The exchange energy of the electron gas: −0.916/rₛ Ry

Units are SI, with energies also in rydbergs, Ry = e²/(8πε₀a₀) = 13.606 eV; the theory is non-relativistic, so the course's metric diag(+1, −1, −1, −1) plays no part. Take jellium from topic 15.01 (quoted): plane waves e^(ik⋅r)/√V in a box of volume V, with the q = 0 Coulomb term cancelled by the background. Unit 10's Fock matrix (quoted) is F_ab = h_ab + Σⱼ (⟨aj|v|bj⟩ − ⟨aj|v|jb⟩) over occupied spin-orbitals j. Between plane waves ⟨k₁k₂|v|k₃k₄⟩ = (1/V) v(k₁ − k₃) δ(k₁ + k₂, k₃ + k₄), with v(q) = e²/(ε₀q²): translation invariance has become momentum conservation. The direct term needs v(0), which is gone. The exchange term ⟨k′p|v|pk⟩ carries δ(k′ + p, p + k) and equal spins, so it vanishes unless k′ = k. F is diagonal in k, and the plane waves that built it are its eigenstates: self-consistent at once, derived. Overhauser showed that a spin-density-wave determinant lies lower still (named, not derived); this unit keeps the symmetric solution.

From topic 15.01 (quoted), V̂ = (1/2V) Σ v(q) c†(k + q, σ) c†(p − q, σ′) c(p, σ′) c(k, σ), summed over k, p, σ, σ′ and q ≠ 0. On the sea |F⟩, filled for |k| ≤ k_F in both spins, Wick's theorem pairs each c† with a c. Pairing c†(k + q, σ) with c(k, σ) is the direct contraction; it needs q = 0, so the Hartree energy of jellium is zero. The only other pairing joins c†(k + q, σ) to c(p, σ′), forcing σ′ = σ and p = k + q at the cost of one fermion sign. So Eₓ = −(1/2V) Σ_σ Σ_q v(q) Σₖ n(k) n(k + q), with n(k) = 1 inside the sphere and 0 outside, derived. Three features are already visible. It is negative, because v(q) and the overlap counts are positive. Only parallel spins contribute, so the unpolarised gas gives two equal spin sums. And it is first order in e²: the Hartree-Fock energy of jellium is exactly the free sea's energy plus first-order perturbation theory, nothing more.

For fixed q, Σₖ n(k) n(k + q) counts states inside both the sphere |k| ≤ k_F and its copy displaced by q (centred on −q; only |q| matters): (V/(2π)³) times their shared volume. Elementary geometry gives that lens as (4π/3)k_F³ f(y), with y = q/2k_F and f = (2 + y)(1 − y)²/2 = 1 − 3y/2 + y³/2, falling from 1 to 0 at q = 2k_F. Turn Σ_q into V∫d³q/(2π)³ with d³q = 4πq² dq: the q² cancels the 1/q² of v(q), so small q is harmless, and Eₓ = −V(e²/ε₀)(4π/3)k_F³ × 4π × 2k_F ∫₀¹ f dy/(2π)⁶. The integral is 1 − 3/4 + 1/8 = 3/8, so Eₓ/V = −(e²/(4πε₀))k_F⁴/(4π³). Divide by n = k_F³/(3π²): εₓ = −(3/(4π))e²k_F/(4πε₀), derived. With e²/(4πε₀) = 2 Ry a₀ and k_F a₀ = (9π/4)¹⁄³/rₛ = 1.919/rₛ, εₓ = −(3/(2π)) × 1.919/rₛ Ry = −0.916/rₛ Ry. A bookkeeping check: the Fock exchange term, averaged over the sea, is 2εₓ, because every pair is counted from both ends.

Ask how likely an electron of spin σ is to sit a distance s from another of the same spin. With ψ_σ(r) = Σₖ c(k, σ)e^(ik⋅r)/√V, the pair density ⟨ψ†_σ(r)ψ†_σ(r′)ψ_σ(r′)ψ_σ(r)⟩ has two Wick pairings and equals n_σ² − |ρ_σ(s)|², where s = r − r′ and ρ_σ(s) = ⟨ψ†_σ(r′)ψ_σ(r)⟩ = (1/V)Σₖ n(k)e^(ik⋅s) is the one-body density matrix. Over a sphere the sine parts cancel, and ρ_σ(s) = (sin x − x cos x)/(2π²s³) = n_σ × 3(sin x − x cos x)/x³, with x = k_F s and n_σ = k_F³/(6π²). So g↑↑ = 1 − 9(sin x − x cos x)²/x⁶, derived. At contact g↑↑ ≈ x²/5: two parallel-spin fermions have an odd relative wavefunction, linear in s, so their probability grows as s². Far out 1 − g↑↑ ≈ 9cos²x/x⁴, ripples from the sharp Fermi surface that touch 1 where tan x = x and never pass it. Opposite spins have no contraction at all, so g↑↓ = 1: one determinant leaves them uncorrelated.

Integrate the deficit: n_σ∫[1 − g↑↑]d³s = (1/n_σ)∫|ρ_σ(s)|²d³s. By Parseval, ∫|ρ_σ|²d³s = (1/V)Σₖ n(k)², and an occupation is 0 or 1, so n(k)² = n(k) and the sum is N_σ/V = n_σ: the hole holds exactly one electron, derived. That is the density matrix of a determinant being a projector, so it holds at every rₛ and contains no e². The energy follows. The background removes the uncorrelated n² part of the pair density, leaving εₓ = −½∫d³s n_σ[1 − g↑↑(s)] e²/(4πε₀s): each electron's pull on its own hole, halved because each pair is shared. With the standard integral ∫₀^∞ j₁(x)² dx/x = 1/4, where j₁ = (sin x − x cos x)/x², this returns −(3/(4π))e²k_F/(4πε₀), the same result by a second route. Since ½ × 2 Ry a₀ = Ry a₀, it reads εₓ = −Ry a₀⟨1/s⟩, so the hole's mean inverse distance is 0.916/(rₛ a₀): one electron's worth of missing charge, spread over a radius of order rₛ a₀.

Dimensions: e²/(4πε₀) is J m and k_F is m⁻¹, so εₓ is in J. Limits: against topic 15.02's kinetic energy 2.21/rₛ² Ry (quoted), |εₓ|/Eₖᵢₙ = 0.415 rₛ, which vanishes as rₛ → 0, so dense electrons are nearly free; switch e² off and εₓ goes while the hole, pure statistics, stays. Measured: sodium's bcc lattice constant, 4.225 Å at 5 K, with two atoms per cubic cell and one electron per atom, gives rₛ = 3.93, so εₓ = −3.17 eV against Eₖᵢₙ = +1.95 eV, while the measured cohesive energy is 1.11 eV per atom. Exchange is the largest attractive term in the gas and nearly three times the binding, so no account of a metal may treat it as small; the balance is topic 15.04. Where it stops: g↑↓ = 1 is wrong in the real gas, where antiparallel electrons avoid each other too, and quantum Monte Carlo puts the correlation energy near −0.064 Ry at rₛ = 4 (quoted), 28% of εₓ there. Overhauser's state is a second warning: the plane-wave determinant is a solution, not the best one.