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, $\bar n\simeq e^{-\beta(\epsilon-\mu)}$.

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

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

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 $T\ll T_F$.

Maxima verification worksheet