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

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  • $\begingroup$ Would you mind defining what $\epsilon_n(\mathbf k)$ means in your notation? Are these the energy levels of some system? $\endgroup$ – joshphysics Feb 27 '13 at 2:15
  • $\begingroup$ Yes, $\epsilon_n(\mathbf{k})$ is the energy eigenvalue. $\endgroup$ – Timothy Feb 27 '13 at 2:30
  • $\begingroup$ Hmm ok well then I'm a bit confused by your expression. Let's consider a quantum system with Hamiltonian having a purely discrete spectrum $\{\epsilon_n\}$, then I would think that the density of states would be $D(E) = \sum_n \delta(E - \epsilon_n)$ since then $\int_{E_1}^{E_2} dE D(E)$ would give the number of states in the interval $(E_1, E_2)$ and that is what the density of states is supposed to do. $\endgroup$ – joshphysics Feb 27 '13 at 2:39
  • $\begingroup$ Also your expression doesn't seem to have the right units. Perhaps I'm just going crazy. $\endgroup$ – joshphysics Feb 27 '13 at 3:41
  • $\begingroup$ @joshphysics: You're right, the formula needs a multiplication with $V$, the system volume. $\endgroup$ – Rafael Reiter Feb 27 '13 at 20:28
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OP's equality involving a delta function is probably easier to appreciate in its equivalent integrated form

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

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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, $$ =\frac{1}{(2\pi)^3}4\pi k'^2\frac{2m}{\hbar^2}\frac{1}{2k'} =\frac{m}{2\pi^2\hbar^2} k' =\frac{m}{2\pi^2\hbar^2} \frac{\sqrt{2mE'}}{\hbar} $$

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