6.3 Density of States

Up to this point, this book has presented energy levels in the form of an energy spectrum. In these spectra, each single-particle energy was shown as a tick mark along the energy axis. The single-particle states with that energy were usually listed next to the tick marks. One example was the energy spectrum of the electron in a hydrogen atom as shown in figure 4.8.

However, the number of states involved in a typical macroscopic system can easily be of the order of 10$\POW9,{20}$ or more. There is no way to show anywhere near that many energy levels in a graph. Even if printing technology was up to it, and it can only dream about it, your eyes would have only about 7 10$\POW9,{6}$ cones and 1.3 10$\POW9,{8}$ rods to see them.

For almost all practical purposes, the energy levels of a macroscopic system of noninteracting particles in a box form a continuum. That is schematically indicated by the hatching in the energy spectrum to the right in figure 6.1. The spacing between energy levels is however very many orders of magnitude tighter than the hatching can indicate.


Table 6.1: Energy of the lowest single-particle state in a cube with 1 cm sides.
\begin{table}\begin{displaymath}
\begin{array}{cccc}
\hline\hline
& \mbox{...
...1} & 0.83 \\
\hline\hline
\end{array}
\end{displaymath}
\end{table}


It can also normally be assumed that the lowest energy is zero for noninteracting particles in a box. While the lowest single particle energy is strictly speaking somewhat greater than zero, it is extremely small. That is numerically illustrated by the values for a 1 cm$\POW9,{3}$ cubic box in table 6.1. The table gives the lowest energy as computed using the formulae given in the previous section. The lowest energy occurs for the state $\pp111////$ with $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. As is common for single-particle energies, the energy has been expressed in terms of electron volts, one eV being about 1.6 10$\POW9,{-19}$ J. The table also shows the same energy in terms of an equivalent temperature, found by dividing it by 1.5 times the Boltzmann constant. These temperatures show that at room temperature, for all practical purposes the lowest energy is zero. However, at very low cryogenic temperatures, photons in the lowest energy state, or “ground state,” may have a relatively more significant energy.

The spacing between the lowest and second lowest energy is comparable to the lowest energy, and similarly negligible. It should be noted, however, that in Bose-Einstein condensation, which is discussed later, there is a macroscopic effect of the finite spacing between the lowest and second-lowest energy states, miniscule as it might be.

The next question is why quantum mechanics is needed here at all. Classical nonquantum physics too would predict a continuum of energies for the particles. And it too would predict the energy to start from zero. The energy of a noninteracting particle is all kinetic energy; classical physics has that zero if the particle is at rest and positive otherwise.

Still, the (anti) symmetrization requirements cannot be accommodated using classical physics. And there is at least one other important quantum effect. Quantum mechanics predicts that there are more single-particle states in a given energy range at high energy than at low energy.

To express that more precisely, physicists define the “density of states” as the number of single-particle states per unit energy range. For particles in a box, the density of states is not that hard to find. First, the number ${\rm d}{N}$ of single-particle states in a small wave number range from $k$ to $k+{\rm d}{k}$ is given by, {D.26},

\begin{displaymath}
{\rm d}N = {\cal V}{\cal D}_k {\,\rm d}k \qquad
{\cal D}_k = \frac{2s+1}{2\pi^2} k^2 %
\end{displaymath} (6.5)

Here ${\cal V}$ is the volume of the box that holds the particles. As you would expect, the bigger the box, the more particles it can hold, all else being the same. Similarly, the larger the wave number range ${\rm d}{k}$, the larger the number of states in it. The factor ${\cal D}_k$ is the density of states on a wave number basis. It depends on the spin $s$ of the particles; that reflects that there are $2s+1$ possible values of the spin for every given spatial state.

(It should be noted that for the above expression for ${\cal D}_k$ to be valid, the wave number range ${\rm d}{k}$ should be small. However, ${\rm d}{k}$ should still be large enough that there are a lot of states in the range ${\rm d}{k}$; otherwise ${\cal D}_k$ cannot be approximated by a simple continuous function. If the spacing ${\rm d}{k}$ truly becomes zero, ${\cal D}_k$ turns into a distribution of infinite spikes.)

To get the density of states on an energy basis, eliminate $k$ in favor of the single-particle energy ${\vphantom' E}^{\rm p}$ using ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2k^2$$\raisebox{.5pt}{$/$}$$2m$, where $m$ is the particle mass. That gives:

\begin{displaymath}
\fbox{$\displaystyle
{\rm d}N = {\cal V}{\cal D}{\,\rm d...
...m}{\hbar^2}\right)^{3/2} \sqrt{{\vphantom' E}^{\rm p}}
$} %
\end{displaymath} (6.6)

The requirements on the energy range ${\rm d}{\vphantom' E}^{\rm p}$ are like those on ${\rm d}{k}$.

The factor ${\cal D}$ is what is conventionally defined as the density of states; it is on a unit energy range and unit volume basis. In the spectrum to the right in figure 6.1, the density of states is indicated by means of the width of the spectrum.

Note that the density of states grows like $\sqrt{{\vphantom' E}^{\rm p}}$: quickly at first, more slowly later, but it continues to grow. There are more states per unit energy range at higher energy than at lower energy. And that means that at nonzero energies, the energy states are spaced many times tighter together still than the ground state spacing of table 6.1 indicates. Assuming that the energies form a continuum is an extremely accurate approximation in most cases.

The given expression for the density of states is not valid if the particle speed becomes comparable to the speed of light. In particular for photons the Planck-Einstein expression for the energy must be used, ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\omega$, where the electromagnetic frequency is $\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $ck$ with $c$ the speed of light. In addition, as mentioned in section 6.2, photons have only two independent spin states, even though their spin is 1.

It is conventional to express the density of states for photons on a frequency basis instead of an energy basis. Replacing $k$ with $\omega$$\raisebox{.5pt}{$/$}$$c$ in (6.5) and $2s+1$ by 2 gives

\begin{displaymath}
\fbox{$\displaystyle
{\rm d}N = {\cal V}{\cal D}_\omega ...
...\qquad
{\cal D}_\omega = \frac{1}{\pi^2c^3} \omega^2
$} %
\end{displaymath} (6.7)

The factor ${\cal D}_\omega$ is commonly called the “density of modes” instead of density of states on a frequency basis.


Key Points
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The spectrum of a macroscopic number of noninteracting particles in a box is practically speaking continuous.

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The lowest single-particle energy can almost always be taken to be zero.

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The density of states ${\cal D}$ is the number of single-particle states per unit energy range and unit volume.

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More precisely, the number of states in an energy range ${\rm d}{\vphantom' E}^{\rm p}$ is ${\cal V}{\cal D}{\,\rm d}{\vphantom' E}^{\rm p}$.

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To use this expression, the energy range ${\rm d}{\vphantom' E}^{\rm p}$ should be small. However, ${\rm d}{\vphantom' E}^{\rm p}$ should still be large enough that there are a lot of states in the range.

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For photons, use the density of modes.