For $T > 0$ the chemical potential is called the Fermi level and is not denoted as $\epsilon_F$.
Another way to describe $\epsilon_F$ is as the energy of the highest occupied state at $T = 0$.
All states occupied up until the Fermi energy are a part of a _Fermi sea_.
In a similar fashion, we define the _Fermi wavevector_ $\mathbf{k}_F$.
This is a wavevector that corresponds to $\epsilon_F$ through the dispersion relation:
$$
\epsilon_F = \frac{\hbar^2 \mathbf{k}_F^2}{2m}.
$$
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.
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 momentum*.
To clarify the relation between these concepts let us take a look at the dispersion relation in 1D:
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.
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...
@@ -222,11 +219,12 @@ $$
## Density of states
Our aim is to compute the density of states (often written as DOS).
Our aim is to compute the density of states.
The density of states is the number of states per energy interval.
Therefore, the density of states is related to the *total* number of states $N(\epsilon)$ through
Therefore, the density of states is related to the *total* number of states $N(\varepsilon)$ through
$$
g(\epsilon) = \frac{dN}{d\epsilon}.
g(\varepsilon) = \frac{dN}{d\varepsilon}.
$$
Let us calculate the density of states for a 3D system.
Because the dispersion of the free electron is isotropic[^2], spherical symmetry can be assumed.
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...
@@ -240,31 +238,31 @@ N &\overset{\mathrm{3D}}{=}2_s \left(\frac{L}{2\pi}\right)^3\int\mathrm{d}{\math
$$
where $\left(\frac{L}{2\pi}\right)^3$ is once again the density of points in k-space.
We rewrite the expression above by substituting $k=\frac{\sqrt{2m\epsilon}}{\hbar}$ and $\mathrm{d}k=\sqrt{\frac{m}{2 \epsilon}}\frac{1}{\hbar}$:
We rewrite the expression above by substituting $k=\frac{\sqrt{2m\varepsilon}}{\hbar}$ and $\mathrm{d}k=\sqrt{\frac{m}{2 \varepsilon}}\frac{1}{\hbar}$:
$$
\begin{align}
N &=\frac{V}{\pi^2}\int\frac{2m \epsilon}{\hbar^3}\sqrt{\frac{m}{2\epsilon}}\mathrm{d}\epsilon\\
## Relationship between the Fermi energy and the system parameters
We would like to relate the Fermi energy $\epsilon_{F}$ to the system parameters the number of electrons in the system ($N$) and the volume of the box $V = L^3$.
We would like to relate the Fermi energy $\varepsilon_{F}$ to the system parameters the number of electrons in the system ($N$) and the volume of the box $V = L^3$.
To do so, we calculate the number of electrons in the system at $T = 0$ using the previously found density of states:
$$
\begin{align}
N &= \int \limits_0^{\infty} n_{F}(\beta(\epsilon-\mu)) g(\epsilon) \mathrm{d}\epsilon\\
The Fermi-Dirac distribution $n_{F}(\beta(\epsilon-\mu))$ for $T = 0$ and $T > 0$ are shown below.
For both cases we have the same chemical potential $\mu = \epsilon_F$.
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()
...
...
@@ -321,20 +319,20 @@ 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'$\epsilon$')
ax.set_ylabel(r'$n_{F}(\epsilon, T)$')
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'$\mu = \epsilon_{F}$'])
ax.set_xticklabels([r'$\mu = \varepsilon_{F}$'])
ax.set_ylim(-.1,1.1)
ax.legend()
draw_classic_axes(ax)
pyplot.tight_layout()
```
We observe that at finite temperature $T>0$, thermal excitations _smear out_ $n_{F}(\beta(\epsilon-\mu))$.
Below we plot $g(\epsilon) n_{F}(\beta(\epsilon-\mu))$ at $T = 0$ (blue + orange shaded area) and $T > 0$ (orange shaded area) alongside the 3D solid density of states to further illustrate the smearing effect.
We observe that at finite temperature $T>0$, thermal excitations _smear out_ $n_{F}(\beta(\varepsilon-\mu))$.
Below we plot $g(\varepsilon) n_{F}(\beta(\varepsilon-\mu))$ at $T = 0$ (blue + orange shaded area) and $T > 0$ (orange shaded area) alongside the 3D solid density of states to further illustrate the smearing effect.
```python
E=np.linspace(0,2,500)
...
...
@@ -362,13 +360,13 @@ 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 $\epsilon_{F}$.
The validity of the approximation depends on the ratio between the thermal energy $E_{\mathrm{T}} = k_{B}T$ and the Fermi energy $\varepsilon_{F}$.
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 $n_{F}(\beta(\epsilon-\mu))$ is approximately a step function.
However, if the ratio is small, there is barely any thermal smearing, and $n_{F}(\beta(\varepsilon-\mu))$ is approximately a step function.
In such a case, the approximation holds.
The approximation was first used by Sommerfeld, hence the name *Sommerfeld expansion*.
**Example**: Copper's Fermi energy is $\epsilon_{F} ~ 7$ eV.
**Example**: Copper's Fermi energy is $\varepsilon_{F} ~ 7$ 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.
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...
@@ -379,15 +377,15 @@ Hence, the Sommerfeld expansion is very accurate in this case.
We now use the Sommerfeld expansion to approximate the electronic contribution to the heat capacity $C_e$.
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_F)$.
Hence the number of excited electrons is $N_\mathrm{exc} \approx g(\epsilon_F)k_\mathrm{B}T$ (neglecting constants not depending on $\epsilon_{F}$).
The base of the triangle scales with $k_\mathrm{B}T$ and the height with $ g(\varepsilon_F)$.
Hence the number of excited electrons is $N_\mathrm{exc} \approx g(\varepsilon_F)k_\mathrm{B}T$ (neglecting constants not depending on $\varepsilon_{F}$).
These electrons gain $k_\mathrm{B}T$ of thermal energy, such that the total extra energy is
where we used $N=\frac{2}{3}\epsilon_Fg(\epsilon_F)$ and defined the _Fermi temperature_ $T_F \equiv \frac{\epsilon_F}{k_\mathrm{B}}$.
where we used $N=\frac{2}{3}\varepsilon_Fg(\varepsilon_F)$ and defined the _Fermi temperature_ $T_F \equiv \frac{\varepsilon_F}{k_\mathrm{B}}$.
How does $C_e$ relate to the phonon contribution $C_\mathrm{p}$?
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@@ -411,25 +409,25 @@ How does $C_e$ relate to the phonon contribution $C_\mathrm{p}$?
## Useful trick: scaling of $C$
The behavior of $C$ can be quickly memorized and understood using the following mnemonic rule:
> Particles within an energy range of ~$k_{B}T$ to the Fermi energy $\epsilon_F$ become thermally excited, and each carry an extra energy $k_{B}T$.
> Particles within an energy range of ~$k_{B}T$ to the Fermi energy $\varepsilon_F$ become thermally excited, and each carry an extra energy $k_{B}T$.
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 $\epsilon_F$.
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 1: 3D free electrons
In 3D, $g(\epsilon_F)$ is roughly constant.
Thus the total energy obtained through thermal excitation is proportional to $T \times \left( T\times g(\epsilon_F) \right)$, from which it follows that $C_e \propto T$.
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 $\epsilon_F = 0$ and a density of states $g(\epsilon) \propto \epsilon$.
Therefore, within the energy range of $k_BT$, $g(\epsilon) \propto k_BT$.
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 free particles with energy dispersion $\epsilon = \frac{\hbar^2k^2}{2m}$.
2. The density of states (DOS) can be derived from the dispersion relation.
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 $T$.
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@@ -440,13 +438,13 @@ Thus the total energy is proportional to $T \times T^2$ and the heat capacity $C
### 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 $g(\epsilon)$ and the occupation number $n_{F}(\beta(\epsilon - \mu))$.
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 (element K), which has a Fermi energy of $\epsilon_{F} = 2.12$ 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.
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@@ -455,7 +453,7 @@ The Sommerfeld model provides a good description of free electrons in alkali met
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.
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. What is the distance between nearest-neighbour points in $\mathbf{k}$-space? Assume periodic boundary conditions.
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@@ -473,14 +471,14 @@ Using the answer for 1, find $g(k)$ for 1D, 2D and 3D.
??? 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 $g(\epsilon) = 2 \times 10^{10} \,\mathrm{eV}^{-\frac{3}{2}} \sqrt{\epsilon}$.
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)(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.
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@@ -496,17 +494,17 @@ Using this expansion, find the difference between the total energy of the electr
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 $\epsilon_F = 0$.
$ \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 DOS of graphene.
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 $|\epsilon|$.
Your result should be linear with $|\varepsilon|$.
??? 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 $\frac{\rm{d} |k|}{\rm{d}k} = \frac{k}{|k|}$.
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 $\frac{\rm{d} |k|}{\rm{d}k} = \frac{k}{|k|}$.
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 $C_e$ as a function of the temperature $T$.
