupied on a 'continuous' energy scale.
Might consider moving it upwards
```python
kf = 3;
extrapol = 1.1;
@@ -80,86 +107,137 @@ 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.
-->
Let us imagine what happens when $T = 0$ in a 2D system.
At $T = 0$, the system occupies the ground state.
Because electrons are fermions, they obey the [Pauli exlusion principle](https://en.wikipedia.org/wiki/Pauli_exclusion_principle).
This means that each $\mathbf{k}$-state can only be occupied by two electrons with oposite spin.
Therefore the electrons fill up all $\mathbf{k}$-states corresponding to the lowest possible energy.
In the reciprocal space, the occupied $\mathbf{k}$-states form a circle (in 1D it becomes a line and in 3D a sphere).
<!---
Still need to make this figure in python.
-->
![](figures/fermi_circle_periodic.svg)
## Density of states
We observe the same behaviour if we look at n_{FD}(\beta(\epsilon-\mu)).
At T = 0, n_{FD}(\beta(\epsilon-\mu)) becomes a step function
$$
n_{FD}(\beta(\epsilon-\mu)) = \Theta(-(\epsilon-\epsilon_F))
$$
Here we have introduced a very important quantity, the _Fermi energy_ $\epsilon_F$, which is the chemical potential $\mu$ of the electrons at $T = 0$ (for $T > 0$ the chemical potential is called the Fermi level).
$$
\epsilon_{\rm F} = \mu, \quad T = 0.
$$
All states occupied up until the Fermi energy is called a _Fermi sea_.
> So far there is no Fermi beach yet.
Another way to describe $\epsilon_F$ is as the energy of the highest occupied state at $T = 0$.
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 a similar fashion we can define the _Fermi wavevector_ $\mathbf{k}_F$, which is related to $\epsilon_F$ via the dispersion relation
$$
\epsilon_F = \frac{\hbar^2 \mathbf{k}_F^2}{2m}
$$
As mentioned earlier, in the reciprocal space the occupied states at $T = 0$ form a circle (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
Using $\mathbf{k}_F$ we can define the _Fermi momentum_ $\mathbf{p}_F = \hbar $\mathbf{k}_F$ and the _Fermi velocity_
$$
\mathbf{k}_F = \frac{\mathbf{p}_F}{m} = \frac{\hbar $\mathbf{k}_F$}{m}
$$
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
<!--
Might be handy to put some bulletpoints with the defined quantities. Let discuss this with Anton.
-->
## Density of states
Our first goal is to compute the density of states (often written as DOS), which we will now do using a more explicit algorithm.
By definition the density of states is the number of states per energy interval.
Therefore if we compute the *total* number of states $N(\epsilon)$ with energy lower than $\epsilon$, then
$$
g(\epsilon) = \frac{dN(\epsilon)}{d\epsilon}.
$$
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,
Let us apply this to the 3D case.
Assuming three dimensions and spherical symmetry (the dispersion in the free electron model is [isotropic](https://en.wikipedia.org/wiki/Isotropic_solid)), 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 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:
Using $k=\frac{\sqrt{2m\epsilon}}{\hbar}$ and $\mathrm{d}k=\frac{1}{\hbar}\sqrt{\frac{m}{2\epsilon}}\mathrm{d}epsilon$ we can rewrite this as:
$$
N=\frac{V}{\pi^2}\int\frac{2mε}{\hbar^3}\sqrt{\frac{m}{2ε}}\mathrm{d}ε=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2ε}\ \mathrm{d}ε
\begin{align}
N &=\frac{V}{\pi^2}\int\frac{2m \epsilon}{\hbar^3}\sqrt{\frac{m}{2\epsilon}}\mathrm{d}\epsilon\\
&=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\epsilon}\ \mathrm{d}\epsilon
\end{align}
$$
So we find for the density of states:
$$
g(ε)=\frac{ \mathrm{d}N}{ \mathrm{d}ε}=\frac{Vm^{3/2}\sqrt{2ε}}{\pi^2\hbar^3}\propto\sqrt{ε}
\begin{align}
g(\epsilon) &= \frac{ \mathrm{d}N}{ \mathrm{d}\epsilon}\\
& =\frac{Vm^{3/2}\sqrt{2\epsilon}}{\pi^2\hbar^3} \propto\sqrt{\epsilon}
\end{align}
$$
Hence, we observe that for a 3D solid, the electronic density of states show a $\sqrt{\epsilon}$ behaviour
$$
g(\epsilon) \propto\sqrt{\epsilon}
$$
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"$g(ε)$")
ax.set_xlabel(r"$ε$")
ax.set_ylabel(r"$g(\epsilon)$")
ax.set_xlabel(r"$\epsilon$")
draw_classic_axes(ax, xlabeloffset=.2)
```
Similarly,
- For 1D: $g(ε) = \frac{2 L}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}ε} \propto 1/\sqrt{ε}$
- For 2D: $g(ε) = \frac{k L^2}{\pi} \frac{ \mathrm{d}k}{ \mathrm{d}ε} \propto \text{constant}$
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_.
## Fermi energy, Fermi wavevector, Fermi wavelength
At $T=0$, the total number of electrons is given by the integral over the density of states up to the _Fermi energy_ $ε_\mathrm{F}$:
## Calculating the Fermi energy and wavevector
Our goal is to find a relation between the Fermi energy $\epsilon_{\rm F}$ and the system parameters $N$ and $V$.
In order to do so, we can calculate the number of electrons in the system at $T = 0$.
$$
N=\int_0^{ε_\mathrm{F}}g(ε)\mathrm{d}ε \overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2mε_\mathrm{F})^{3/2}.
\begin{align}
N &= \int_0^{\inf} \mathrm{d}\epsilon n_{FD}(\beta(\epsilon-\mu)) g(\epsilon)
&\overset{\mathrm{T = 0}}{=}\int_0^{\epsilon_\mathrm{F}}g(\epsilon)\mathrm{d}\epsilon \\
&\overset{\mathrm{3D}}{=} \frac{V}{3\pi^2\hbar^3}(2m\epsilon_\mathrm{F})^{3/2}.
\end{align}
$$
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 $k_\mathrm{F}=\sqrt{2mε_\mathrm{F}}/\hbar$
This can be rewritten to
$$
N \overset{\mathrm{3D}}{=} 2\frac{L^3}{(2\pi)^3}\int_0^{k_\mathrm{F}} 4\pi k^2\mathrm{d}k= 2\frac{V}{(2\pi)^3} \frac{4}{3}\pi k_\mathrm{F}^3
\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{\frac{2}{3}}.
$$
These equations allow us to relate $ε_\mathrm{F}$ and $k_\mathrm{F}$ to the electron density $N/V$:
We can use the dispersion relation ($\epsilon = \frac{\hbar^2 \mathbf{k}^2}{2m}$) to find an expression for Fermi wavevector $k_{\rm F}$
$$
ε_\mathrm{F}=\frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{2/3}, \mathrm{and} \quad k_\mathrm{F}=\left( 3\pi^2\frac{N}{V} \right)^{1/3}.
k_\mathrm{F} = \left( 3\pi^2\frac{N}{V} \right)^{\frac{1}{3}}.
$$
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.
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[^3].
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.
<!--
Move this image upwards such that it nicely illustrates the effect of the Fermi energy.
This might be nice after introducing the Fermi wavevector through the means of the dispersion relation.
Also remove the colouring as it misleading. This makes it look like that all coloured states are occupied. But this is not the case for a 1D system.
```python
kf = 3.0;
@@ -190,26 +268,24 @@ ax.set_xlim(-kf*extrapol, kf*extrapol);
ax.set_ylim(0.0, kf*kf*extrapol);
draw_classic_axes(ax, xlabeloffset=.6);
```
The bold line indicates all occupied k states. This is another way of visualizing the Fermi sea.
-->
The bold line represents all filled states at $T=0$. This is called the _Fermi sea_.
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)
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 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 including a temperature dependent occupation function $n_F(ε,T)$ into our expression for the total number of electrons:
## Finite temperature, heat capacity
We now extend our discussion to $T > 0$ by taking a close look to the Fermi-Dirac distribution $n_{FD}(\beta(\epsilon-\mu))$.
$$
N=\int_0^\infty n_F(ε,T)g(ε)\mathrm{d}ε,
$$
where the probability for a certain electron state to be occupied is given by the Fermi-Dirac distribution
$$
n_F(ε,T)=\frac{1}{ \mathrm{e}^{(ε-\mu)/k_\mathrm{B}T}+1}
n_{FD}(\beta(\epsilon-\mu)) = \frac{1}{e^{\beta(\epsilon-\mu)}+1}
$$
$n_{FD}(\beta(\epsilon-\mu))$ for $T = 0$ and $T > 0$ is shown below. Both curves have the same chemical potential $\mu$.
```python
fig = pyplot.figure()
@@ -220,20 +296,21 @@ 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'$\mu = \epsilon_{\rm F}$'])
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 for $T>0$, n_{FD}(\beta(\epsilon-\mu)) _smears out_.
In order to further illustrate this, $g(\epsilon)\dot n_{FD}(\beta(\epsilon-\mu))$ is plotted for $T = 0$ and $T > 0$ for a 3D solid, in which $g(\epsilon)\propto\sqrt{\epsilon}$.
```python
E = np.linspace(0, 2, 500)
@@ -260,52 +337,79 @@ ax.set_ylabel(r"$g(ε)$")
ax.set_xlabel(r"$ε$")
draw_classic_axes(ax, xlabeloffset=.2)
```
We can approximate the blue and orange areas as triangles.
The validity of this approximation depends on the ratio between the thermal energy $E_{\rm T} = k_{\rm B}T$ and the Fermi energy $\epsilon_{\rm F}$.
If this ration is very large, there is a lot of thermal smearing and the approximation breaks down as the area can not be accurately approximated as triangles anymore.
However, for a small ratio, there is barely any thermal smearing and n_{FD}(\beta(\epsilon-\mu)) is almost a step function.
In this case the area can be accurately approximated as a triangle and the approximation holds.
For electronic structures, this approximation was first used by Sommerfeld, hence the name _Sommerfeld model_.
**Example**: For copper, the Fermi energy $\epsilon_{\rm F} ~ 7$ eV.
It would take a temperature of $\sim 70 000$K for electrons to gain such energy through a thermal excitation!
Therefore at room temperature ($T = 300 \: \rm{K}$) there is barely any thermal smearing.
Hence, the Sommerfeld model is very accurate in this case.
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).
??? question "Suppose we have a material with a Fermi energy of 5 m$eV$, can we still apply the Sommerfeld model at room temperarure?"
No, the thermal energy at room temperature is approximately 25 m$eV$. Therefore, a lot of thermal smearing occurs and the Sommerfeld model is not accurate.
We will now use the Sommerfeld model to approximate the electronic contribution to the heat capacity $C_\mathrm{e}$.
At finite temperature, the electrons in the top triangle excite to the bottom triangle.
Because the base of this triangle scales with $k_\mathrm{B}T$ and its height with $ g(\epsilon_\mathrm{F})$, it follows that the number of excited electrons $N_\mathrm{exc} \approx g(\epsilon_\mathrm{F})k_\mathrm{B}T$ (neglecting constants not depending on $\epsilon_{\rm F}$).
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 we 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_{\rm F}, else the Sommerfeld model 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.
-->
Behavior of $C$ can be very quickly memorized or 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 ~$k_{\rm B}T$ become thermally excited, and each carries extra energy $k_{\rm B}T$.
#### 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 propertional 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 $E_{\rm F}=0$
$g(ε) \propto ε$ ⇒ 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 waves with 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 derived quickly by estimating the number of particles in an energy range $k_\mathrm{B}T$, using the DOS.
## Exercises
@@ -367,3 +471,10 @@ $$ \epsilon(\mathbf{k}) = \pm c|\mathbf{k}|.$$ Note that the $\pm$ here means th
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 $T$.
[^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.
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