= Fermi momentum
{c}
{title2=$p_F$}
Fermi momentum is the magnitude of the <momentum> at the boundary of filled one-particle states in an isotropic zero-temperature <Fermi gas>. For the uniform three-dimensional two-component <electron> model, a short state count gives
$$
N\simeq 2\,\frac{V}{(2\pi\hbar)^3}\,\frac{4\pi p_F^3}{3},\qquad
n=\frac{p_F^3}{3\pi^2\hbar^3},\qquad
p_F=\hbar(3\pi^2n)^{1/3}.
$$
To see the factors, impose <periodic boundary conditions> on a cubic box of side $L$, so $V=L^3$ and each <momentum> component is $p_i=2\pi\hbar j_i/L$ for an <integer> $j_i$. Each allowed <momentum> point occupies a cell of <volume> $(2\pi\hbar/L)^3$ in <momentum space>. Divide the <sphere> <volume> $4\pi p_F^3/3$ by this cell <volume> and multiply by two for the two <spin> states. Replacing the discrete count with a <volume> is the large-box approximation; surface and shell effects are neglected.<Statistical Physics by David Tong/3 Quantum Gases>[Statistical Physics by David Tong; §3.1, periodic wavevectors and momentum-space integration measure; §3.6.1, spin degeneracy]{ref}
\Image[/fermi-momentum-sphere-state-count.png]
{title=The occupied momentum sphere counts two spin states per momentum point; a slice shows the state lattice}
The factor two is not automatic for every <Fermi gas>. If only one internal state is populated, its <number density> $n_\sigma$ instead satisfies $p_{F,\sigma}=\hbar(6\pi^2n_\sigma)^{1/3}$. For two equally populated states, $n_\sigma=n/2$ recovers the displayed result. A <Fermi surface> in a crystal can have a more complicated shape, so a single spherical <Fermi momentum> is a property of this isotropic model.
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