1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
|
---
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.
2. B. Van Zeghbroeck,
[Principles of semiconductor devices](https://ecee.colorado.edu/~bart/book/book/chapter2/ch2_4.htm), 2011,
University of Colorado.
|
