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Density of States (1d, 2d, 3d) of a Free Electron Gas

Table of contents

Density of states (2d) Here you will learn how to derive the two-dimensional density of states of a free electron gas using the quadratic dispersion relation.
Density of states (3d) Here you will learn how to derive the three-dimensional density of states of an electron gas using the quadratic dispersion relation.
Density of states (1d) Here you will learn how to derive the one-dimensional density of states of a free electron gas from the quadratic dispersion relation.

The density of states $ D(W) $ is the derivative of the number of states $ N $ with respect to the energy $W$:

$$ \begin{align} D(W) ~=~ \frac{\text{d}N}{\text{d}W} \end{align} $$

Sometimes we consider the density of states $ g(W) $ per volume $ V $ of the crystal:

$$ \begin{align} g(W) ~=~ \frac{1}{V} \, \frac{\text{d}N}{\text{d}W} \end{align} $$

In the following, we will consider the density of states of a free electron gas that is spatially confined:

Density of state of a two-dimensional electron gas. This occurs in 2d materials, such as graphene or in the quantum Hall effect.
Density of state of a three-dimensional electron gas. This occurs, for example, in metals. The density of states can be used to determine the charge carrier density in the metal.
Density of state of a one-dimensional electron gas. 1d density of states is useful for materials where electrons can propagate in only one dimension. 1d materials include, for example, nanotubes and nanowires.

What is a density of states?

The density of states of a free electron gas indicates how many available states an electron with a certain energy can occupy.

Density of states (2d)

Allowed k-states (dots) of the free electrons in the lattice in reciprocal 2d-space.

The smallest reciprocal area $ \mathcal{A}_1 $ (in k-space) occupied by one single state is:

$$ \begin{align} \mathcal{A}_1 ~=~ \frac{(2\pi)^2}{A} \end{align} $$

Here $ A $ is the area of a real two-dimensional crystal lattice.

To find out the 2d density of states, according to the definition 1, we must first find out the number of available states $ N^{\small 2\text d} $ in the 2d electron gas. In most cases, however, the exact number $ N^{\small 2\text d} $ of states is not known. We can determine it, however, if in addition to Eq. 3 we provide the reciprocal area $ \mathcal{A}_N $ occupied by these $ N^{\small 2\text d} $ states.

Since the states are very close to each other (see the dimension of the 1st Brillouin zone), you can approximate the reciprocal area $ \mathcal{A}_N $ occupied by states $ N^{\small 2\text d} $ by a circular area. Each circle inside of this circular area is a line of constant energy. Why? Because the distance from the coordinate origin to all $k$ values on the circle line is given by the radius $ k $. The area of a circle with radius $k$ is given by the following formula:

$$ \begin{align} \mathcal{A}_N ~=~ \pi \, k^2 \end{align} $$

To find out the number $ N^{\small 2\text d} $ of states, you have to find out how often the area $ \mathcal{A}_1 $ fits into $ \mathcal{A}_N $. So divide eq. 4 by eq. 3:

$$ \begin{align} N^{\small 2\text d} &~=~ \frac{ \mathcal{A}_N }{ \mathcal{A}_1 } \\\\
&~=~ \pi \, k^2 ~\cdot~ \frac{ A }{ (2\pi)^2 } \\\\
&~=~ \frac{A}{4\pi} \, k^2 \end{align} $$

Because the electrons have a spin (spin-up and spin-down), not only one state fits into the area $ \mathcal{A}_1 $, but two! Equation 5 gives only half of the states (e.g. only spin-up states). So multiply equation 5 by 2 to consider also the other half of the spin states:

$$ \begin{align} N^{\small 2\text d} &~=~ 2 \cdot \frac{A}{4\pi} \, k^2 \\\\
&~=~ \frac{A}{2\pi} \, k^2 \end{align} $$

Remember: Our goal is to find out the density of states using the derivative 1. We have now determined the number of states $ N^{\small 2\text d}(k) $, but as a function of the wavenumber $k$. But we need a number of states $ N^{\small 2\text d}(W) $ expressed with the energy $W$. Only then we can differentiate $ N^{\small 2\text d}(W) $ with respect to the energy.

To express the number of states $ N^{\small 2\text d}(W) $ using the energy, we need to know the relation between the energy $ W $ and the wavenumber $ k $. This relation is given by the dispersion relation $ W(k) $. Depending on the system, it can be linear in $ k $, quadratic in $ k $, or have a complicated form that can be determined by experiment. We look here at the quadratic dispersion relation of a free electron gas.

For a free electron gas the following quadratic dispersion relation holds:

$$ \begin{align} W(k) ~=~ \frac{\hbar^2}{2m} \, k^2 \end{align} $$

Here $ m $ is the mass of the electron (in practice the effective mass $ m^* $) and $ \hbar $ is the reduced Planck's constant. Wavenumber $ k $ is the magnitude of the wave vector $ \boldsymbol{k} $. Rearrange the dispersion relation 7 for $ k $:

$$ \begin{align} k ~=~ \sqrt{ \frac{2m \, W}{\hbar^2} } \end{align} $$

Put Eq. 8 into Eq. 6. Thus you make the number of states $ N^{\small 2\text d} $ independent of the wavenumber $ k $, but dependent on the energy $W$:

$$ \begin{align} N^{\small 2\text d} &~=~ \frac{A}{2\pi} \, \left(\sqrt{ \frac{2m \, W}{\hbar^2} }\right)^2 \\\\
&~=~ \frac{A}{2\pi} \, \frac{2m \, W}{\hbar^2} \end{align} $$

Differentiate the number of states 9 with respect to the energy $W$ to get the density of states $ D(W) $ of a two-dimensional free electron gas:

$$ \begin{align} D(W) ~=~ \frac{A \, \class{brown}{m}}{\pi \, \hbar^2} ~=~ \text{const.} \end{align} $$

Constant density of states of a free 2d electron gas, which has a quadratic dispersion relation.

As you can see in the formula 10, the 2d density of states is independent of the energy $ W $. Thus, the 2d density of states is constant for all energies that an electron can have.

Density of states (3d)

Fermi sphere in reciprocal space with drawn in Fermi wave vector as the radius.

The smallest reciprocal volume $ \mathcal{V}_1 $ occupied by a single state is:

$$ \begin{align} \mathcal{V}_1 ~=~ \frac{(2\pi)^3}{V} \end{align} $$

The reciprocal volume of a sphere occupied by $ N^{\small 3\text d} $ states is:

$$ \begin{align} \mathcal{V}_N ~=~ \frac{4}{3} \, \pi \, k^3 \end{align} $$

Here the wavenumber $ k $ is the radius of the 3d sphere and $ V $ is the volume of a real crystal lattice.

You get the number of states analogously to the two-dimensional case by calculating the ratio of the reciprocal volumes:

$$ \begin{align} N^{\small 3\text d} &~=~ \frac{\mathcal{V}_N}{\mathcal{V}_1} \\\\
&~=~ \frac{4}{3} \, \pi \, k^3 ~\cdot~ \frac{V}{(2\pi)^3} \\\\
&~=~ \frac{V}{6\pi^2} \, k^3 \end{align} $$

Let us also take into account the spin of the electron by multiplying Eq. 13 by 2:

$$ \begin{align} N^{\small 3\text d} ~=~ \frac{V}{3\pi^2} \, k^3 \end{align} $$

Now you can substitute wavenumber 8 from the quadratic dispersion relation into Eq. 14 to express it as a function of energy $W$:

$$ \begin{align} N^{\small 3\text d} &~=~ \frac{V}{3\pi^2} \, \left( \sqrt{\frac{2m \, W}{\hbar^2} } \right)^3 \\\\
&~=~ \frac{V}{3\pi^2} \, \left( \frac{2m \, W}{\hbar^2} \right)^{3/2} \\\\
&~=~ \frac{V}{3\pi^2} \, \left( \frac{2m}{\hbar^2} \right)^{3/2} \, W^{3/2} \end{align} $$

To determine the density of states, differentiate the number of states 15 with respect to the energy $W$:

$$ \begin{align} D(W) ~=~ \frac{V}{2\pi^2} \, \left(\frac{2\class{brown}{m}}{\hbar^2}\right)^{3/2} \, \sqrt{W} \end{align} $$

Density of states for a free electron gas (quadratic dispersion relation) in three dimensions.

Sometimes you are interested in the density of states $ g(W) $ per volume. If so, then divide 16 by the volume $ V $ to get the number of states per energy AND per volume:

$$ \begin{align} g(W) ~=~ \frac{1}{2\pi^2} \, \left( \frac{2\class{brown}{m}}{\hbar^2} \right)^{3/2} \, \sqrt{W} \end{align} $$

Density of states (1d)

A single state occupies the following reciprocal length $ \mathcal{L}_1 $:

$$ \begin{align} \mathcal{L}_1 ~=~ \frac{2\pi}{L} \end{align} $$

Here $ L $ is the length of a one-dimensional system (e.g. of a quantum wire) in which there is a free electron gas.

In one-dimensional systems, $ N^{\small 1\text d} $ states take the following reciprocal length:

$$ \begin{align} \mathcal{L}_N ~=~ 2k \end{align} $$

Why exactly $ 2k $? Because $ k $ is the radius. The total reciprocal length is therefore $ 2k$! Still unclear? Draw a circle with the radius $ k $. What is then its diameter?

Divide Eq. 19 by Eq. 18 to determine the number $ N^{\small 1\text d} $ of states that fit into the length section $ \mathcal{L}_N $:

$$ \begin{align} N^{\small 1\text d} &~=~ 2k \cdot \frac{L}{2\pi} \\\\
&~=~ \frac{L}{\pi} \, k \end{align} $$

Substitute the wavenumber 8 of the quadratic dispersion relation into Eq. 20:

$$ \begin{align} N^{\small 1\text d} &~=~ \frac{L}{\pi} \, k \\\\
&~=~ \frac{L}{\pi} \, \sqrt{ \frac{2m \, W}{\hbar^2} } \\\\
&~=~ \frac{L}{\pi} \, \left(\frac{2m}{\hbar^2}\right)^{1/2} \, W^{1/2} \end{align} $$

Differentiate the number of states 21 with respect to the energy $W$ to get the 1d density of states $ D(W) $:

$$ \begin{align} D^{\small 1\text d}(W) ~=~ \frac{L}{2\pi} \, \left(\frac{2m}{\hbar^2}\right)^{1/2} \, \frac{1}{\sqrt{W}} \end{align} $$

If the spin is to be taken into account, we have to multiply Eq. 22 by 2:

$$ \begin{align} D(W) ~=~ \frac{L}{\pi} \, \left(\frac{2\class{brown}{m}}{\hbar^2}\right)^{1/2} \, \frac{1}{\sqrt{W}} \end{align} $$

Density of states of a 1d electron gas decreases with electron energy.

Now you have learned what the density of states is and how to find it for a free electron gas in different dimensions. In further lessons, we will look at how the density of states can be used to determine the charge carrier density and heat capacity of a material.