Sommerfeld model
from matplotlib import pyplot import numpy as np from common import draw_classic_axes, configure_plotting configure_plotting()
(based on chapter 4 of the book)
Expected prerequisites
Before the start of this lecture, you should be able to:
- Given an elementary (e.g. linear) dispersion relation, compute the density of states.
- Write down the Fermi-Dirac distribution.
- Write down the Schrödinger equation and solve it in free space to obtain the parabolic dispersion relation of free electrons.
- Describe the relation between the wavevector, wavenumber, and wavelength of a plane wave.
Learning goals
After this lecture you will be able to:
- Calculate the electron density of states in 1D, 2D, and 3D for the parabolic dispersion of free electrons.
- Use the density of states to express the number and energy of electrons in a system as an integral over energy for
. - Use the Fermi-Dirac distribution to extend the previous learning goal to
. - Calculate the electron contribution to the specific heat of a solid.
- Describe what is the Fermi energy, the Fermi temperature, the Fermi surface, the Fermi wavevector, and the Fermi wavelength.
The free electron dispersion¶
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?"
In the Debye model, we studied 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
with
kf = 3; extrapol = 1.1; ks = np.arange(-kf, kf+1); kcont = np.linspace(-extrapol*kf, extrapol*kf, 200); Edis = ks**2; Econt = kcont**2; fig = pyplot.figure(); ax = fig.add_subplot(111); ax.plot(kcont, Econt); ax.plot(ks, Edis, 'k.', markersize=10); for i in range(2*kf + 1): ax.plot([ks[i], ks[i]], [0.0, Edis[i]], 'k:'); ax.set_xlim(-3.75, 3.75); ax.set_ylim(0.0, 11); 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);
Here, each black dot is a possible electron state.
Comparing the analyses of electrons and phonons¶
In contrast with the linear dispersion of the phonon modes in the Debye model, 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
Here
Just like with phonons, we replaced the discrete sum over
Check: why is there a degeneracy of 2? Weren't there 3 polarizations?
The factor
Similar to the total energy of phonons in the Debye model, the expression for the total energy of electrons is
There are only two differences: the distribution is
Phonons | Electrons | |
---|---|---|
Dispersion relation | ||
Statistics | Bose-Einstein | Fermi-Dirac |
Degeneracy per |
3 (polarization) | 2 (spin) |
Total particle number | temperature-dependent | constant |
It is important to note the last difference in this table: 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
About
Within quantum mechanics energy and frequency are related by Planck's constant:
The Fermi sea, energy, surface, wavevector, and wavelength¶
To determine the chemical potential
The chemical potential at
# creating grid N = 10 x = np.linspace(-N//2, N//2, N+1) xx, yy = np.meshgrid(x,x) # Initialzing figure fig = pyplot.figure(figsize = (10,10)); ax = fig.add_subplot(111); # 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);
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 the magnitude of the Fermi wavevector (i.e., the Fermi wavenumber). To clarify the relation between these concepts let us take a look at the dispersion relation in 1D:
kf = 3.0; extrapol = 4.0/3.0; kfilled = np.linspace(-extrapol*kf, extrapol*kf, 100); kstates = np.linspace(-extrapol*kf, extrapol*kf, 500); Efilled = kfilled**2; Estates = kstates**2; fig = pyplot.figure(); ax = fig.add_subplot(111); # Creating plot trans = 1 ax.plot([kf, kf], [0.0, kf*kf], '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]); ax.set_xticklabels([r"$k_F$"]); 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_ylim(0.0, kf*kf*extrapol); draw_classic_axes(ax, xlabeloffset=.6);
By using the dispersion relation, we obtain the relation between the Fermi energy and the Fermi wavevector
The Fermi wavevector
The Fermi energy of copper is ~7 eV. What is the corresponding Fermi velocity?
The Fermi velocity
Deriving the density of states for the parabolic dispersion of the free electron model¶
Like before, to compute the energy and 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
Let us calculate the density of states for a 3D system.
Because the free-electron dispersion is isotropic2, we use spherical coordinates.
The total number of states below energy
We rewrite this expression into an integral over energy using the dispersion relation, substituting
We thus find the density of states:
We observe that the density of states for the 3D parabolic free-electron dispersion is proportional to the square root of energy:
Repeating similar derivations, we find the density of states of 1D and 2D systems:
- 1D:
- 2D:
We plot these three behaviors of
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() ax.set_xticks([]) ax.set_yticks([]) draw_classic_axes(ax, xlabeloffset=.2)
Relation between the Fermi energy and the number of electrons¶
The Fermi energy sets the number of electrons in the system
Solving this equation for the Fermi energy yields:
Having found the Fermi energy, we can use the dispersion relation (
Using the Fermi wavenumber, we calculate the Fermi wavelength
The chemical potential and the Fermi energy¶
We now extend our discussion to
The Fermi-Dirac distribution
fig = pyplot.figure() ax = fig.add_subplot(1,1,1) xvals = np.linspace(0, 2, 200) mu = .75 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'$\varepsilon$') 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'$\mu = \varepsilon_{F}$']) ax.set_ylim(-.1, 1.1) ax.legend() draw_classic_axes(ax) pyplot.tight_layout()
At a finite temperature
The electron heat capacity¶
To calculate the electron heat capacity, we compare the occupied electron states
E = np.linspace(0, 2, 500) fig, ax = pyplot.subplots() ax.plot(E, np.sqrt(E), linestyle='dashed') ax.text(1.7, 1.4, r'$g(ε)\propto \sqrt{ε}$', ha='center') ax.fill_between(E, np.sqrt(E) * (E < 1), alpha=.3) n = np.sqrt(E) / (1 + np.exp(20*(E-1))) ax.plot(E, n) ax.fill_between(E, n, alpha=.5) w = .17 ax.annotate('', xy=(1, 1), xytext=(1-w, 1), arrowprops=dict(arrowstyle='<->', shrinkA=0, shrinkB=0)) ax.text(1-w/2, 1.1, r'$\sim k_BT$', ha='center') ax.plot([1-w, 1+w], [1, 0], c='k', linestyle='dashed') ax.annotate('', xy=(1, 0), xytext=(1, 1), arrowprops=dict(arrowstyle='<->', shrinkA=0, shrinkB=0)) ax.text(1.2, .7, r'$g(ε_F)$', ha='center') ax.set_xticks([1]) ax.set_xticklabels([r'$ε_F$']) ax.set_yticks([]) ax.set_ylabel(r"$g(ε)$") ax.set_xlabel(r"$ε$") draw_classic_axes(ax, xlabeloffset=.2)
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
Sommerfeld expansion
A more rigorous way to estimate the energy of electrons at a finite temperature is to apply the Sommerfeld expansion.
It still uses on the smallness of
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
These electrons have gained
Therefore, the electron heat capacity
where we used
How does the electron heat capacity
- At room temperature
, because . - Near
, the phonon heat capacity , and it becomes smaller than the electron heat capacity at
Useful trick: scaling of ¶
Similar to how we understood the low temperature heat capacity of the Debye model, the behavior of
Particles within an energy range of
to the Fermi energy become thermally excited, and each carries an extra energy .
Example 1: 3D free electrons¶
In 3D,
Example 2: graphene¶
In exercise 2, we will analyze graphene. Unlike in metals, electrons in graphene cannot be treated as 'free'. As a result, graphene
has a density of states
Conclusions¶
- The Sommerfeld free electron model treats electrons as free particles with energy dispersion
. - The density of states
follows from the dispersion relation by using a general procedure that is analogous to that for phonons. - The Fermi-Dirac distribution gives the probability of an electron state at energy
to be occupied. - The electron contribution to the heat capacity is proportional to
. It is much lower than phonon heat capacity at high temperatures, and much higher at low temperatures. - The scaling of heat capacity with
can be quickly estimated by estimating the number of particles in an energy range from the Fermi energy.
Exercises¶
Warm-up questions*¶
- List the differences between electrons and phonons from your memory.
- Write down the dispersion of free electrons.
- Write down an integral expression for the total energy of particles with density of states
and occupation number . - Describe what is the Fermi surface. For the free-electron model, what does the Fermi surface look like in 1D, 2D, and 3D?
- Argue if the heat capacity of a solid at temperatures near
is dominated by electrons or phonons.
Exercise 1*: Deriving the density of states for a parabolic dispersion relation.¶
In this lecture, we found that the parabolic free-electron dispersion yields a density of states
- Write down the free-electron dispersion
- What is the distance between nearest-neighbour points in
-space? Assume periodic boundary conditions. What is the density of -points in 1, 2, and 3 dimensions? - Express the number of states between energies
as an integral over k-space. Do so for 1D, 2D and 3D. Do not forget spin degeneracy. - Transform these integrals into integrals over energy for 1D, 2D and 3D. What relation do you need to do so? Indicate the integral boundaries. Extract the density of states. Are the integral boundaries important for the result?
Exercise 2*: Applying the free electron model to 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
- Check the Fermi surface database. Explain why potassium and (most) other alkali metals can be described well with the Sommerfeld model.
- Calculate the Fermi temperature, Fermi wave vector and Fermi velocity for potassium. Sketch the Fermi distribution at room temperature and indicate the role of the Fermi temperature.
- Calculate the free electron density
in potassium. - 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 3: The electron dispersion, density of states, and heat capacity of graphene¶
One of the most famous recently discovered materials is graphene. It consists of carbon atoms arranged in a 2D honeycomb structure.
Unlike in metals, electrons in graphene cannot be treated as 'free'. Instead, close to the Fermi level, the dispersion relation can be approximated by a linear relation:
- Make a sketch of the dispersion relation. Include both positive and negative energies. What other well-known particles have a linear dispersion relation?
-
Using the dispersion relation and assuming periodic boundary conditions, calculate 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 .Hint
It is convenient to first start by only considering the positive energy contributions
and calculate the density of states for it. Then account for the negative energy contributions by realizing it should have the same density of states (why?), but now for negative energies. -
At finite temperatures, assume that electrons close to the Fermi level (i.e. not more than
below the Fermi level) will get thermally excited, thereby increasing their energy by . Calculate the difference between the energy of the thermally excited state and that of the ground state . To do so, show first that the number of electrons that will get excited is (approximately) given by -
Calculate the heat capacity
as a function of the temperature .
Exercise 4: Two energy bands¶
An 'energy band' is a range of energies within which there are states available for the particles in the system. An energy band is therefore closely related to the dispersion relation. For instance, for the free electron dispersion
The dispersion of energy band 1 is
- Sketch the two dispersions in one plot. Indicate the Fermi energy.
- Calculate the density of states and sketch it as a function of energy. Hint: the total density of states is obtained by adding the density of states associated with the individual bands.
- Express the number of electrons in the system in terms of the Fermi energy
. - Express the number of electrons in the energy range
, for some energy , as an integral over energy, for . - Assuming
, explicitly calculate the integral of the previous subquestion.
-
This is not completely true, as we will see when learning about semiconductors ↩
-
An isotropic material means that the material is the same in all directions. ↩
-
The mean inter-particle distance is related to the electron density
as . The exact proportionality constant depends on the properties of the system. The Fermi wavelength sets the scale at which quantum interference effects of the electronic waves become important. In some materials (e.g. graphene) it can be on the 100 nm scale - accessible to nanofabrication techniques. Striking images of electron interference at the atomic scale are visible with a scanning tunneling microscope. ↩