The Hartree-Fock spectrum and its singularity at k_F

Work in SI units, with energies also in rydbergs (1 Ry = 13.606 eV) and lengths in Bohr radii (a₀ = 52.92 pm); everything here is non-relativistic, so no metric signature enters. Built from occupied plane waves, the Fock operator commutes with every translation, so its eigenfunctions are plane waves again: they are self-consistent from the start and nothing needs iterating. The Hartree potential of the uniform electron density is the q = 0 term that topic 15.01 cancelled against the background, so what is left is exchange. Momentum conservation makes the exchange operator diagonal in k, and the Fock operator returns e^(ik⋅r) multiplied by εₖ = ħ²k²/2m + Σₓ(k), with Σₓ(k) = −(1/V) Σ e²/(ε₀|k − k′|²) summed over the occupied k′ of the same spin. By Koopmans' theorem, quoted from Unit 10, −εₖ is the frozen-orbital cost of removing the electron in state k, the quantity photoemission measures.

Replace (1/V)Σ by ∫d³k′/(2π)³ over the sphere k′ ≤ k_F. The angular integral is elementary: ∫dΩ′/(k² + k′² − 2kk′ cos θ) = (2π/(kk′)) ln|(k + k′)/(k − k′)|. The radial one, from 0 to k_F, is ∫k′ ln|(k + k′)/(k − k′)| dk′ = kk_F + ((k_F² − k²)/2) ln|(k + k_F)/(k − k_F)|, and together they give Σₓ(k) = −(2e²k_F/(4π²ε₀)) F(k/k_F). Dimensions: e²/ε₀ is J m and k_F is m⁻¹, so the prefactor is an energy, (4/π)(k_F a₀) Ry = 2.444/rₛ Ry, or 8.46 eV for sodium. Limit: as x → ∞, F → 1/(3x²) and Σₓ → −n_σe²/(ε₀k²): a fast electron finds every same-spin electron nearly at rest, each exchange transfers its whole momentum k, and each contributes −e²/(ε₀k²V). Topic 15.03: F averages 3/4 over the sea, and half the average Σₓ, half because summing εₖ counts every pair twice, is −0.916/rₛ Ry.

Differentiate: dF/dx = 1/(2x) − ((1 + x²)/(4x²)) ln|(1 + x)/(1 − x)|, which runs to −∞ as x → 1. Near the Fermi surface the group velocity v = (1/ħ) dεₖ/dk is therefore v/v_F = 1 + 0.166 rₛ [ln(2k_F/|k − k_F|) − 1], in units of the free v_F = ħk_F/m. For sodium the coefficient is 0.652: v = 3.82 v_F at 0.99 k_F, 5.31 v_F at 0.999 k_F, and 1.50 more for each further decade. The origin is geometric. ∂Σₓ/∂k weights each occupied k′ by the gradient of 1/q², which falls as 1/q³. Deep inside the sea the occupied neighbourhood of k is a whole ball, and the small-q contributions cancel in opposite pairs; on the Fermi surface half that ball is empty, and the uncancelled half gives ∫q² dq/q³ = ∫dq/q, a logarithm. An interaction that stays finite at q = 0 makes the same integral converge.

Define m* by dεₖ/dk = ħ²k/m*. The density of states per unit energy is g(ε) = Vk²/(π² dεₖ/dk) for both spins, so g/g_free = m*/m at the same k, and an infinite slope makes g(ε_F) = 0, vanishing as 1/ln. Thermal excitations sample states within k_BT of ε_F, which lie at |k − k_F| ≈ k_BT/(ħv), so the effective density of states falls as 1/ln(T₀/T), with T₀ of order ε_F/k_B, and the electronic specific heat goes as T/ln(T₀/T). Every simple metal shows instead C = γT with γ constant at low temperature; sodium's γ is 1.38 mJ mol⁻¹ K⁻², quoted, while Hartree-Fock predicts about a tenth of the free 1.09 near 1 K. The band fares no better. The exchange widening Σₓ(k_F) − Σₓ(0) = 1.222/rₛ Ry is derived, so sodium's band is 3.24 + 4.23 = 7.47 eV wide; angle-resolved photoemission (Jensen and Plummer, 1985) measures 2.65 eV, quoted: narrower than free, not 2.3 times wider.

Replace the bare interaction by e²/(ε₀(q² + kₛ²)). The angular integral becomes (π/(kk′)) ln[((k + k′)² + kₛ²)/((k − k′)² + kₛ²)], finite at k′ = k, and the radial integral closes again: Σₓ = −(2e²k_F/(4π²ε₀)) G(x, y), with y = kₛ/k_F and G = 1/2 + ((1 − x² + y²)/(8x)) ln[((1 + x)² + y²)/((1 − x)² + y²)] − (y/2)[arctan((1 + x)/y) + arctan((1 − x)/y)], which is F at y = 0. Its slope at k_F is finite: v/v_F = 1 + 0.332 rₛ [((2 + y²)/8) ln(1 + 4/y²) − 1/2]. For small y it reduces to the bare formula with |k − k_F| replaced by kₛ, so screening does not remove the logarithm's growth, it stops it at the screening wavenumber. Weak screening cures slowly: with kₛ = 0.05 k_F, sodium's m*/m at k_F is still only 0.36.

Why should the interaction be screened? In the exchange term two electrons interact through the bare e²/(ε₀q²), as though the rest of the gas stayed frozen while they exchanged. Real electrons of both spins move in response, so any slowly varying charge is surrounded, within a distance of order 1/kₛ, by a compensating cloud, and for q ≪ kₛ the interaction is cut off. That rearrangement is correlation, a mixing of many determinants, so one determinant cannot express it and the cure lies outside Hartree-Fock. Thomas-Fermi screening, the static long-wavelength response of the gas, is quoted here: kₛ² = 4k_F/(πa₀), or kₛ = 1.61 k_F for sodium. With it, sodium's m*/m at k_F is 0.956, independent of temperature, so the linear specific heat returns, and the bandwidth falls from 7.47 eV to 3.43 eV. That is still above the measured 2.65 eV: the narrowing below free electrons is a further correlation effect, quoted and not derived. Total energies were never in danger, because ln|k − k_F| is integrable, so the −0.916/rₛ Ry of topic 15.03 stands.