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

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

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:

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:

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:

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:

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 \( k \):

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 \(W\):

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:

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.

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

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

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:

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

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

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

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:

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

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:

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 \( \mathcal{L}_N \):

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

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

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