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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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Would you mind defining what $\epsilon_n(\mathbf k)$ means in your notation? Are these the energy levels of some system? –  joshphysics Feb 27 '13 at 2:15
    
Yes, $\epsilon_n(\mathbf{k})$ is the energy eigenvalue. –  Jeremy Feb 27 '13 at 2:30
    
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. –  joshphysics Feb 27 '13 at 2:39
    
Also your expression doesn't seem to have the right units. Perhaps I'm just going crazy. –  joshphysics Feb 27 '13 at 3:41
    
@joshphysics: You're right, the formula needs a multiplication with $V$, the system volume. –  Rafael Reiter Feb 27 '13 at 20:28
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1 Answer 1

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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