Thermodynamic limit

In the thermodynamic limit, the total number of particles N are so large that the quantum number n may be treated as a continuous variable. In this case, the overall number density profile in the box is indeed uniform.

The number of quantum states in the range ${\displaystyle n_{1}<n<n_{1}+dn}$ is:

$${\displaystyle D_{n}(n_{1})\,dn=2\,dn\,.}$$

Without loss of generality, the zero-point energy is chosen to be zero, with the following result:

$${\displaystyle E_{n}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}n^{2}\implies dE={\frac {\hbar ^{2}\pi ^{2}}{mL^{2}}}n\,dn={\frac {\hbar \pi }{L}}{\sqrt {\frac {2E}{m}}}dn\,.}$$

Therefore, in the range:

$${\displaystyle E_{1}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}n_{1}^{2}<E<E_{1}+dE\,,}$$

the number of quantum states is:

$${\displaystyle D_{n}(n_{1})\,dn=2{\frac {dE}{dE/dn}}={\frac {2}{{\frac {\hbar ^{2}\pi ^{2}}{mL^{2}}}n}}\,dE\equiv D(E_{1})\,dE\,.}$$

Here, the degree of degeneracy is:

$${\displaystyle D(E)={\frac {2}{dE/dn}}={\frac {2L}{\hbar \pi }}{\sqrt {\frac {m}{2E}}}\,.}$$

And the density of states is:

$${\displaystyle g(E)\equiv {\frac {1}{L}}D(E)={\frac {2}{\hbar \pi }}{\sqrt {\frac {m}{2E}}}\,.}$$

In modern literature,^[5] the above ${\displaystyle D(E)}$ is sometimes also called the "density of states". However, ${\displaystyle g(E)}$ differs from ${\displaystyle D(E)}$ by a factor of the system's volume (which is ${\displaystyle L}$ in this 1D case).

Based on the following formula:

$${\displaystyle \int _{0}^{E_{\mathrm {F} }}D(E)\,dE=N\,,}$$

the Fermi energy in the thermodynamic limit can be calculated to be:

$${\displaystyle E_{\mathrm {F} }^{({\text{1D}})}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}\left({\frac {N}{2}}\right)^{2}\,.}$$

Thermodynamic limit

When the box contains N non-interacting fermions of spin ½, it is interesting to calculate the energy in the thermodynamic limit, where N is so large that the quantum numbers n_x, n_y, n_z can be treated as continuous variables.

With the vector ${\displaystyle \mathbf {n} =(n_{x},n_{y},n_{z})}$, each quantum state corresponds to a point in 'n-space' with energy

$${\displaystyle E_{\mathbf {n} }=E_{0}+{\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}|\mathbf {n} |^{2}\,}$$

With ${\displaystyle |\mathbf {n} |^{2}}$denoting the square of the usual Euclidean length ${\displaystyle |\mathbf {n} |={\sqrt {n_{x}^{2}+n_{y}^{2}+n_{z}^{2}}}}$. The number of states with energy less than E_F +  E₀ is equal to the number of states that lie within a sphere of radius ${\displaystyle |\mathbf {n} _{\mathrm {F} }|}$ in the region of n-space where n_x, n_y, n_z are positive. In the ground state this number equals the number of fermions in the system:

$${\displaystyle N=2\times {\frac {1}{8}}\times {\frac {4}{3}}\pi n_{\mathrm {F} }^{3}}$$

The free fermions that occupy the lowest energy states form a sphere 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.

$${\displaystyle n_{\mathrm {F} }=\left({\frac {3N}{\pi }}\right)^{1/3}}$$

The Fermi energy is given by

$${\displaystyle E_{\mathrm {F} }={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}n_{\mathrm {F} }^{2}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}\left({\frac {3N}{\pi }}\right)^{2/3}}$$

Which results in a relationship between the Fermi energy and the number of particles per volume (when L² is replaced with V^2/3):

${\displaystyle E_{\mathrm {F} }={\frac {\hbar ^{2}}{2m}}\left({\frac {3\pi ^{2}N}{V}}\right)^{2/3}}$

This is also the energy of the highest-energy particle (the ${\displaystyle N}$th particle), above the zero point energy ${\displaystyle E_{0}}$. The ${\displaystyle N'}$th particle has an energy of

$${\displaystyle E_{N'}=E_{0}+{\frac {\hbar ^{2}}{2m}}\left({\frac {3\pi ^{2}N'}{V}}\right)^{2/3}\,=E_{0}+E_{\mathrm {F} }{\big |}_{N'}}$$

The total energy of a Fermi sphere of ${\displaystyle N}$ fermions (which occupy all ${\displaystyle N}$ energy states within the Fermi sphere) is given by:

$${\displaystyle E_{\rm {T}}=NE_{0}+\int _{0}^{N}E_{\mathrm {F} }{\big |}_{N'}\,dN'=\left({\frac {3}{5}}E_{\mathrm {F} }+E_{0}\right)N}$$

Therefore, the average energy per particle is given by:

$${\displaystyle E_{\mathrm {av} }=E_{0}+{\frac {3}{5}}E_{\mathrm {F} }}$$

Density of states

Density of states (DOS) of a Fermi gas in 3-dimensions

For the 3D uniform Fermi gas, with fermions of spin-½, the number of particles as a function of the energy ${\textstyle N(E)}$ is obtained by substituting the Fermi energy by a variable energy ${\textstyle (E-E_{0})}$:

$${\displaystyle N(E)={\frac {V}{3\pi ^{2}}}\left[{\frac {2m}{\hbar ^{2}}}(E-E_{0})\right]^{3/2},}$$

from which the density of states (number of energy states per energy per volume) ${\displaystyle g(E)}$ can be obtained. It can be calculated by differentiating the number of particles with respect to the energy:

$${\displaystyle g(E)={\frac {1}{V}}{\frac {\partial N(E)}{\partial E}}={\frac {1}{2\pi ^{2}}}\left({\frac {2m}{\hbar ^{2}}}\right)^{3/2}{\sqrt {E-E_{0}}}.}$$

This result provides an alternative way to calculate the total energy of a Fermi sphere of ${\displaystyle N}$ fermions (which occupy all ${\displaystyle N}$ energy states within the Fermi sphere):

$${\displaystyle {\begin{aligned}E_{T}&=\int _{0}^{N}E\mathrm {d} N(E)=EN(E){\big |}_{0}^{N}-\int _{E_{0}}^{E_{0}+E_{F}}N(E)\mathrm {d} E\\&=(E_{0}+E_{F})N-\int _{0}^{E_{F}}N(E)\mathrm {d} (E-E_{0})\\&=(E_{0}+E_{F})N-{\frac {2}{5}}E_{F}N(E_{F})=\left(E_{0}+{\frac {3}{5}}E_{\mathrm {F} }\right)N\end{aligned}}}$$

Thermodynamic quantities

Degeneracy pressure

Pressure vs temperature curves of classical and quantum ideal gases (Fermi gas, Bose gas) 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

$${\displaystyle P=-{\frac {\partial E_{\rm {T}}}{\partial V}}={\frac {2}{5}}{\frac {N}{V}}E_{\mathrm {F} }={\frac {(3\pi ^{2})^{2/3}\hbar ^{2}}{5m}}\left({\frac {N}{V}}\right)^{5/3},}$$

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

See also: Fermi level

Assuming that the concentration of fermions does not change with temperature, then the total chemical potential µ (Fermi level) of the three-dimensional ideal Fermi gas is related to the zero temperature Fermi energy E_F by a Sommerfeld expansion (assuming ${\displaystyle k_{\rm {B}}T\ll E_{\mathrm {F} }}$):

$${\displaystyle \mu (T)=E_{0}+E_{\mathrm {F} }\left[1-{\frac {\pi ^{2}}{12}}\left({\frac {k_{\rm {B}}T}{E_{\mathrm {F} }}}\right)^{2}-{\frac {\pi ^{4}}{80}}\left({\frac {k_{\rm {B}}T}{E_{\mathrm {F} }}}\right)^{4}+\cdots \right],}$$

where T is the temperature.^[6][7]

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 T_F. 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 ${\displaystyle N/V}$ 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:

$${\displaystyle E_{\mathrm {F} }={\frac {\hbar ^{2}}{2m_{e}}}\left(3\pi ^{2}\ 10^{28\ \sim \ 29}\ \mathrm {m^{-3}} \right)^{2/3}\approx 2\ \sim \ 10\ \mathrm {eV} ,}$$

where m_e is the electron rest mass.^[8] This Fermi energy corresponds to a Fermi temperature of the order of 10⁶ kelvins, much higher than the temperature of the sun surface. Any metal will boil before reaching this temperature under atmospheric pressure. Thus for any practical purpose, a metal can be considered as a Fermi gas at zero temperature as a first approximation (normal temperatures are small compared to T_F).

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:

$${\displaystyle E_{\mathrm {F} }={\frac {\hbar ^{2}}{2m_{e}}}\left({\frac {3\pi ^{2}(10^{36})}{1\ \mathrm {m^{3}} }}\right)^{2/3}\approx 3\times 10^{5}\ \mathrm {eV} =0.3\ \mathrm {MeV} }$$

Nucleus

Another typical example is that of the particles in a nucleus of an atom. The radius of the nucleus is roughly:

$${\displaystyle R=\left(1.25\times 10^{-15}\mathrm {m} \right)\times A^{1/3}}$$

where A is the number of nucleons.

The number density of nucleons in a nucleus is therefore:

$${\displaystyle \rho ={\frac {A}{{\frac {4}{3}}\pi R^{3}}}\approx 1.2\times 10^{44}\ \mathrm {m^{-3}} }$$

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:

$${\displaystyle E_{\mathrm {F} }={\frac {\hbar ^{2}}{2m_{\rm {p}}}}\left({\frac {3\pi ^{2}(6\times 10^{43})}{1\ \mathrm {m} ^{3}}}\right)^{2/3}\approx 3\times 10^{7}\ \mathrm {eV} =30\ \mathrm {MeV} ,}$$

where m_p 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.

Density of states

Using a volume integral on ${\textstyle d}$ dimensions, the density of states is:

$${\displaystyle g^{(d)}(E)=g_{s}\int {\frac {\mathrm {d} ^{d}\mathbf {k} }{(2\pi )^{d}}}\delta \left(E-E_{0}-{\frac {\hbar ^{2}|\mathbf {k} |^{2}}{2m}}\right)=g_{s}\ \left({\frac {m}{2\pi \hbar ^{2}}}\right)^{d/2}{\frac {(E-E_{0})^{d/2-1}}{\Gamma (d/2)}}}$$

The Fermi energy is obtained by looking for the number density of particles:

$${\displaystyle \rho ={\frac {N}{V}}=\int _{E_{0}}^{E_{0}+E_{\mathrm {F} }^{(d)}}g^{(d)}(E)\,dE}$$

To get:

$${\displaystyle E_{\mathrm {F} }^{(d)}={\frac {2\pi \hbar ^{2}}{m}}\left({\tfrac {1}{g_{s}}}\Gamma \left({\tfrac {d}{2}}+1\right){\frac {N}{V}}\right)^{2/d}}$$

where ${\textstyle V}$ is the corresponding d-dimensional volume, ${\textstyle g_{s}}$ is the dimension for the internal Hilbert space. For the case of spin-½, every energy is twice-degenerate, so in this case ${\textstyle g_{s}=2}$.

A particular result is obtained for ${\displaystyle d=2}$, where the density of states becomes a constant (does not depend on the energy):

$${\displaystyle g^{(2\mathrm {D} )}(E)={\frac {g_{s}}{2}}{\frac {m}{\pi \hbar ^{2}}}.}$$

Grand canonical ensemble

Most of the calculations above are exact at zero temperature, yet remain as good approximations for temperatures lower than the Fermi temperature. For other thermodynamics variables it is necessary to write a thermodynamic potential. For an ensemble of identical fermions, the best way to derive a potential is from the grand canonical ensemble with fixed temperature, volume and chemical potential µ. The reason is due to Pauli exclusion principle, as the occupation numbers of each quantum state are given by either 1 or 0 (either there is an electron occupying the state or not), so the (grand) partition function ${\displaystyle {\mathcal {Z}}}$ can be written as

${\displaystyle {\mathcal {Z}}(T,V,\mu )=\sum _{\{q\}}e^{-\beta (E_{q}-\mu N_{q})}=\prod _{q}\sum _{n_{q}=0}^{1}e^{-\beta (\varepsilon _{q}-\mu )n_{q}}=\prod _{q}\left(1+e^{-\beta (\varepsilon _{q}-\mu )}\right),}$

where ${\displaystyle \beta ^{-1}=k_{\rm {B}}T}$, ${\displaystyle \{q\}}$ indexes the ensembles of all possible microstates that give the same total energy ${\textstyle E_{q}=\sum _{q}\varepsilon _{q}n_{q}}$ and number of particles ${\textstyle N_{q}=\sum _{q}n_{q}}$, ${\textstyle \varepsilon _{q}}$ is the single particle energy of the state ${\textstyle q}$ (it counts twice if the energy of the state is degenerate) and ${\textstyle n_{q}=0,1}$, its occupancy. Thus the grand potential is written as

${\displaystyle \Omega (T,V,\mu )=-k_{\rm {B}}T\ln \left({\mathcal {Z}}\right)=-k_{\rm {B}}T\sum _{q}\ln \left(1+e^{\beta (\mu -\varepsilon _{q})}\right).}$

The same result can be obtained in the canonical and microcanonical ensemble, as the result of every ensemble must give the same value at thermodynamic limit ${\textstyle (N/V\rightarrow \infty )}$. The grand canonical ensemble is recommended here as it avoids the use of combinatorics and factorials.

As explored in previous sections, in the macroscopic limit we may use a continuous approximation (Thomas–Fermi approximation) to convert this sum to an integral:

$${\displaystyle \Omega (T,V,\mu )=-k_{\rm {B}}T\int _{-\infty }^{\infty }D(\varepsilon )\ln \left(1+e^{\beta (\mu -\varepsilon )}\right)\,d\varepsilon }$$

where D(ε) is the total density of states.

Relation to Fermi-Dirac distribution

The grand potential is related to the number of particles at finite temperature in the following way

$${\displaystyle N=-\left({\frac {\partial \Omega }{\partial \mu }}\right)_{T,V}=\int _{-\infty }^{\infty }D(\varepsilon ){\mathcal {f}}\left({\frac {\varepsilon -\mu }{k_{\rm {B}}T}}\right)\,\mathrm {d} \varepsilon }$$

where the derivative is taken at fixed temperature and volume, and it appears

$${\displaystyle {\mathcal {f}}(x)={\frac {1}{e^{x}+1}}}$$

also known as the Fermi–Dirac distribution.

Similarly, the total internal energy is

$${\displaystyle U=\Omega -T\left({\frac {\partial \Omega }{\partial T}}\right)_{V,\mu }-\mu \left({\frac {\partial \Omega }{\partial \mu }}\right)_{T,V}=\int _{-\infty }^{\infty }D(\varepsilon ){\mathcal {f}}\!\left({\frac {\varepsilon -\mu }{k_{\rm {B}}T}}\right)\varepsilon \,d\varepsilon .}$$

Exact solution for power-law density-of-states

Many systems of interest have a total density of states with the power-law form:

$${\displaystyle D(\varepsilon )=Vg(\varepsilon )={\frac {Vg_{0}}{\Gamma (\alpha )}}(\varepsilon -\varepsilon _{0})^{\alpha -1},\qquad \varepsilon \geq \varepsilon _{0}}$$

for some values of g₀, α, ε₀. The results of preceding sections generalize to d dimensions, giving a power law with:

• α = d/2 for non-relativistic particles in a d-dimensional box,
• α = d for non-relativistic particles in a d-dimensional harmonic potential well,
• α = d for hyper-relativistic particles in a d-dimensional box.

For such a power-law density of states, the grand potential integral evaluates exactly to:^[12]

$${\displaystyle \Omega (T,V,\mu )=-Vg_{0}(k_{\rm {B}}T)^{\alpha +1}F_{\alpha }\left({\frac {\mu -\varepsilon _{0}}{k_{\rm {B}}T}}\right),}$$

where ${\displaystyle F_{\alpha }(x)}$ is the complete Fermi–Dirac integral (related to the polylogarithm). From this grand potential and its derivatives, all thermodynamic quantities of interest can be recovered.

Relativistic Fermi gas

Radius–mass relations for a model white dwarf, relativistic relation vs non-relativistic. The Chandrasekhar limit is indicated as M_Ch.

The article has only treated the case in which particles have a parabolic relation between energy and momentum, as is the case in non-relativistic mechanics. For particles with energies close to their respective rest mass, the equations of special relativity are applicable. Where single-particle energy is given by:

$${\displaystyle E={\sqrt {(pc)^{2}+(mc^{2})^{2}}}.}$$

For this system, the Fermi energy is given by:

$${\displaystyle E_{\mathrm {F} }={\sqrt {(p_{\mathrm {F} }c)^{2}+(mc^{2})^{2}}}-mc^{2}\approx p_{\mathrm {F} }c,}$$

where the ${\displaystyle \approx }$ equality is only valid in the ultrarelativistic limit, and^[13]

$${\displaystyle p_{\mathrm {F} }=\hbar \left({\frac {1}{g_{s}}}6\pi ^{2}{\frac {N}{V}}\right)^{1/3}.}$$

The relativistic Fermi gas model is also used for the description of massive white dwarfs which are close to the Chandrasekhar limit. For the ultrarelativistic case, the degeneracy pressure is proportional to ${\displaystyle (N/V)^{4/3}}$.

Fermi liquid

In 1956, Lev Landau developed the Fermi liquid theory, where he treated the case of a Fermi liquid, i.e., a system with repulsive, not necessarily small, interactions between fermions. The theory shows that the thermodynamic properties of an ideal Fermi gas and a Fermi liquid do not differ that much. It can be shown that the Fermi liquid is equivalent to a Fermi gas composed of collective excitations or quasiparticles, each with a different effective mass and magnetic moment.