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\begin{document}
\title{The Density of States (DoS):\\ Mode Counting vs. Phase Space}
\author{Yosun Chang}
\maketitle 
\date 

In its practical usage, the density of states replaces the degeneracy (in the continuum limit) in the appropriate partition function. 

The (usual definition of the) density of states (DoS) is the number of states in the range $E$ to $E+dE$. Alternate definitions 
may use, for example, $\omega$ to $\omega+d\omega$, where frequency is a more prudent parameter to use. 

Define $N$ as the total number of states or modes, and also define: 
\begin{equation}
G:=\frac{number\; of \; states}{unit \; volume}=\frac{N}{V}
\end{equation}
and let $g$ denote the DoS, defined as: 
\begin{equation}
g=\frac{dG}{dE}=number\;of\;states\;in\;E\;to\;E+dE
\end{equation}

Note that per unit volume refers to the spacing between intervals or points, and thus may be a length or area in lower dimensions. 

There are, in general, two equivalent ways to calculating the density of state. 

Method one involves counting modes (i.e., the number 
of states). After deducing $G$, one substitute the dispersion relation to obtain $G=G(E)$, then one takes the derivative 
with respect to $E$ to get $g$, the DoS. Note that in this version, one has to divide G by the number of quadrants, since 
only the energy values in the first quadrant are of matter. 

Method two involves calculating the momentum integral in phase space. $n$ is the dimension number.
\begin{equation}
G=\frac{1}{h^n}\int d^n p,
\end{equation} whence after finding $G$, one subs in the dispersion relation, then differentiate with respect to $E$ to obtain $g$.

 Both methods are easy.  Both utilize the definitions 
of $G$ (total number of states per unit volume) and $g$ above (number of states ver unit volume in a thin shell of energy). 

\section{Non-relativistic particle in a box:}
The non-relativistic particle in a box has the following dispersion relation: 

\begin{equation}
E=\frac{\hbar^2 k^2}{2m},
\end{equation}
where $p=\hbar k$, and $k$ is the wave number. Solving for $k$ in terms of $E$, one gets: 
\begin{equation}
k=\sqrt{\frac{2mE}{\hbar^2}} \Rightarrow \frac{dk}{dE}=\sqrt{\frac{m}{2\hbar^2 E}}
\end{equation}

A factor of $2S+1$ multiplies the density of state, where $S$ is the spin of the particle.

Values of the DoS varies depending on dimension. The following examples should make things crystal clear:

\subsection{1 Dimension} Box of Size $L$:
\subsubsection{Mode Counting}
\begin{equation}
N(k)=\frac{kL}{\pi}=\frac{length\;of\;line\;in\;k\;space}{length\;of\;interval \;between \;points}
\end{equation}
\begin{equation}
G=\frac{N}{L}=\frac{k}{\pi}=\sqrt{\frac{2mE}{\hbar^2}}\frac{1}{\pi}
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{d}{dE}\left(\sqrt{\frac{2mE}{\hbar^2}}\frac{1}{\pi}\right)=\sqrt{\frac{2m}{\hbar^2 E}}\frac{1}{\pi}
\end{equation}
\subsubsection{Phase Space}
\begin{equation}
G=\frac{1}{h}\int dp = \frac{1}{h}p = \frac{1}{h}\hbar k = \frac{1}{2\pi}\sqrt{\frac{2mE}{\hbar^2}}=\frac{1}{\pi}\sqrt{\frac{mE}{2\hbar^2}}
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{d}{dE}\left(\sqrt{\frac{2mE}{\hbar^2}}\frac{1}{\pi}\right)=\sqrt{\frac{2m}{\hbar^2 E}}\frac{1}{\pi}
\end{equation}
\subsubsection{General Comments}
Note that in the above, the dispersion relation $k=\sqrt{\frac{2mE}{\hbar^2}}$ has been used in glorious abundance in order to 
change from $G(k)$ to $G(E)$. Finally, note that both mode counting and phase space approaches yield the same value for both the number of states $G$ 
and the 
density of state $g$. The phase space approach does not require worrying about multiplying a factor to reduce the calculation to that 
of just one quadrant, while in the mode counting approach, one must be wary of this, lest one is off by a non-trivial factor. 

\subsection{2 Dimension} Box of Width $L_x$ and Length $L_y$: 
\subsubsection{Mode Counting}
\begin{equation}
N(k)=\frac{\pi k^2}{4}\frac{L_x L_y}{\pi^2}=\frac{area\;of\;quarter\;circle\;in\;k\;space}{area\;per\;point}
\end{equation}
\begin{equation}
G=\frac{k^2}{4\pi}=\frac{1}{4\pi}\frac{2mE}{\hbar^2}
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{d}{dE}\left(\frac{1}{4\pi}\frac{2mE}{\hbar^2}\right)=\frac{1}{4\pi}\frac{2m}{\hbar^2}
\end{equation}
\subsubsection{Phase Space}
\begin{equation}
G=\frac{1}{h^2}\int d^2 p=\frac{1}{h^2} \int p dp \int_0^{2\pi} d\theta=\frac{1}{h^2}\pi p^2=\frac{\pi \hbar^2}{h^2}k^2=\frac{1}{4\pi}\frac{2mE}{\hbar^2}
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{d}{dE}\left(\frac{1}{4\pi}\frac{2mE}{\hbar^2}\right)=\frac{1}{4\pi}\frac{2m}{\hbar^2}
\end{equation}

Indeed, both $G$ and $g$ are the same for both methods.\footnote{Rather obvious, if you realize that the only difference between $G$ and $g$ is the derivative with respect to $E$.} From now on, $g$ will be calculated just once. 

\subsection{3 Dimension}
\subsubsection{Mode Counting}
\begin{equation}
N(k)=\frac{1}{8}\frac{4\pi k^3}{3}\frac{L_x L_y L_z}{\pi^3}=\frac{volume\;of\;quadrant\;of\;sphere}{volume\;per\;point}
\end{equation}
\begin{equation}
G=\frac{N}{L_x L_y L_z}=\frac{k^2}{6}\frac{1}{\pi^2} =\frac{1}{6}\left(\frac{2mE}{\hbar^2}\right)^{3/2}\frac{1}{\pi^2}
\end{equation}
\subsubsection{Phase Space}
\begin{equation}
G=\frac{1}{h^3}\int d^3 p=\frac{4\pi}{h^3}\int p^2 dp=\frac{4\pi}{h^3}\int(\hbar k)^2\hbar dk=\frac{1}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\int E^{1/2}dE
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{1}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2} E^{1/2}
\end{equation}
\subsection{n Dimension}
\subsubsection{Mode Counting}
\subsubsection{Phase Space}

\section{Relativistic particle in a box:}
The dispersion relation for the relativistic particle is elegant: 
\begin{equation}
E=pc=\hbar k c,
\end{equation}
where $p=\hbar k$, as usual. 
\subsection{1 Dimension}
\begin{equation}
G(k)=\frac{1}{h}\int dp=\frac{p}{h}=\frac{E}{hc}=\frac{E}{2\pi \hbar c}
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{1}{2\pi \hbar c}
\end{equation}
\subsection{2 Dimension}
\begin{equation}
G(k)=\frac{1}{h^2}\int d^2 p = \frac{1}{h^2}\int p dp d\theta = \frac{\pi p^2}{h^2}=\frac{\pi}{h^2}\left(\frac{E}{c}\right)^2
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{2\pi E}{h^2}\left(\frac{1}{c}\right)^2
\end{equation}
\subsection{3 Dimension}
\begin{equation}
G(k)=\frac{1}{h^3}\int d^3 p = \frac{4\pi}{h^3}\int p^2 dp= \frac{4\pi}{3 h^3}\left(\frac{E}{c}\right)^3
\end{equation}
\begin{equation}
g=\frac{dG}{dE}=\frac{1}{h^3}\int d^3 p = \frac{4\pi}{h^3}\int p^2 dp= \frac{4\pi E^2}{ h^3}\left(\frac{1}{c}\right)^3
\end{equation}

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