= Three-dimensional free-electron density of states
{title2=$g(E)=\frac{V}{2\pi^2}(2m_e/\hbar^2)^{3/2}\sqrt E$}
For the uniform noninteracting <Electron> model with two <spin> states and <kinetic energy> of <nonrelativistic particles> $E=p^2/(2m_e)$, count all available one-particle states up to an <energy> $E\geq0$:
$$
\mathcal N(E)=\frac{V}{3\pi^2}\left(\frac{2m_eE}{\hbar^2}\right)^{3/2},\qquad
g(E)=\frac{d\mathcal N}{dE}=\frac{V}{2\pi^2}\left(\frac{2m_e}{\hbar^2}\right)^{3/2}\sqrt E.
$$
This extensive <density of states> has units of inverse <energy>; $g(E)/V$ is the <density of states> per unit <volume>. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero <temperature> those states are empty. The occupied <energy> distribution is $g(E)f(E)$, with the <Fermi-Dirac distribution> reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum <integral>.
\Image[/fermi-sea-density-of-states-occupation.png]
{title=At zero temperature occupation cuts off at the Fermi energy although the density of available states keeps growing}
With the <dimensionless quantity> $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt x/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the <internal energy> and its <mean> per particle:
$$
N=\int_0^{E_F}g(E)\,dE,\qquad
U=\int_0^{E_F}Eg(E)\,dE=\frac35NE_F,\qquad
\frac UN=\frac35E_F.
$$
Thus the highest occupied <energy> and the <mean> <energy> differ. For the <Fermi energy of a model electron gas>, $U/N\simeq1.0155\,\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the <derivative> $P=-(\partial U/\partial V)_N$ gives the <electron degeneracy pressure> $P=2nE_F/5$. The same state-counting geometry therefore predicts a nonzero <pressure> at zero <temperature>.
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