26 Jul 2025

Bose-Einstein and Fermi-Dirac Statistics

Quantum distributions, degenerate gases, condensation, photon thermodynamics, electron heat capacity, emissions, and Pauli paramagnetism.

bsc semester-vi statistical-mechanics bose-einstein fermi-dirac

For one-particle state $i$, put $x_i=e^{-\beta(\epsilon_i-\mu)}$. A bosonic state permits $n_i=0,1,2,\ldots$, so its grand partition factor and mean occupation are

\[\Xi_i^{BE}=\sum_{n_i=0}^{\infty}x_i^{n_i}=\frac1{1-x_i},\qquad \bar n_i=x_i\frac{\partial\ln\Xi_i}{\partial x_i}=\frac{x_i}{1-x_i}.\]

A fermionic state permits only $n_i=0,1$, hence

\[\Xi_i^{FD}=1+x_i,\qquad \bar n_i=x_i\frac{\partial\ln\Xi_i}{\partial x_i}=\frac{x_i}{1+x_i}.\]

Therefore

\[\boxed{\bar n_{BE}=\frac1{e^{\beta(\epsilon-\mu)}-1}},\qquad \boxed{\bar n_{FD}=\frac1{e^{\beta(\epsilon-\mu)}+1}}.\]

The minus sign permits any boson occupation; the plus sign enforces fermion occupation $0$ or $1$. When $e^{\beta(\epsilon-\mu)}\gg1$, both reduce to Maxwell-Boltzmann statistics, $\bar n\simeq e^{-\beta(\epsilon-\mu)}$.

The degeneracy parameter $n\lambda_T^3$ locates this crossover: $n\lambda_T^3\ll1$ is classical, whereas values of order unity require quantum statistics. For bosons the denominator must remain positive, so $\mu$ cannot exceed the one-particle ground-state energy; choosing that energy as zero gives $\mu\le0$. Fermions have no analogous upper bound because Pauli exclusion already limits each state occupation.

Bose gas and condensation

For free particles in three dimensions, the density of states is

\[g(\epsilon)=\frac{V}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\epsilon^{1/2}.\]

The excited population is

\[N_{ex}=\int_0^\infty\frac{g(\epsilon)d\epsilon}{z^{-1}e^{\beta\epsilon}-1},\qquad z=e^{\beta\mu}\le1.\]

With the thermal wavelength $\lambda_T=h/\sqrt{2\pi mk_BT}$ and Bose function

\[g_s(z)=\frac1{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{z^{-1}e^x-1}\,dx,\]

the particle number, pressure, and energy are

\[N_{ex}=\frac{V}{\lambda_T^3}g_{3/2}(z),\qquad P=\frac{k_BT}{\lambda_T^3}g_{5/2}(z),\qquad U=\frac32PV.\]

At fixed $T$, this is largest at $z=1$:

\[N_{ex}^{max}=\frac{V}{\lambda_T^3}\zeta(3/2).\]

When $N>N_{ex}^{max}$, the excess occupies the ground state. The critical temperature and condensed fraction are

\[T_c=\frac{2\pi\hbar^2}{mk_B}\left[\frac{N}{V\zeta(3/2)}\right]^{2/3},\] \[\frac{N_0}{N}=1-\left(\frac{T}{T_c}\right)^{3/2}\quad(T<T_c).\]

Below $T_c$, $z=1$ and $g_s(1)=\zeta(s)$. Therefore the thermodynamic functions of the strongly degenerate ideal Bose gas are

\[U=\frac32\frac{Vk_BT}{\lambda_T^3}\zeta(5/2),\qquad S=\frac{U+PV}{T}=\frac52\frac{Vk_B}{\lambda_T^3}\zeta(5/2),\] \[C_V=\left(\frac{\partial U}{\partial T}\right)_{V,N} =\frac{15}{4}Nk_B\frac{\zeta(5/2)}{\zeta(3/2)} \left(\frac{T}{T_c}\right)^{3/2}.\]

Liquid helium-4 is a strongly interacting Bose liquid, so the ideal-gas formula is not quantitatively exact; nevertheless, macroscopic quantum occupation is central to its superfluid behavior.

Photon gas and Bose derivation of Planck’s law

Photon number is not conserved, hence $\mu=0$. Each mode has

\[\bar n=\frac1{e^{\beta h\nu}-1},\]

and multiplying $h\nu\bar n$ by the electromagnetic mode density $8\pi\nu^2/c^3$ reproduces Planck’s law. Integrating gives

\[U=aVT^4,\qquad P=\frac{U}{3V},\qquad F=-PV=-\frac13aVT^4,\]

and from $S=(U+PV)/T$,

\[S=\frac43aVT^3,\qquad C_V=4aVT^3.\]

Degenerate Fermi gas

At $T=0$, $f(\epsilon)=1$ below the Fermi energy and $0$ above it. For spin-$1/2$ particles,

\[N=2\frac{V}{(2\pi)^3}\frac{4\pi k_F^3}{3} =\frac{V}{3\pi^2}k_F^3,\]

so

\[k_F=(3\pi^2n)^{1/3},\qquad E_F=\frac{\hbar^2k_F^2}{2m}.\]

The factor $2$ counts spin. Differentiating the state count gives the total density of states

\[g_F(\epsilon)=\frac{V}{2\pi^2} \left(\frac{2m}{\hbar^2}\right)^{3/2}\epsilon^{1/2}.\]

The zero-temperature energy follows directly:

\[U=\int_0^{E_F}\epsilon g_F(\epsilon)d\epsilon =\frac35NE_F.\]

For a nonrelativistic gas the kinetic relation $PV=2U/3$ then gives

\[P_0=\frac25\frac{N}{V}E_F.\]

Only electrons within about $k_BT$ of $E_F$ can change occupation. The Sommerfeld expansion

\[\int_0^\infty\Phi(\epsilon)f(\epsilon)d\epsilon =\int_0^\mu\Phi(\epsilon)d\epsilon +\frac{\pi^2}{6}(k_BT)^2\Phi^{\prime}(\mu)+\cdots\]

first applied to $N$ fixes the chemical potential, and then applied to $U$ gives

\[\mu(T)=E_F\left[1-\frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2\right],\] \[U=\frac35NE_F\left[1+\frac{5\pi^2}{12} \left(\frac{T}{T_F}\right)^2\right].\]

Therefore

\[C_V=\frac{\pi^2}{2}Nk_B\frac{T}{T_F},\qquad S=\frac{\pi^2}{2}Nk_B\frac{T}{T_F},\qquad T_F=E_F/k_B,\]

which is linear in $T$ and much smaller than the classical value.

Equation-generated Bose-Einstein, Fermi-Dirac, and Maxwell-Boltzmann occupations versus $(\epsilon-\mu)/(k_BT)$, together with $N_0/N=1-(T/T_c)^{3/2}$.

Thermionic and photoelectric emission

Electrons escaping a metal must overcome the work function $\phi$. In the high-energy tail the Fermi factor becomes Boltzmann-like. Integrating the outward normal flux gives

\[J=e\frac{2}{h^3}\int_{p_z>p_0}\frac{p_z}{m} e^{-[p^2/(2m)-E_F]/k_BT}\,d^3p,\]

where $p_0^2/(2m)=E_F+\phi$. The two transverse Gaussian integrals give $2\pi mk_BT$, while

\[\int_{p_0}^{\infty}\frac{p_z}{m}e^{-p_z^2/(2mk_BT)}dp_z =k_BT\,e^{-p_0^2/(2mk_BT)}.\]

Thus

\[J=A_RT^2e^{-\phi/k_BT},\qquad A_R=\frac{4\pi em_ek_B^2}{h^3},\]

for an ideal free-electron surface; real materials modify the prefactor. In photoelectric emission, one photon supplies $h\nu$ and

\[K_{max}=h\nu-\phi.\]

Thermionic emission is controlled by the thermal high-energy tail; photoemission is controlled by photon energy.

Pauli spin paramagnetism

A weak field shifts spin energies by $\mp\mu_BB$. Only states near $E_F$ can repopulate. With $g_F(E_F)$ denoting the total two-spin density of states, each spin direction has density $g_F(E_F)/2$, so

\[N_+-N_- =\frac{g_F(E_F)}2(2\mu_BB) =g_F(E_F)\mu_BB.\]

Thus

\[M=\mu_B(N_+-N_-)/V,\qquad \chi_P=\mu_0\mu_B^2\frac{g_F(E_F)}{V}=\frac{3\mu_0n\mu_B^2}{2E_F}.\]

Unlike Curie paramagnetism, $\chi_P$ is nearly temperature independent for $T\ll T_F$.

Solved Problems

1. Bose condensation temperature and condensate fraction

An ideal gas of $^{87}\mathrm{Rb}$ atoms has number density $n=1.00\times10^{20}\ \mathrm{m^{-3}}$. Using $m=86.9091805\ \mathrm u$ and $\zeta(3/2)=2.61237535$, find $T_c$ and the condensate fraction at $T=T_c/2$.

At the transition, the excited states hold all particles:

\[n=\frac{\zeta(3/2)}{\lambda_{T_c}^3}.\]

Solving for temperature gives

\[T_c=\frac{2\pi\hbar^2}{mk_B} \left[\frac{n}{\zeta(3/2)}\right]^{2/3} =3.98331\times10^{-7}\ \mathrm K.\]

Thus $T_c=398.331\ \mathrm{nK}$. Below $T_c$,

\[\frac{N_0}{N}=1-\left(\frac{T}{T_c}\right)^{3/2} =1-\left(\frac12\right)^{3/2}=0.646447.\]

About $64.6\%$ is condensed in the ideal-gas model. The fraction approaches one as $T\to0$ and vanishes continuously as $T\to T_c^-$.

2. Fermi scales and degeneracy pressure of conduction electrons

A metal has conduction-electron density $n=8.50\times10^{28}\ \mathrm{m^{-3}}$. Treat the electrons as a zero-temperature free Fermi gas and find $k_F$, $E_F$, $T_F$, and $P_0$.

State counting gives

\[k_F=(3\pi^2n)^{1/3}=1.36023\times10^{10}\ \mathrm{m^{-1}}.\]

Therefore

\[E_F=\frac{\hbar^2k_F^2}{2m_e} =1.12941\times10^{-18}\ \mathrm J=7.04936\ \mathrm{eV},\] \[T_F=\frac{E_F}{k_B}=8.18044\times10^4\ \mathrm K,\]

and

\[P_0=\frac25nE_F=3.84007\times10^{10}\ \mathrm{Pa}.\]

Room temperature is far below $T_F$, so the electrons are strongly degenerate. The pressure remains nonzero at $T=0$ because Pauli exclusion, not thermal motion, fills momentum states up to $k_F$.

Descriptive Questions

1. Derive the Bose-Einstein and Fermi-Dirac occupation functions from their single-state grand partition factors, identifying where the sign difference enters.
2. Explain why a conserved ideal Bose gas condenses in three dimensions and derive the $T^{3/2}$ excited fraction below $T_c$.
3. Use the Sommerfeld expansion to explain why the electronic heat capacity is linear in $T$ and much smaller than the classical equipartition value.
4. Derive Pauli spin susceptibility from the field-induced transfer of electrons at the Fermi surface and contrast its temperature dependence with Curie’s law.

Numerical Problems

1. At an energy $\epsilon-\mu=2k_BT$, calculate the Bose-Einstein, Fermi-Dirac, and Maxwell-Boltzmann mean occupations of one state.

   Answer: $\bar n_{BE}=0.156518$, $\bar n_{FD}=0.119203$, and $\bar n_{MB}=0.135335$.

2. Estimate the ideal Richardson-Dushman current density from a metal with work function $4.50\ \mathrm{eV}$ at $2500\ \mathrm K$. Use $A_R=1.20173229\times10^6\ \mathrm{A\,m^{-2}\,K^{-2}}$.

   Answer: $J=6.36923\times10^3\ \mathrm{A\,m^{-2}}$.

3. Light of wavelength $400\ \mathrm{nm}$ illuminates a metal with work function $2.30\ \mathrm{eV}$. Find the maximum photoelectron energy and stopping potential.

   Answer: $K_{max}=0.799605\ \mathrm{eV}$ and $V_s=0.799605\ \mathrm V$.

4. Estimate the Pauli spin susceptibility for $n=5.00\times10^{28}\ \mathrm{m^{-3}}$ and $E_F=5.50\ \mathrm{eV}$.

   Answer: $\chi_P=9.19884\times10^{-6}$ in SI units.

Maxima verification worksheet

References

1. Particle statistics — Wikipedia
2. R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., chapters 6–8 on ideal Bose and Fermi gases.
3. C. Kittel and H. Kroemer, Thermal Physics, 2nd ed., chapters 6–8 on ideal gases, Fermi systems, and Bose systems.
4. N. W. Ashcroft and N. D. Mermin, Solid State Physics, chapters 2 and 31 on the free-electron gas and conduction-electron paramagnetism.
5. C. J. Pethick and H. Smith, Bose-Einstein Condensation in Dilute Gases, 2nd ed., chapter 2 on the ideal Bose gas.