The Fermi sea: k_F and a kinetic energy of 2.21/rₛ² Ry

Units are SI, with energies also in rydbergs, 1 Ry = ħ²/(2ma₀²) = 13.606 eV, and lengths in bohr radii, a₀ = 0.5292 Å; nothing here needs the course's metric (+, −, −, −). Topic 15.01 put N electrons in a periodic box of side L, where the plane waves e^(ik⋅r)/√V have k = (2π/L)(n₁, n₂, n₃): one k per (2π)³/V of k space, two spin states each. With the interaction off, H = Σₖσ εₖc†ₖσcₖσ with εₖ = ħ²k²/2m, and the cheapest way to place N fermions, one per spin-orbital, takes the N smallest |k|, a sphere: |FS⟩ = Π over |k| ≤ k_F of c†ₖ↑c†ₖ↓|0⟩. Then N = 2[V/(2π)³](4π/3)k_F³, so n = k_F³/(3π²), derived. Every occupation ⟨c†ₖσcₖσ⟩ is 1 inside and 0 outside, and the state carries zero momentum and zero spin. In a finite box the sphere is unique only for closed shells, N = 2, 14, 38, 54, …; the distinction vanishes as V → ∞. This |FS⟩ is the determinant whose exchange energy topic 15.03 computes and the vacuum of topic 15.06's holes. Sodium's n = 2.65 × 10²⁸ m⁻³ gives k_F = 9.22 × 10⁹ m⁻¹.

Metals are compared by rₛ, the radius of the sphere each electron owns, in bohr radii: (4π/3)(rₛa₀)³ = 1/n. Putting n = 3/(4πrₛ³a₀³) into k_F³ = 3π²n gives k_F³ = 9π/(4rₛ³a₀³), so k_F = (9π/4)¹⁄³/(rₛa₀) = 1.919/(rₛa₀), derived, with an inverse length on both sides. The 1.919 says something physical: the Fermi wavelength 2π/k_F = 3.27rₛa₀ is 1.6 times the diameter of an electron's own sphere, so even the shortest waves in the sea overlap their neighbours, which is why exclusion, not thermal motion, sets their energy. Simple metals run from rₛ = 2.07 in aluminium, three electrons per atom, to 5.62 in caesium. For sodium, rₛ = 3.93 gives k_F = 0.922 Å⁻¹, and de Haas–van Alphen measurements find its Fermi surface a sphere, round to about one part in 10³, of that radius. The agreement survives the Coulomb interaction for a reason quoted, not derived: Luttinger's theorem fixes the k-space volume inside the Fermi surface at 4π³n per spin, whatever the interaction.

The kinetic energy is an expectation value in |FS⟩: ⟨T⟩ = Σₖσ εₖ⟨c†ₖσcₖσ⟩, twice the sum of ħ²k²/2m over |k| ≤ k_F. Replace the sum by [V/(2π)³]∫d³k: ⟨T⟩ = 2[V/(2π)³](ħ²/2m) × 4π∫k⁴dk from 0 to k_F = (V/π²)(ħ²/2m)k_F⁵/5. Dividing by N = Vk_F³/(3π²) leaves ⟨T⟩/N = (3/5)ħ²k_F²/2m = (3/5)E_F, derived. The 3/5 is geometry: shells hold states in proportion to k², so half the electrons lie beyond 2⁻¹⁄³k_F = 0.794k_F, and the rms wavevector is √(3/5)k_F = 0.775k_F. In rydbergs E_F = (k_Fa₀)² Ry = 3.683/rₛ² Ry, so ⟨T⟩/N = 2.21/rₛ² Ry, derived, and ħ²k²/m carries J² s² m⁻² kg⁻¹ = J, as an energy must. Sodium: E_F = 3.24 eV, T_F = 3.76 × 10⁴ K and ⟨T⟩/N = 0.143 Ry = 1.95 eV. Statistical physics supplies the thermal correction, used here and not re-derived: the energy rises by (5π²/12)(T/T_F)² = 2.5 × 10⁻⁴ of itself at 293 K, so at room temperature a metal's electrons carry, to that accuracy, the energy of |FS⟩.

At fixed N, E = (3/5)NE_F with E_F ∝ (N/V)²⁄³, so E ∝ V⁻²⁄³ and P = −(∂E/∂V)_N = (2/3)E/V = (2/5)nE_F, derived. That is the non-relativistic relation for εₖ ∝ k²; an ultra-relativistic sea, with εₖ ∝ k, would give E/(3V). Differentiating again, B = −V(∂P/∂V)_N = (5/3)P = (2/3)nE_F. In atomic units P = 0.352/rₛ⁵ Ry a₀⁻³ = 5.17 × 10¹² Pa/rₛ⁵. For sodium P = 5.5 GPa and B = 9.2 GPa, against a measured bulk modulus near 6.4 GPa (quoted): the free sea sets the scale of a simple metal's stiffness, about 40% high, though not its value, which depends on the attractions still to come. A classical gas at the same density and 293 K would push with nk_BT = 0.11 GPa, fifty times less. Yet sodium rests at zero pressure, so something pulls inward with 5.5 GPa. Exchange supplies most of it, about −4.5 GPa at this density by topic 15.03's result, and correlation and the ions the rest. That is the sense in which (2/5)nE_F is what any binding must overcome.

Let N↑ = N(1 + ζ)/2 and N↓ = N(1 − ζ)/2. Each spin fills its own sphere, with N_σ = Vk_Fσ³/(6π²), so k_Fσ = k_F(1 ± ζ)¹⁄³, and each contributes (3/5)N_σħ²k_Fσ²/2m. Adding them, ⟨T⟩/N = (2.21/rₛ²)[(1 + ζ)⁵⁄³ + (1 − ζ)⁵⁄³]/2 Ry, derived. Fully polarised, ζ = 1, the single sphere is 2¹⁄³ = 1.260 times wider and the energy 2²⁄³ = 1.587 times higher, 3.51/rₛ² Ry: the second spin state halves the crowding. For small ζ the bracket is 1 + (5/9)ζ², so the energy rises by NE_Fζ²/3. Add the Zeeman energy −μ_BBNζ of a field B, with ζ now counting moments along B, and the minimum lies at ζ = 3μ_BB/(2E_F): a magnetisation nμ_Bζ = μ_B²g(E_F)B with g(E_F) = 3n/(2E_F), which is Pauli's susceptibility from statistical physics, here a check on the formula rather than a new result. For sodium μ₀μ_B²g(E_F) = 8.3 × 10⁻⁶, dimensionless in SI. Exchange favours parallel spins, and topic 15.04 asks at what rₛ it can pay this kinetic bill.

|FS⟩ is an eigenstate of the kinetic operator, not of jellium's Hamiltonian: the q ≠ 0 Coulomb terms of topic 15.01 scatter pairs of electrons from inside the sphere to outside. Because |FS⟩ is T's ground state, any other N-electron state has more kinetic energy, so 2.21/rₛ² Ry is a lower bound on the interacting gas's ⟨T⟩/N, derived. The rest is quoted: the true occupation falls below 1 inside k_F and rises above 0 outside, with a jump Z smaller than 1 at k_F, while the sphere's volume is unchanged. The neglected terms grow relative to the kinetic one as rₛ, since kinetic energy scales as 1/rₛ² and every Coulomb energy as 1/rₛ: dense metals are the clean limit. Relativity sets a second edge. v_F/c = ħk_F/(mc) = 1.919α/rₛ = 0.0140/rₛ is 3.6 × 10⁻³ in sodium, so the leading correction, −(5/28)(v_F/c)² of the energy, is 2 × 10⁻⁶; near rₛ = 0.014, in white-dwarf matter, the Fermi momentum reaches mc. The lattice is the third: sodium's ions barely dent the sphere, while aluminium's cut it into pieces.