Categories: Physics, Statistics.

# Density of states

The **density of states** $g(E)$ of a physical system is defined such that
$g(E) \dd{E}$ is the number of states which could be occupied
with an energy in the interval $[E, E + \dd{E}]$.
In fact, $E$ need not be an energy;
it should just be something that effectively identifies the state.

In its simplest form, the density of states is as follows, where $\Gamma(E)$ is the number of states with energy less than or equal to the argument $E$:

$\begin{aligned} g(E) = \dv{\Gamma}{E} \end{aligned}$If the states can be treated as waves, which is often the case, then we can calculate the density of states $g(k)$ in $k$-space, i.e. as a function of the wavenumber $k = |\vb{k}|$. Once we have $g(k)$, we use the dispersion relation $E(k)$ to find $g(E)$, by demanding that:

$\begin{aligned} g(k) \dd{k} = g(E) \dd{E} \quad \implies \quad g(E) = g(k) \dv{k}{E} \end{aligned}$Inverting the dispersion relation $E(k)$ to get $k(E)$ might be difficult, in which case the left-hand equation can be satisfied numerically.

Define $\Omega_n(k)$ as the number of states with a $k$-value less than or equal to the argument, or in other words, the volume of a hypersphere with radius $k$. Then the $n$-dimensional density of states $g_n(k)$ has the following general form:

$\begin{aligned} \boxed{ g_n(k) = \frac{D}{2^n k_{\mathrm{min}}^n} \: \dv{\Omega_n}{k} } \end{aligned}$Where $D$ is each state’s degeneracy (e.g. due to spin), and $k_{\mathrm{min}}$ is the smallest allowed $k$-value, according to the characteristic length $L$ of the system. We divide by $2^n$ to limit ourselves to the sector where all axes are positive, because we are only considering the magnitude of $k$.

In one dimension $n = 1$, the number of states within a distance $k$ from the origin is the distance from $k$ to $-k$ (we let it run negative, since its meaning does not matter here), given by:

$\begin{aligned} \Omega_1(k) = 2 k \end{aligned}$To get $k_{\mathrm{min}}$, we choose to look at a rod of length $L$, across which the function is a standing wave, meaning that the allowed values of $k$ must be as follows, where $m \in \mathbb{N}$:

$\begin{aligned} \lambda = \frac{2 L}{m} \quad \implies \quad k = \frac{2 \pi}{\lambda} = \frac{m \pi}{L} \end{aligned}$Take the smallest option $m = 1$, such that $k_{\mathrm{min}} = \pi / L$, the 1D density of states $g_1(k)$ is:

$\begin{aligned} \boxed{ g_1(k) = \frac{D L}{2 \pi} \: 2 = \frac{D L}{\pi} } \end{aligned}$In 2D, the number of states within a range $k$ of the origin is the area of a circle with radius $k$:

$\begin{aligned} \Omega_2(k) = \pi k^2 \end{aligned}$Analogously to the 1D case, we take the system to be a square of side $L$, so $k_{\mathrm{min}} = \pi / L$ again. The density of states then becomes:

$\begin{aligned} \boxed{ g_2(k) = \frac{D L^2}{4 \pi^2} \:2 \pi k = \frac{D L^2 k}{2 \pi} } \end{aligned}$In 3D, the number of states is the volume of a sphere with radius $k$:

$\begin{aligned} \Omega_3(k) = \frac{4 \pi}{3} k^3 \end{aligned}$For a cube with side $L$, we once again find $k_{\mathrm{min}} = \pi / L$. We thus get:

$\begin{aligned} \boxed{ g_3(k) = \frac{D L^3}{8 \pi^3} \:4 \pi k^2 = \frac{D L^3 k^2}{2 \pi^2} } \end{aligned}$All these expressions contain the characteristic length/area/volume $L^n$, and therefore give the number of states in that region only. Keep in mind that $L$ is free to choose; it need not be the physical size of the system. In fact, we typically want the density of states per unit length/area/volume, so we can just set $L = 1$ in our preferred unit of distance.

If the system is infinitely large, or if it has periodic boundaries, then $k$ becomes a continuous variable and $k_\mathrm{min} \to 0$. But again, $L$ is arbitrary, so a finite value can be chosen.

## References

- H. Gould, J. Tobochnik,
*Statistical and thermal physics*, 2nd edition, Princeton. - B. Van Zeghbroeck, Principles of semiconductor devices, 2011, University of Colorado.