Here, each black dot is a possible electron state.
## Comparing the description of electrons and phonons
## Comparing the analyses of electrons and phonons
In contrast with the linear dispersion of the phonon modes in the [Debye model](2_debye_model.md), the dispersion of free electrons in the Sommerfeld model is quadratic. In addition, the electrons obey fermionic instead of bosonic statistics.
As a result, the occupation of the electron states is described by the Fermi-Dirac distribution
## Relation between the Fermi energy and the number of electrons
What is the relation between the Fermi energy $\varepsilon_{F}$ and the number of electrons in the system ($N$)?
To answer this, we calculate the number of electrons in the system at $T = 0$ using the density of states:
The Fermi energy sets the number of electrons in the system $N$. We can see this by expressing $N$ at $T = 0$ as an integral over the density of states:
\begin{align}
N &= \int \limits_0^{\infty} n_{F}(\beta(\varepsilon-\mu)) g(\varepsilon) \mathrm{d}\varepsilon\\
We use the dispersion relation ($\varepsilon(\mathrm{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}$) to find the length of the Fermi wavevector (i.e., the Fermi wavenumber) $k_F = |\mathbf{k_{F}}|$:
Having found the Fermi energy, we can use the dispersion relation ($\varepsilon(\mathrm{k}) = \frac{\hbar^2 \mathbf{k}^2}{2m}$) to find the length of the Fermi wavevector (i.e., the Fermi wavenumber) $k_F = |\mathbf{k_{F}}|$:
Using the Fermi wavenumber, we calculate the _Fermi wavelength_ $\lambda_F\equiv 2\pi/k_F$. Plugging in numbers (see exercise) shows that it is on the order of the atomic spacing for typical free electron densities in metals[^3].
Using the Fermi wavenumber, we calculate the _Fermi wavelength_ $\lambda_F\equiv 2\pi/k_F$. Plugging in numbers (see exercise) shows that the Fermi wavelength is on the order of the atomic spacing for typical free electron densities in metals[^3].
## The chemical potential and the Fermi energy
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@@ -343,11 +342,10 @@ pyplot.tight_layout()
```
At a finite temperature $T>0$, thermal excitations _smear out_ the sharp change in the number of occupied electrons near $\varepsilon_F$.
The Fermi energy is typically in the range of electronvolts, which corresponds to a Fermi temperature of $T_F = \varepsilon_F/k_B \approx 10 000$K. As such, a temperature of approximately 10000 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 $T=0$. We can thus safely approximate the chemical potential at all temperatures by the Fermi energy: $\mu = \varepsilon_F$.
The Fermi energy is typically in the range of electronvolts, which corresponds to a Fermi temperature of $T_F = \varepsilon_F/k_B \approx 10 000$K. This would therefore be the temperature required for thermal excitations to give an electron a similar amount of energy as the Fermi energy itself! Therefore, at room temperature ($T = 300$K), the electron distribution over energies is very similar to that at $T=0$, and we can safely approximate the chemical potential at all temperatures by the Fermi energy: $\mu = \varepsilon_F$.
## The electron heat capacity
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).
To calculate the electron heat capacity, we compare the occupied electron states $g(\varepsilon) n_{F}(\beta(\varepsilon-\mu))$ at $T = 0$ (blue shaded area) to the occupied states at $T > 0$ (orange shaded area).
```python
E=np.linspace(0,2,500)
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@@ -375,19 +373,18 @@ ax.set_xlabel(r"$ε$")
draw_classic_axes(ax,xlabeloffset=.2)
```
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 over an energy range $\Delta E \approx k_B T$, which is much smaller than the Fermi energy $\varepsilon_{F}$.
To estimate the increase in the electron energy caused by the increase in temperature, we approximate the difference between the blue and orange areas by triangles, as shown in the figure.
This approximation is appropriate because the thermal smearing happens over an energy range $\Delta E \approx k_B T$ that is much smaller than the Fermi energy $\varepsilon_{F}$.
!!! note "Sommerfeld expansion"
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.
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 $\varepsilon_{F}$).
At a finite temperature, the electrons occupying the top triangle (blue) become thermally excited to occupy the bottom triangle (orange).
Since the base of the triangle is proportional to $k_BT$ and the height is $\sim g(\varepsilon_F)$, the number of excited electrons is $N_\mathrm{exc} \approx g(\varepsilon_F)k_BT$ (neglecting constants not depending on $\varepsilon_{F}$).
These electrons gain $k_BT$ of thermal energy, such that the total extra energy is
These electrons have gained $k_BT$ of thermal energy, such that the total extra energy is
where we used $N=\frac{2}{3}\varepsilon_Fg(\varepsilon_F)$ and defined the _Fermi temperature_ $T_F \equiv \varepsilon_F/k_B$.
where we used $N=\frac{2}{3}\varepsilon_Fg(\varepsilon_F)$ and the Fermi temperature $T_F \equiv \varepsilon_F/k_B$.
How does $C_e$ relate to the phonon contribution $C_p$?
How does the electron heat capacity $C_e$ compare to the phonon heat capacity $C_p$?
- At room temperature $C_\mathrm{p}\approx 3Nk_B\gg C_e \propto k_B T / T_F$, because $T \ll T_F$.
- Near $T=0$, the phonon heat capacity $C_p\propto k_B (T/T_D)^3$, and it becomes smaller than the electron heat capacity at $T \lesssim \sqrt{T_D^3/T_F}$
## Useful trick: scaling of $C$
## Useful trick: scaling of $C_e$
Similar to how we understood the low temperature heat capacity of the Debye model, the behavior of $C_e$ can be quickly memorized and understood using the following argument:
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@@ -422,8 +419,8 @@ 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 $\varepsilon_F = 0$ and a density of states $g(\varepsilon) \propto \varepsilon$.
In exercise 2, we will analyze [graphene](https://en.wikipedia.org/wiki/Graphene). Unlike in metals, electrons in graphene cannot be treated as 'free'. As a result, graphene
has a density of states $g(\varepsilon) \propto \varepsilon$, and Fermi energy $\varepsilon_F = 0$.
Therefore, within the energy range of $k_BT$, $g(\varepsilon) \propto k_BT$.
Thus the total energy is proportional to $T \times T^2$ and the heat capacity $C_e \propto T^2$.
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@@ -463,7 +460,7 @@ The Sommerfeld model provides a good description of free electrons in alkali met
### Exercise 3*: graphene
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'.
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 $\mathbf{k}$.
