# The definition of Density of States

The density of states (DOS) is generally defined as $$D(E)=\frac{d\Omega(E)}{dE},$$ where $$\Omega(E)$$ is the number of states in a volume $$V$$. But why DOS can also be defined using delta function, as $$D(E)~=~\sum\limits_{n} \int \frac{d^3k}{(2\pi)^3}\delta(E-\epsilon_n(\mathbf{k}))~?$$

• Yes, $\epsilon_n(\mathbf{k})$ is the energy eigenvalue. Commented Feb 27, 2013 at 2:30

We can use both definitions to determine the density of state, $D(E')$, for a free particle of mass $m$ confined to a volume $V$ with energy $E'$.

The free particle has wave vector $\vec{k}'$ and energy $E'=\frac{\hbar^2k'^2}{2m}$.

$\Omega(E')$ is defined to be the number of states with energy $E<E'$ over the total position-space volume, $L^3$. In $k$-space, this is the $k$-space volume of the sphere of radius $k'$ divided by the $k$-space volume of the cell taken up by one state in discretized $k$-space, or $\Delta k^3=\left(\frac{2\pi}{L}\right)^3$.

$$\Omega(E')=\frac{1}{L^3}\frac{\frac{4}{3}\pi k'^3}{\left(\frac{2\pi}{L}\right)^3} =\frac{k'^3}{6\pi^2}$$ Therefore, by your first definition, $$D(E')=\frac{d\Omega(E')}{dE'}=\frac{1}{2\pi^2}k'^2\frac{dk'}{dE'}$$ with $$dE'=\frac{\hbar^2k'dk'}{m}\Rightarrow \frac{dk'}{dE'}=\frac{m}{\hbar^2k'}$$ $$D(E')=\frac{1}{2\pi^2}k'^2\frac{m}{\hbar^2k'}=\frac{m}{2\pi^2\hbar^2}k' =\frac{m}{2\pi^2\hbar^2}\frac{\sqrt{2mE'}}{\hbar}$$

From your second definition, $$D(E')=\sum_n\int\frac{d^3k}{(2\pi)^3}\delta(E'-E_n(\vec{k}))$$ Since we are only considering states in the continuum, we do not need to sum over $n$ (though if, say you are considering electrons with degenerate spin states, this summation would account for this degeneracy).

In inelastic scattering problems, in which the density of states appears in Fermi's Golden Rule, you might have $\vec{k}$ states representing the free particle alongside bound $n$ states of atoms, in which case you would need both the integration over $k$ as well as summation over $n$. A state denoted $|\vec{k},n\rangle=|\vec{k}\rangle|n\rangle$ would have energy $E_n(\vec{k})=\frac{\hbar^2k^2}{2m}+E_n$.

The subspace in 3D $k$-space which gives a nonzero value for the integral is the spherical shell with radius $k'$. Converting to spherical coordinates,

$$D(E')=\frac{1}{(2\pi)^3}\int d\Omega\int_0^\infty dk k^2\delta\left(\frac{\hbar^2k'^2}{2m}-\frac{\hbar^2k^2}{2m}\right)$$ integrating over $d\Omega$ and rescaling the delta-function, $$=\frac{1}{(2\pi)^3}4\pi\int_0^\infty dk k^2\frac{2m}{\hbar^2}\delta\left(k'^2-k^2\right)$$ using a delta function identity, $$=\frac{1}{(2\pi)^3}4\pi\int_0^\infty dk k^2\frac{2m}{\hbar^2}\frac{1}{2k'}[\delta(k+k')+\delta(k-k')]$$ and noting only one of the delta functions contributes to a nonzero integral, $$=\frac{1}{(2\pi)^3}4\pi k'^2\frac{2m}{\hbar^2}\frac{1}{2k'} =\frac{m}{2\pi^2\hbar^2} k' =\frac{m}{2\pi^2\hbar^2} \frac{\sqrt{2mE'}}{\hbar}$$

OP's equality involving a delta function is probably easier to appreciate in its equivalent integrated form

\begin{align}\int \!d\Omega(E) ~f(E) ~=~ &\int \!dE~D(E) ~f(E)\cr ~=~&\sum_{n} \int \frac{d^3k}{(2\pi)^3}f(\epsilon_n(\mathbf{k})),\end{align} where $$f(E)$$ is an arbitrary function.

• Is there any intuition behind using the dirac-delta function in the micro-canonical partition function in the first place (omega)? Or is it just a mathematically convenient way to count states (and if the latter is true, I'm struggling to convince myself that's what it does). For reference I'm talking about this particular formulation of it: $$\Omega(E)=\int{d^{3N}}q\int{d^{3N}p}\ \delta(E-H(\{\vec{q_i},\vec{p_i}\}))$$ Edit: might be worth mentioning that my understanding of the dirac-delta function comes from a Systems and Signals course (EE), so may be a gap there for me Commented May 24, 2021 at 16:52