Motivation behind definition of density of state The definition of density of state per unit volume stated in Girvin and Yang's Modern Condensed Matter Physics is $$\rho(E)=\int \frac{d^3k'}{(2\pi)^3} \delta(E-E')$$
I would like to gain more intuition on this definition. Why is it defined this way?
On a related note: naively, I would think that $\int dE \space\rho(E)$ (I am not sure what the integration bound should be) will give me the number of particles per unit volume, but it doesn't seem to be the case. For example, if I treat the free particle case $\left(E'=\frac{\hbar ^2k'^2}{2m}\right)$, I have $$\int dE \space\rho(E) =\int \frac{d^3k'}{(2\pi)^3}\int dE \space \delta(E-E')=\int \frac{d^3k'}{(2\pi)^3}E'=\int \frac{d^3k'}{(2\pi)^3}\frac{\hbar ^2k'^2}{2m}=\frac{\hbar^2}{(2\pi)^32m}\int (4\pi)k'^4 dk'=\frac{\hbar^2k^5}{5(2\pi)^2m}$$
which doesn't even seem dimensionally correct.
 A: You'll generally be interested in the density of states of a system with a non-trivial dispersion, like the bands in a crystal $\epsilon_{n}(\vec k)$. For physical intuition, I think it helps to write the density of states per unit volume a little more suggestively (G&Y eq. 7.94):
$$\rho_n(E)=\int_{BZ}\frac{d^3\vec {k'}}{(2\pi)^3}\delta\left(E-\epsilon_n(\vec {k'})\right)$$
This equation answers the question, "in the $n^{th}$ band, how many states are there per unit volume at energy $E$?" The integral probes all momenta in BZ, and when it comes across a $\vec k$ mapped to $E$ by $\epsilon_n(\vec k)$, the delta function "clicks," thereby counting the contribution of that state to the total number of states at energy $E$. In this way, you count the degeneracy at energy $E$.
Since $\rho_n(E)$ counts the states in band $n$ at a given energy $E$, integrating over $E$ and summing over $n$ gives the total number of states per unit volume.
To flesh out your example a little more, consider a free Fermi gas at $T=0$ with the dispersion you cite. We have:
$$\int dE' g_s\rho(E')f(E')=2\int dE'\int_{BZ}\frac{d^3\vec {k'}}{(2\pi)^3}\delta\left(E'-\frac{\hbar ^2k'^2}{2m}\right)\Theta(E_F-E')$$
The result of the $E'$ integral is:
$$=2\int_{BZ}\frac{d^3\vec {k'}}{(2\pi)^3}\Theta\left(E_F-\frac{\hbar ^2k'^2}{2m}\right)=2\int_{BZ}\frac{d^3\vec {k'}}{(2\pi)^3}\Theta\left(\frac{2 m E_F}{\hbar^2}-k'^2\right)$$
Assuming isotropy in $\vec k$ and defining $k_\text{max}\equiv\sqrt{\frac{2 m E_F}{\hbar^2}}$:
$$=2\int_0^{k_\text{max}}\frac{d{k'}}{(2\pi)^3} 4\pi k'^2 =\frac{k_\text{max}^3}{3 \pi^2}$$
Substituting the canonical result $E_F=\frac{\hbar^2}{2m}\left(\frac{3\pi^2N}{V}\right)^{2/3}$, we finally arrive at $\boxed{\frac{N}{V}}$, as expected.
