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Wyatt's Notes — University

Quantum Statistics in Detail

15.1 Fermi—Dirac and Bose—Einstein Distributions

For non-interacting quantum particles:

⟨ni⟩=1eβ(ϵi−μ)±1\langle n_i \rangle = \frac{1}{e^{\beta(\epsilon_i - \mu)} \pm 1}

Where ++ is for fermions (Fermi—Dirac) and −- for bosons (Bose—Einstein).

Fermions (half-integer spin): Pauli exclusion limits ⟨ni⟩≤1\langle n_i \rangle \leq 1.

Bosons (integer spin): No restriction on occupation number; ⟨ni⟩\langle n_i \rangle can diverge when ϵi=μ\epsilon_i = \mu.

15.2 The Fermi Gas

For a 3D gas of NN non-interacting fermions in volume VV:

N=∑k1eβ(ℏ2k2/2m−μ)+1→continuumV(2π)3∫d3k f(ϵk)N = \sum_{\mathbf{k}} \frac{1}{e^{\beta(\hbar^2 k^2/2m - \mu)} + 1} \xrightarrow{\text{continuum} \frac{V}{(2\pi)^3}\int d^3k\, f(\epsilon_k)}

The Fermi energy at T=0T = 0:

ϵF=ℏ22m(3π2n)2/3\epsilon_F = \frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}

Where n=N/Vn = N/V is the number density. The Fermi temperature is TF=ϵF/kBT_F = \epsilon_F/k_B.

At low temperature (T≪TFT \ll T_F), the Sommerfeld expansion gives:

E=35NϵF[1+5π212(TTF)2+⋯ ]E = \frac{3}{5}N\epsilon_F\left[1 + \frac{5\pi^2}{12}\left(\frac{T}{T_F}\right)^2 + \cdots\right]

CV=NkBπ22TTF+⋯C_V = Nk_B\frac{\pi^2}{2}\frac{T}{T_F} + \cdots

The linear specific heat is a hallmark of degenerate Fermi systems.

15.3 The Bose Gas and Bose—Einstein Condensation

For bosons, the chemical potential must satisfy μ≤ϵ0\mu \leq \epsilon_0 (ground state energy). When μ→ϵ0\mu \to \epsilon_0A macroscopic fraction of particles condenses into the ground state.

The critical temperature for BEC in 3D:

Tc=2πℏ2mkB(nζ(3/2))2/3T_c = \frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}

Where ζ(3/2)≈2.612\zeta(3/2) \approx 2.612.

Below TcT_cThe condensate fraction is:

N0N=1−(TTc)3/2\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}

Worked Example 15.1: Fermi Energy of Copper

Copper has one conduction electron per atom, atomic mass 63.563.5 g/mol, density 8.968.96 g/cm3^3.

n=8.96×103 kg/m363.5×10−3 kg/mol×NA=1.41×1029 m−3×6.022×1023=8.49×1028 m−3n = \frac{8.96 \times 10^3 \text{ kg/m}^3}{63.5 \times 10^{-3} \text{ kg/mol} \times N_A = 1.41 \times 10^{29} \text{ m}^{-3} \times 6.022 \times 10^{23} = 8.49 \times 10^{28} \text{ m}^{-3}}

ϵF=ℏ22me(3π2×8.49×1028)2/3\epsilon_F = \frac{\hbar^2}{2m_e}(3\pi^2 \times 8.49 \times 10^{28})^{2/3}

=(1.055×10−34)22×9.109×10−31×(2.52×1030)2/3= \frac{(1.055 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31}} \times (2.52 \times 10^{30})^{2/3}

=1.113×10−681.822×10−30×1.846×1020= \frac{1.113 \times 10^{-68}}{1.822 \times 10^{-30}} \times 1.846 \times 10^{20}

=6.11×10−39×1.846×1020=1.13×10−18 J= 6.11 \times 10^{-39} \times 1.846 \times 10^{20} = 1.13 \times 10^{-18} \text{ J}

TF=1.13×10−181.38×10−23≈81 900 KT_F = \frac{1.13 \times 10^{-18}}{1.38 \times 10^{-23}} \approx 81\,900 \text{ K}

This is enormously higher than room temperature, confirming that conduction electrons in metals form a highly degenerate Fermi gas.

Worked Example 15.2: BEC in a Trap

For N=106N = 10^6 rubidium-87 atoms in a harmonic trap with frequency ωˉ/(2π)=100\bar{\omega}/(2\pi) = 100 Hz:

In a harmonic trap, the density of states is g(ϵ)=ϵ2/(2ℏ3ωˉ3)g(\epsilon) = \epsilon^2/(2\hbar^3\bar{\omega}^3)Giving:

Tc=ℏωˉkB(6Nπ2ζ(3))1/3T_c = \frac{\hbar\bar{\omega}}{k_B}\left(\frac{6N}{\pi^2\zeta(3)}\right)^{1/3}

=1.055×10−34×2π×1001.38×10−23(6×106π2×1.202)1/3= \frac{1.055 \times 10^{-34} \times 2\pi \times 100}{1.38 \times 10^{-23}}\left(\frac{6 \times 10^6}{\pi^2 \times 1.202}\right)^{1/3}

=6.63×10−321.38×10−23×(5.07×105)1/3= \frac{6.63 \times 10^{-32}}{1.38 \times 10^{-23}} \times (5.07 \times 10^5)^{1/3}

=4.81×10−9×79.7≈383 nK= 4.81 \times 10^{-9} \times 79.7 \approx 383 \text{ nK}

This is consistent with the 1995 Cornell—Wieman BEC experiment.