🎲Quantum Statistics - Fermi-Dirac, Bose-Einstein, Fermi energy, BEC, Debye model

How to use Quantum statistics & heat capacity of solids

1. 1On Three statistics, drag kT and μ and see where the curves merge (the classical limit)
2. 2Run the level animation and change N and temperature to compare fermions and bosons
3. 3On Electron gas, pick a metal or enter an electron density to read E_F, T_F and v_F, then raise T to see excitations near the Fermi surface
4. 4On BEC, choose an atom and density for T_c; on Heat capacity, pick a solid and compare the Debye and Einstein curves

FAQ

How do the three distributions differ?

All three have mean occupation 1/(e^((ε−μ)/kT) + a): classical MB corresponds to a = 0 (i.e. e^(−(ε−μ)/kT)), fermions to a = +1 and bosons to a = −1. The +1 caps each state at one particle (Pauli exclusion), while the −1 lets the boson occupation diverge as ε approaches μ, so bosons bunch together. When ε − μ is many kT, all three agree. That is the classical limit, which covers ordinary dilute gases at room temperature.

What does a Fermi energy of 7 eV for copper mean?

Copper has about 8.47×10²⁸ conduction electrons per cubic metre, so the free-electron model gives E_F = ħ²(3π²n)^{2/3}/2m ≈ 7.0 eV, a Fermi temperature of about 82,000 K and a Fermi velocity near 1.6×10⁶ m/s. Even at absolute zero the fastest electrons move at about 0.5% of light speed. Room-temperature kT ≈ 0.026 eV is tiny next to E_F, so the gas is highly degenerate and only a thin shell near the Fermi surface can be excited.

Why do electrons add so little to a metal's heat capacity?

Classical equipartition would give each free electron (3/2)k. Because of the Pauli principle, only electrons within about kT of E_F can absorb heat, a fraction of roughly T/T_F. The exact result is C_el = (π²/2)(T/T_F)R, about 0.6% of the classical value for copper at room temperature. That is why lattice vibrations dominate, except below a few kelvin, where the electronic term (∝ T) overtakes the phonon term (∝ T³).

How is the BEC critical temperature calculated?

For a uniform ideal Bose gas, T_c = (2πħ²/mk)(n/2.612)^{2/3}, which is the same as the phase-space density nλ³ reaching ζ(3/2) ≈ 2.612, where λ is the thermal de Broglie wavelength. For rubidium-87 at about 3×10¹⁹ m⁻³ this gives roughly 170 nK, the scale of the 1995 JILA experiment. For atoms in a harmonic trap, T_c ≈ 0.94ħω̄N^{1/3}/k and the condensate fraction follows 1 − (T/T_c)³.

Which is more accurate, the Debye or the Einstein model?

Both approach the Dulong–Petit value 3R at high temperature. The Einstein model gives every atom the same frequency, so the heat capacity drops exponentially at low T, which is too fast. Debye treats lattice vibrations as a phonon gas with a cutoff frequency and gets C = (12π⁴/5)R(T/θ_D)³ at low T, which matches insulators well. Real phonon spectra lie between the two, and the fitted Debye temperature varies slightly with temperature.