The free-electron Fermi gas

Sommerfeld keeps Drude's box and replaces his statistics. Solve the free-particle Schrödinger equation in a cube of side L with periodic boundaries: the allowed wavevectors form a grid with one state per (2π/L)³ of k-space, a volume (2π)³/V that depends on the sample's size and not its shape. Each state holds two electrons, spin up and spin down. At T = 0 exclusion forces you to fill outward from the origin, so the occupied region is a sphere. Setting N = 2 × (4πk_F³/3) ÷ ((2π)³/V) gives N = V k_F³/3π², that is k_F = (3π²n)¹⁄³ with n = N/V. Copper's one free electron per atom gives n = 8.47 × 10²⁸ m⁻³, so k_F = 1.36 × 10¹⁰ m⁻¹ and E_F = ℏ²k_F²/2m = 7.0 eV. Notice what E_F does not depend on: not the chemistry of the metal, not the size of the crystal, not the temperature. Density alone.

Count states below an energy instead. A sphere of radius k = √(2mE)/ℏ holds N(E) = V k³/3π² = (V/3π²)(2mE/ℏ²)³⁄², so g(E) = dN/dE = (V/2π²)(2m/ℏ²)³⁄² √E. That square root is the three-dimensional signature: in two dimensions g is constant, and in one it diverges as E^(−1/2). Eliminating the constants in favour of N and E_F gives a tidier form, g(E) = (3N/2E_F)√(E/E_F), so g(E_F) = 3N/2E_F — about 0.21 states per electronvolt per electron in copper. Now the structural point. √E is nonzero for every energy above the band bottom, so wherever you put the Fermi level there are empty states immediately above it. This model cannot produce a forbidden band, which means it describes metals and nothing else; silicon, diamond and every insulator lie outside it. Opening a gap needs the periodic lattice potential Sommerfeld deleted, which is the business of Bloch's theorem in the next topic.

The occupancy of a state of energy E is f(E) = 1/[exp((E − μ)/k_B T) + 1], with the chemical potential μ tending to E_F as T tends to zero. At absolute zero it is a step: every state below E_F full, every state above it empty. Warm the metal and the step feathers, but only over a few k_B T — f = 0.73 at μ − k_B T, exactly 0.5 at μ, 0.27 at μ + k_B T, and 0.12 at μ + 2k_B T, while μ itself sits below E_F by a fraction (π²/12)(T/T_F)², about 10⁻⁵ for copper at 300 K. Put numbers on it. At 300 K, k_B T = 0.0259 eV; against copper's E_F = 7.03 eV that is a feather 0.37% of the height of the cliff. Equivalently T_F = E_F/k_B = 8.2 × 10⁴ K, so across the whole range in which copper is a solid, T/T_F stays under 2%. That is what degenerate means: Fermi-Dirac and Maxwell-Boltzmann bear no resemblance to each other here, and the classical limit would need temperatures no metal survives.

The argument is one line long. An electron well below E_F cannot absorb k_B T of energy, because every state k_B T above it is already occupied — Pauli blocking, not any shortage of energy. Only those within a few k_B T of the Fermi surface have somewhere to go, and they are a fraction of order T/T_F of the whole. Each gains about k_B T, so the thermal energy is U ≈ N k_B T²/T_F and Cₑₗ = dU/dT is proportional to T. The Sommerfeld expansion supplies the coefficient: Cₑₗ = (π²/3)k_B²T g(E_F) = (π²/2)N k_B (T/T_F). Take two things from that. The heat capacity is linear in T, not constant, so it is a different function altogether from the classical (3/2)Nk_B. And it is small: the ratio is (π²/3)(T/T_F), which for copper at 300 K is 1.2%. That is why Dulong-Petit's 3R survived the discovery that every copper atom donates a mobile electron — those electrons contribute 0.15 J mol⁻¹ K⁻¹ against the lattice's 24.9.

Far below the Debye temperature, T ≪ θ_D, the lattice contributes Cₗₐₜ = (12π⁴/5)R(T/θ_D)³, so a measured molar heat capacity is C = γT + AT³. Divide through by T: C/T = γ + AT². Plot C/T against T² and you get a straight line whose intercept is the electronic coefficient γ and whose slope A gives θ_D. The two terms are equal where γT = AT³, that is at T = √(γ/A), which for sodium is 1.5 K — so the linear law is a liquid-helium measurement, never a room-temperature one. What comes back is close to, but not equal to, the free-electron prediction. Copper: 0.50 mJ mol⁻¹ K⁻² predicted against 0.695 measured. Potassium: 1.67 against 2.08. The usual bookkeeping is a thermal effective mass, m*/m = γ/γ_free, giving 1.38 and 1.25. That ratio is the receipt for what the model discarded — the lattice potential, and the interaction between the electrons themselves.

The filled sphere carries real energy and real pressure. Averaging E over it gives U/N = (3/5)E_F, which is 4.2 eV per electron in copper at absolute zero, and the degeneracy pressure P = (2/5)nE_F is 38 GPa, with a bulk modulus B = (2/3)nE_F of 64 GPa against copper's measured 137 GPa. A factor of two, from a model with no lattice and no Coulomb interaction, is a real success. So is Wiedemann-Franz: the same Fermi-surface electrons carry charge and heat, so κ/σT = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W Ω K⁻², a constant most metals obey near room temperature. The failures are all one failure. There is no gap, so the model predicts every solid is a metal and cannot say why diamond insulates. Aluminium and beryllium show positive Hall coefficients, as though their carriers were positive. And fitting copper's conductivity demands a mean free path of 39 nm, some 150 atomic spacings, which is absurd for a particle bouncing off a dense array of ions. Each of those needs the periodic potential back.