From 4780106a4f191c41d3b82ca9d1327a1c95a72055 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 27 May 2021 20:46:01 +0200
Subject: Expand knowledge base

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+---
+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.
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