---
title: The Fermi Gas Model
module: Nuclear Models
moduleNumber: 3
lessonNumber: 1
order: 301
summary: >
  Treating the nucleus as two degenerate Fermi gases of protons and neutrons
  confined in a common well fixes the Fermi momentum near 250 MeV/c and the
  Fermi energy near 33 MeV from the nuclear density alone. The average kinetic
  energy per nucleon is about 20 MeV, the well depth is the Fermi energy plus
  the separation energy, and unequal proton and neutron Fermi levels reproduce
  the asymmetry term of the mass formula.
topics: [Nuclear Models]
sources:
  - book: Krane
    ref: "Ch. 5 — Nuclear Models; §5.1 The Shell Model (Fermi gas)"
  - book: Wong
    ref: "Ch. 6 — Nuclear Bulk Properties; §6-2 The Fermi Gas Model"
  - book: Povh
    ref: "Ch. 17 — The Nuclear Force"
draft: false
---

The nucleus binds $A$ nucleons at a nearly constant density, and each nucleon
moves through the volume with only rare collisions because the exclusion
principle blocks scattering into occupied states. The simplest quantitative
model that respects both facts treats the protons and the neutrons as two
independent gases of free spin-$\tfrac12$ fermions confined to a spherical
well of volume set by the measured
[nuclear radius](/nuclear-physics/nuclear-properties/nuclear-size-charge-distributions).
The gas is fully degenerate: at the low temperatures relevant to a ground-state
nucleus every level up to a sharp Fermi surface is filled and every level above
it is empty. From the density alone this fixes the Fermi momentum, the average
kinetic energy, the depth of the confining well, and, through the difference of
the proton and neutron Fermi levels, the asymmetry term of the
[semi-empirical mass formula](/nuclear-physics/nuclear-properties/semi-empirical-mass-formula).[^krane-fg]

## Counting states in the well

A nucleon confined to a cubical box of side $L$ and volume $V = L^3$ has
plane-wave states labelled by wavevectors $\vec k = (\pi/L)(n_x, n_y, n_z)$ with
positive integers $n_i$. Each triple occupies a cell of volume $(\pi/L)^3$ in
$\vec k$-space, so the number of spatial states with wavenumber below $k$ is the
positive octant of a sphere divided by the cell,

$$
n(k) = \frac{1}{8}\cdot\frac{\tfrac{4}{3}\pi k^3}{(\pi/L)^3}
     = \frac{V k^3}{6\pi^2}.
$$

Each spatial state holds two nucleons of one kind, spin up and spin down. In
terms of momentum $p = \hbar k$, the number of neutrons filling every level up
to the Fermi momentum $p_{F}$ is

$$
N = 2\,n(k_F) = \frac{V\,p_{F,n}^3}{3\pi^2 \hbar^3},
$$

and the same relation with $Z$ and $p_{F,p}$ counts the protons. Inverting for a
single species gives the Fermi momentum directly from its number density
$\rho_i = N_i/V$,

$$
p_{F,i} = \hbar\bigl(3\pi^2 \rho_i\bigr)^{1/3}.
$$

The Fermi momentum is set by the density, not by the size or the charge of the
nucleus.

$$
% caption: Protons and neutrons fill two independent ladders of levels up to
% separate Fermi surfaces inside a common well of depth V0; a heavier neutron
% population fills to a higher Fermi level than the protons.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % common well walls
  \draw[black, very thick] (0.2,4.4) -- (0.2,0.2) -- (5.6,0.2) -- (5.6,4.4);
  \node[black, anchor=south] at (2.9,4.45) {common well, depth $V_0$};
  % proton stack: reference species, plain
  \foreach \y in {0.5,0.9,1.3,1.7,2.1} \draw[very thick] (0.7,\y) -- (2.5,\y);
  \node[black, anchor=north] at (1.6,0.4) {protons};
  \draw[black, dashed] (0.7,2.35) -- (2.5,2.35);
  \node[black, anchor=south, font=\scriptsize] at (1.6,2.35) {proton Fermi level};
  % neutron stack (fills higher)
  \fill[acc!12] (3.3,0.42) rectangle (5.1,2.98);
  \foreach \y in {0.5,0.9,1.3,1.7,2.1,2.5,2.9} \draw[very thick] (3.3,\y) -- (5.1,\y);
  \node[black, anchor=north] at (4.2,0.4) {neutrons};
  \draw[acc, dashed] (3.3,3.15) -- (5.1,3.15);
  \node[acc, anchor=south, font=\scriptsize] at (4.2,3.15) {neutron Fermi level};
  % energy axis
  \draw[->, black] (6.0,0.2) -- (6.0,4.2) node[above, black] {energy};
\end{tikzpicture}
$$

## Fermi energy and momentum for nuclear matter

Charge symmetry of the nuclear force makes symmetric matter, $N = Z = A/2$, the
reference case. Each species then has density $\rho_i = \tfrac12\rho$, where the
total density follows from packing $A$ nucleons into a sphere of radius
$R = R_0 A^{1/3}$,

$$
\rho = \frac{A}{\tfrac{4}{3}\pi R_0^3 A} = \frac{3}{4\pi R_0^3}.
$$

The density is independent of $A$, the empirical fact of nuclear saturation.
With $R_0 = 1.2\ \mathrm{fm}$ this is $\rho \approx 0.138\ \mathrm{fm^{-3}}$, so
$\rho_i \approx 0.069\ \mathrm{fm^{-3}}$ for each species. The Fermi momentum is

$$
p_{F} = \hbar\left(3\pi^2\cdot\frac{\rho}{2}\right)^{1/3}
      = \frac{\hbar}{R_0}\left(\frac{9\pi}{8}\right)^{1/3}
      \approx 250\ \mathrm{MeV}/c,
$$

using $\hbar c = 197.3\ \mathrm{MeV{\cdot}fm}$. Nucleons are non-relativistic
here, $p_F c \approx 0.27\,Mc^2$, so the Fermi energy is the kinetic energy at
the surface,

$$
E_{F} = \frac{p_F^2}{2M}
      = \frac{(p_F c)^2}{2 M c^2}
      = \frac{(250\ \mathrm{MeV})^2}{2(939\ \mathrm{MeV})}
      \approx 33\ \mathrm{MeV}.
$$

Both numbers come entirely from the saturation density: every nucleus, light or
heavy, has the same Fermi surface.

> **Worked example (Fermi energy of nuclear matter).** With
> $R_0 = 1.2\ \mathrm{fm}$,
> $$
> E_F = \frac{\hbar^2}{2M R_0^2}\left(\frac{9\pi}{8}\right)^{2/3}
>     = \frac{(197.3\ \mathrm{MeV{\cdot}fm})^2}{2(939\ \mathrm{MeV})(1.2\ \mathrm{fm})^2}
>       (3.534)^{2/3}
>     \approx 33\ \mathrm{MeV}.
> $$
> Increasing $R_0$ to $1.25\ \mathrm{fm}$ lowers $E_F$ to about
> $31\ \mathrm{MeV}$; the model is only as sharp as the assumed radius.

## The density of states

Turning the count $n(k)$ into a distribution over energy $E = p^2/2M$ gives the
number of levels per unit energy. Writing the state count for one species with
spin as $N(E) = V(2ME)^{3/2}/(3\pi^2\hbar^3)$ and differentiating,

$$
g(E) = \frac{\mathrm{d}N}{\mathrm{d}E}
     = \frac{V}{2\pi^2}\left(\frac{2M}{\hbar^2}\right)^{3/2}\!E^{1/2}
     \propto E^{1/2}.
$$

The density of states rises as $\sqrt{E}$ and is filled up to $E_F$; every state
below the Fermi energy is occupied, none above. The total kinetic energy of one
species is the first moment of this distribution,

$$
E_{\text{kin}} = \int_0^{E_F} E\,g(E)\,\mathrm{d}E
              = \frac{3}{5}N E_F,
$$

so the average kinetic energy per nucleon is

$$
\langle E\rangle = \frac{E_{\text{kin}}}{N} = \frac{3}{5}E_F \approx 20\ \mathrm{MeV}.
$$

Even in its lowest state a nucleus carries roughly $20\ \mathrm{MeV}$ of kinetic
energy per nucleon, a purely quantum zero-point motion forced by the exclusion
principle in a small volume.

$$
% caption: The single-particle density of states rises as the square root of
% energy; every level below the Fermi energy is occupied (shaded), every level
% above it is empty.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.2,0) node[right, black!70] {energy};
  \draw[->, black] (0,0) -- (0,4.2) node[above, black!70] {states per unit energy};
  % sqrt curve
  \draw[acc, very thick] (0,0)
    .. controls (1.2,1.6) and (2.6,2.4) .. (4.3,3.0)
    .. controls (5.4,3.4) and (6.2,3.65) .. (6.9,3.85);
  % Fermi cutoff at x=4.3
  \draw[black, dashed] (4.3,0) -- (4.3,3.0);
  \node[anchor=north, black!70] at (4.3,-0.03) {Fermi energy};
  % shade filled region under curve up to cutoff
  \fill[acc!14] (0,0)
    .. controls (1.2,1.6) and (2.6,2.4) .. (4.3,3.0)
    -- (4.3,0) -- cycle;
  \node[acc, anchor=center, font=\scriptsize] at (2.2,1.0) {occupied};
  \node[black, anchor=west, font=\scriptsize] at (5.0,2.3) {empty};
\end{tikzpicture}
$$

## The depth of the well

The Fermi energy is measured from the bottom of the well, not from the free-particle
zero. The most weakly bound nucleon sits at the Fermi surface, and removing it
costs the separation energy $S$, the binding energy of the last nucleon, about
$7$–$8\ \mathrm{MeV}$ across stable nuclei. The well depth is therefore the sum

$$
V_0 = E_F + S \approx 33\ \mathrm{MeV} + 7\ \mathrm{MeV} \approx 40\ \mathrm{MeV}.
$$

This $40\ \mathrm{MeV}$ is the depth of the average single-particle potential
that each nucleon moves in, in agreement with the well depths extracted from
optical-model fits to nucleon scattering. The Fermi-gas picture thus turns the
measured density and separation energy into the mean field itself.

$$
% caption: The confining well of depth V0 measured from its floor; nucleons fill
% up to the Fermi energy, and the topmost nucleon is unbound by the separation
% energy S, so that V0 equals the Fermi energy plus S.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % well
  \draw[black, very thick] (0.6,0.4) -- (0.6,3.8);
  \draw[black, very thick] (0.6,0.4) -- (5.4,0.4);
  \draw[black, very thick] (5.4,0.4) -- (5.4,3.8);
  % zero (free) level
  \draw[black, dashed] (0.2,3.6) -- (5.8,3.6);
  \node[black, anchor=west, font=\scriptsize] at (5.85,3.6) {free nucleon (zero)};
  % Fermi level
  \draw[acc, very thick] (0.6,2.9) -- (5.4,2.9);
  \node[acc, anchor=west, font=\scriptsize] at (5.85,2.9) {Fermi level};
  \fill[acc!12] (0.62,0.42) rectangle (5.38,2.88);
  % V0 bracket
  \draw[black] (0.35,0.4) -- (0.35,3.6);
  \node[black, anchor=east] at (0.32,2.0) {$V_0$};
  % E_F bracket
  \draw[acc] (3.0,0.4) -- (3.0,2.9);
  \node[acc, anchor=west, font=\scriptsize] at (3.05,1.6) {Fermi energy};
  % S bracket
  \draw[black] (2.0,2.9) -- (2.0,3.6);
  \node[black, anchor=east, font=\scriptsize] at (1.95,3.25) {$S$};
\end{tikzpicture}
$$

## The asymmetry energy

The two Fermi seas fill to different levels once $N \neq Z$. Adding neutrons
beyond $Z$ raises the neutron Fermi energy while the proton sea stays lower, so
the total kinetic energy climbs above its symmetric value. This excess is the
microscopic origin of the asymmetry term.

At fixed volume $V$, each species contributes total kinetic energy
$\tfrac35 N_i E_{F,i}$ with $E_{F,i} = C N_i^{2/3}$ and
$C = (\hbar^2/2M)(3\pi^2/V)^{2/3}$, so

$$
E_{\text{kin}}(N,Z) = \frac{3}{5}C\bigl(N^{5/3} + Z^{5/3}\bigr).
$$

Write $N = \tfrac12(A + D)$ and $Z = \tfrac12(A - D)$ with $D = N - Z$, and expand
for $D \ll A$. Using $(1+x)^{5/3} + (1-x)^{5/3} \approx 2 + \tfrac{10}{9}x^2$ with
$x = D/A$,

$$
N^{5/3} + Z^{5/3} \approx 2\left(\frac{A}{2}\right)^{5/3}
   \left[1 + \frac{5}{9}\left(\frac{D}{A}\right)^2\right].
$$

The constant term is the symmetric kinetic energy; the $D^2$ term is the
asymmetry energy. Substituting $C(A/2)^{2/3} = E_F$ collapses the prefactor,

$$
\Delta E_{\text{kin}} = \frac{3}{5}C\cdot 2\left(\frac{A}{2}\right)^{5/3}
   \cdot\frac{5}{9}\frac{D^2}{A^2}
   = \frac{E_F}{3}\,\frac{(N-Z)^2}{A}.
$$

The kinetic energy alone gives an asymmetry coefficient
$a_{\text{sym}}^{\text{kin}} = E_F/3 \approx 11\ \mathrm{MeV}$. The empirical
coefficient in the mass formula is roughly $23\ \mathrm{MeV}$, so the kinetic
Fermi-gas contribution accounts for about half. The remainder is a potential
effect: the $n$-$p$ force is more attractive than $n$-$n$ or $p$-$p$, so trading
a proton for a neutron loses attractive $n$-$p$ bonds as well as raising the
Fermi energy. Both push the same way, toward $N = Z$.[^wong-fg]

$$
% caption: The kinetic energy of the two Fermi gases is minimized at N equals Z;
% moving away in either direction raises the neutron or proton Fermi level and
% adds energy quadratically in the neutron excess.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0.4) -- (7.0,0.4) node[right, black!70] {neutron excess};
  \draw[->, black] (3.5,0.2) -- (3.5,4.2) node[above, black!70] {asymmetry energy};
  % parabola centered at x=3.5
  \draw[acc, very thick] (0.6,3.9)
    .. controls (2.0,1.4) and (2.9,0.55) .. (3.5,0.55)
    .. controls (4.1,0.55) and (5.0,1.4) .. (6.4,3.9);
  \node[black, anchor=south, font=\scriptsize] at (5.7,3.4) {rises quadratically};
  \node[black, anchor=north, font=\scriptsize] at (3.5,0.38) {$N=Z$};
\end{tikzpicture}
$$

## Reach and limits

The Fermi-gas model is the crudest mean-field picture and its successes are
therefore instructive. From nothing but the saturation density it delivers a
Fermi momentum of $250\ \mathrm{MeV}/c$, a Fermi energy of $33\ \mathrm{MeV}$, an
average kinetic energy of $20\ \mathrm{MeV}$ per nucleon, a well depth of
$40\ \mathrm{MeV}$, and half of the asymmetry coefficient. The high internal
momenta it predicts are confirmed by quasi-elastic electron scattering, where the
knocked-out nucleon carries a momentum distribution that extends to the Fermi
surface.

The model ignores everything that distinguishes one nucleus from the next. It has
no shell structure, no magic numbers, and no correlation between nucleons beyond
the exclusion principle; it treats the mean field as a structureless box. The
next refinement keeps the independent-particle idea but replaces the box with a
realistic
[central potential plus a strong spin-orbit term](/nuclear-physics/nuclear-models/shell-model-single-particle),
which quantizes the levels into shells and reproduces the magic numbers the Fermi
gas cannot see.

| Quantity | Fermi-gas prediction | Source of the number |
| --- | --- | --- |
| Fermi momentum $p_F$ | $\approx 250\ \mathrm{MeV}/c$ | saturation density |
| Fermi energy $E_F$ | $\approx 33\ \mathrm{MeV}$ | $p_F^2/2M$ |
| $\langle E\rangle$ per nucleon | $\approx 20\ \mathrm{MeV}$ | $\tfrac35 E_F$ |
| Well depth $V_0$ | $\approx 40\ \mathrm{MeV}$ | $E_F + S$ |
| Asymmetry coefficient (kinetic) | $\approx 11\ \mathrm{MeV}$ | $E_F/3$ |

[^krane-fg]: Krane, "Introductory Nuclear Physics," Wiley (1988), §5.1. Evaluated
    masses and densities from NNDC, [https://www.nndc.bnl.gov/](https://www.nndc.bnl.gov/).
[^wong-fg]: Wong, "Introductory Nuclear Physics," 2nd ed., Wiley-VCH (1998),
    §6-2; Povh et al., "Particles and Nuclei," Springer, Ch. 17.