s: protons (red) and neutrons (blue). As a first approximation, the nucleus can be treated as composed of non-interacting proton and neutron gases. The three-dimensional
isotropic and non-
relativistic uniform Fermi gas case is known as the
Fermi sphere. A three-dimensional infinite square well, (i.e. a cubical box that has a side length
L) has the potential energy V(x,y,z) = \begin{cases} 0, & -\frac{L}{2} The states are now labelled by three quantum numbers
nx,
ny, and
nz. The single particle energies are E_{n_x,n_y,n_z} = E_0 + \frac{\hbar^2 \pi^2}{2m L^2} \left( n_x^2 + n_y^2 + n_z^2\right) \,, where
nx,
ny,
nz are positive integers. In this case, multiple states have the same energy (known as
degenerate energy levels), for example E_{211}=E_{121}=E_{112}.
Thermodynamic limit When the box contains
N non-interacting fermions of spin-, it is interesting to calculate the energy in the thermodynamic limit, where
N is so large that the quantum numbers
nx,
ny,
nz can be treated as continuous variables. With the vector \mathbf{n}=(n_x,n_y,n_z), each quantum state corresponds to a point in 'n-space' with energy E_{\mathbf{n}} = E_0 + \frac{\hbar^2 \pi^2}{2m L^2} |\mathbf{n}|^2 \, With |\mathbf{n}|^2 denoting the square of the usual Euclidean length \left|\mathbf{n}\right| = \sqrt{n_x^2+n_y^2+n_z^2} . The number of states with energy less than
EF +
E0 is equal to the number of states that lie within a sphere of radius |\mathbf{n}_{\mathrm{F}}| in the region of n-space where
nx,
ny,
nz are positive. In the ground state this number equals the number of fermions in the system: N =2\times\frac{1}{8}\times\frac{4}{3} \pi n_{\mathrm{F}}^3 in
reciprocal space. The surface of this sphere is the
Fermi surface. The factor of two expresses the two spin states, and the factor of 1/8 expresses the fraction of the sphere that lies in the region where all
n are positive. n_{\mathrm{F}}=\left(\frac{3 N}{\pi}\right)^{1/3} The
Fermi energy is given by E_{\mathrm{F}} = \frac{\hbar^2 \pi^2}{2m L^2} n_{\mathrm{F}}^2 = \frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{3 N}{\pi} \right)^{2/3} Which results in a relationship between the Fermi energy and the
number of particles per volume (when
L2 is replaced with
V2/3): :{{box|border color=#ccccff|E_{\mathrm{F}} = \frac{\hbar^2}{2m} \left( \frac{3 \pi^2 N}{V} \right)^{2/3} }} This is also the energy of the highest-energy particle (the Nth particle), above the zero point energy E_0. The N'th particle has an energy of E_{N'} = E_0 + \frac{\hbar^2}{2m} \left( \frac{3 \pi^2 N'}{V} \right)^{2/3} \,=E_0 + E_{\mathrm{F}} \big |_{N'} The total energy of a Fermi sphere of N fermions (which occupy all N energy states within the Fermi sphere) is given by: E_{\rm T} = N E_0 + \int_0^N E_{\mathrm{F}}\big |_{N'} \, dN' = \left(\frac{3}{5} E_{\mathrm{F}} + E_0\right)N Therefore, the average energy per particle is given by: E_\mathrm{av} = E_0 + \frac{3}{5} E_{\mathrm{F}}
Density of states For the 3D uniform Fermi gas, with fermions of spin-, the number of particles as a function of the energy N(E) is obtained by substituting the Fermi energy by a variable energy (E-E_0): N(E)=\frac{V}{3\pi^2}\left[\frac{2m}{\hbar^2}(E-E_0)\right]^{3/2}, from which the
density of states (number of energy states per energy per volume) g(E) can be obtained. It can be calculated by differentiating the number of particles with respect to the energy: g(E) =\frac{1}{V}\frac{\partial N(E)}{\partial E}= \frac {1}{2\pi^2} \left(\frac {2m}{\hbar^2}\right)^{3/2}\sqrt{E-E_0}. This result provides an alternative way to calculate the total energy of a Fermi sphere of N fermions (which occupy all N energy states within the Fermi sphere): \begin{align} E_T &= \int_0^N E \, \mathrm{d} N(E) = E N(E)\big |_0^N-\int_{E_0}^{E_0+E_F} N(E) \, \mathrm{d} E \\ &=(E_0+E_F) N - \int_{0}^{E_F} N(E) \, \mathrm{d} (E-E_0) \\ &=(E_0+E_F)N - \frac{2}{5}E_FN(E_F) = \left(E_0 + \frac{3}{5} E_{\mathrm{F}}\right)N \end{align}
Thermodynamic quantities Degeneracy pressure ) in three dimensions. Pauli repulsion in fermions (such as electrons) gives them an additional pressure over an equivalent classical gas, most significantly at low temperature. By using the
first law of thermodynamics, this internal energy can be expressed as a pressure, that is P = -\frac{\partial E_{\rm T}}{\partial V} = \frac{2}{5}\frac{N}{V}E_{\mathrm{F}}= \frac{(3\pi^2)^{2/3}\hbar^2}{5m}\left(\frac{N}{V}\right)^{5/3}, where this expression remains valid for temperatures much smaller than the Fermi temperature. This pressure is known as the
degeneracy pressure. In this sense, systems composed of fermions are also referred as
degenerate matter. Standard
stars avoid collapse by balancing thermal pressure (
plasma and radiation) against gravitational forces. At the end of the star lifetime, when thermal processes are weaker, some stars may become white dwarfs, which are only sustained against gravity by
electron degeneracy pressure. Using the Fermi gas as a model, it is possible to calculate the
Chandrasekhar limit, i.e. the maximum mass any star may acquire (without significant thermally generated pressure) before collapsing into a black hole or a neutron star. The latter, is a star mainly composed of neutrons, where the collapse is also avoided by neutron degeneracy pressure. For the case of metals, the electron degeneracy pressure contributes to the compressibility or
bulk modulus of the material.
Chemical potential ) in three dimensions. Assuming that the concentration of fermions does not change with temperature, the total chemical potential
μ (Fermi level) of the three-dimensional ideal Fermi gas is related to the zero temperature Fermi energy
EF by a
Sommerfeld expansion (assuming k_{\rm B}T \ll E_{\mathrm{F}}): \mu(T) = E_0 + E_{\mathrm{F}} \left[ 1- \frac{\pi ^2}{12} \left(\frac{k_{\rm B}T}{E_{\mathrm{F}}}\right) ^2 - \frac{\pi^4}{80} \left(\frac{k_{\rm B}T}{E_{\mathrm{F}}}\right)^4 + \cdots \right], where
T is the
temperature. Hence, the
internal chemical potential,
μ-
E0, is approximately equal to the Fermi energy at temperatures that are much lower than the characteristic Fermi temperature
TF. This characteristic temperature is on the order of 105
K for a metal, hence at room temperature (300 K), the Fermi energy and internal chemical potential are essentially equivalent.
Typical values Metals Under the
free electron model, the electrons in a metal can be considered to form a uniform Fermi gas. The number density N/V of conduction electrons in metals ranges between approximately 1028 and 1029 electrons per m3, which is also the typical density of atoms in ordinary solid matter. This number density produces a Fermi energy of the order: E_{\mathrm{F}} = \frac{\hbar^2}{2m_e} \left( 3 \pi^2 \ 10^{28 \ \sim \ 29} \ \mathrm{m^{-3}} \right)^{2/3} \approx 2 \ \sim \ 10 \ \mathrm{eV}, where
me is the
electron rest mass. This Fermi energy corresponds to a Fermi temperature of the order of 106 kelvins, much higher than the temperature of the
Sun's surface. Any metal will boil before reaching this temperature under atmospheric pressure. Thus for any practical purpose, free electrons in a metal can be considered as a Fermi gas at zero temperature as an approximation (normal temperatures are small compared to
TF).
White dwarfs Stars known as
white dwarfs have mass comparable to the
Sun, but have about a hundredth of its radius. The high densities mean that the electrons are no longer bound to single nuclei and instead form a degenerate electron gas. The number density of electrons in a white dwarf is of the order of 1036 electrons/m3. This means their Fermi energy is: E_{\mathrm{F}} = \frac{\hbar^2}{2m_e} \left( \frac{3 \pi^2 (10^{36})}{1 \ \mathrm{m^3}} \right)^{2/3} \approx 3 \times 10^5 \ \mathrm{eV} = 0.3 \ \mathrm{MeV}
Nucleus Another typical example is that of the particles in a nucleus of an atom. The
radius of the nucleus is roughly: R = \left(1.25 \times 10^{-15} \mathrm{m} \right) \times A^{1/3} where
A is the number of
nucleons. The number density of nucleons in a nucleus is therefore: \rho = \frac{A}{ \frac{4}{3} \pi R^3} \approx 1.2 \times 10^{44} \ \mathrm{m^{-3}} This density must be divided by two, because the Fermi energy only applies to fermions of the same type. The presence of
neutrons does not affect the Fermi energy of the
protons in the nucleus, and vice versa. The Fermi energy of a nucleus is approximately: E_{\mathrm{F}} = \frac{\hbar^2}{2m_{\rm p}} \left( \frac{3 \pi^2 (6 \times 10^{43})}{1 \ \mathrm{m}^3} \right)^{2/3} \approx 3 \times 10^7 \ \mathrm{eV} = 30 \ \mathrm{MeV} , where
mp is the proton mass. The
radius of the nucleus admits deviations around the value mentioned above, so a typical value for the Fermi energy is usually given as 38
MeV. == Arbitrary-dimensional uniform gas ==