---
title: "Density of states"
firstLetter: "D"
publishDate: 2021-05-08
categories:
- Physics
- Statistics

date: 2021-05-08T18:35:46+02:00
draft: false
markup: pandoc
---

# 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
1.  H. Gould, J. Tobochnik,
    *Statistical and thermal physics*, 2nd edition,
    Princeton.