Holes in the Fermi sea: a non-relativistic Dirac sea

Units are SI with ħ kept, jellium energies also in rydbergs (1 Ry = 13.606 eV, rₛ in Bohr radii a₀ = 52.92 pm); the metric diag(1, −1, −1, −1) enters only where Unit 8 is recalled. Quoted from topic 15.02: |F⟩ fills every plane wave with |k| ≤ k_F in both spins, k_F = 1.919/(rₛ a₀) and ε_F = 3.683/rₛ² Ry, 3.24 eV for sodium at rₛ = 3.93. Write cₖσ for the electron annihilator, or c(k, σ) when the label carries a sign or a sum. |F⟩ is not their vacuum: cₖσ|F⟩ = 0 only outside the sphere, and inside it is c†ₖσ that gives zero. Unit 9 met this in the Dirac field and solved it by renaming. Inside the sphere define bₖσ = σ c†(−k, −σ), σ = +1 for ↑ and −1 for ↓. Since σ² = 1, bb† + b†b inherits δₖₖ′δσσ′ from the c's, and each b anticommutes with every c outside, a different mode: fermion operators, derived. Now cₖσ|F⟩ = 0 outside, where nothing is, and bₖσ|F⟩ = 0 inside, where Pauli forbids a second electron. |F⟩ is a vacuum.

Momentum, spin and number are sums over occupied states, P = Σ ħk c†ₖσcₖσ, S_z = Σ (σħ/2)c†ₖσcₖσ and N = Σ c†ₖσcₖσ, and |F⟩ has P = 0, S_z = 0. Empty (k, σ) and each sum loses one term: P = −ħk, S_z = −σħ/2, N = −1. The electron's charge is −e and the background +Ne of topic 15.01 is untouched, so the metal now carries +e. Inside the sphere c†c = 1 − b†b turns P into Σ′ ħk c†c + Σ″ ħk b†ₖσbₖσ once −k is relabelled k, so b†ₖσ makes momentum +ħk and spin +σħ/2: the minus signs make the labels the hole's own. The factor σ fixes phases, not quantum numbers. Annihilators rotate as the conjugate spinor, and σ, the iσ_y of time reversal and of Unit 8's ψ^c = iγ²ψ*, restores a standard doublet. Test it: the spin-0 density fluctuation Σ_σ c†(k + q, σ)c(k, σ) becomes −[c†(k + q, ↑)b†(−k, ↓) − c†(k + q, ↓)b†(−k, ↑)], the Clebsch-Gordan singlet; without σ it would read as the m = 0 triplet. Derived.

At fixed N, μN is a constant, so subtracting it changes no physics; it measures energies from the Fermi level. Inside the sphere (εₖ − μ)c†ₖσcₖσ = (εₖ − μ) − (εₖ − μ)b†(−k, −σ)b(−k, −σ), and relabelling gives H₀ − μN = Ω₀ + Σ′(εₖ − μ)c†c + Σ″(μ − εₖ)b†b with μ = ε_F at T = 0, derived. Every operator is now normal-ordered against |F⟩ and both coefficients are |εₖ − μ| ≥ 0: in sodium a particle at 1.2k_F costs 0.44ε_F, a hole at 0.5k_F costs 0.75ε_F, and the two together 1.19ε_F = 3.86 eV, a sum in which μ cancels. The constant is Ω₀ = Σ″(εₖ − ε_F) over both spins, (3/5 − 1)Nε_F = −(2/5)Nε_F, −1.30 eV per electron in sodium. Every term is an energy, and as k_F → 0 the inner sum empties, Ω₀ → 0 and the free particles of an empty box return. Unlike Unit 9's −Σ Eₚ, Ω₀ is finite and physical: it is the T = 0 grand potential −PV, giving the free gas a pressure (2/5)nε_F = 5.5 GPa at sodium's density, which the exchange energy of topic 15.03 partly offsets.

A single hole changes N, so it is what photoemission makes: the photon ejects (k, σ) and the spectrum records μ − εₖ = ε_F(1 − k²/k_F²) as a binding energy below the Fermi level, from 0 at the Fermi surface to ε_F = 3.24 eV at the bottom of sodium's sea. Photoemission finds that bottom near 2.65 eV (quoted): the interactions this model omits narrow the band. In its own momentum K = −k the hole's energy is ε_F − ħ²K²/2m, falling as K grows, so its group velocity −ħK/m = ħk/m is the missing electron's. The absence moves with the state it lacks, as in Unit 8, but that state moves along its momentum, so this hole moves against its own. In fields ħdK/dt = +e(ℰ + v × B), a charge +e, while v changes as an electron's would: charge +e with mass −m responds exactly as charge −e with mass +m. Derived. So sodium's Hall coefficient is negative and close to −1/(ne) = −2.35 × 10⁻¹⁰ m³ C⁻¹ (quoted), whichever carriers one counts.

With N fixed, particles and holes come in equal numbers; the simplest excitation is c†(k + q, σ)c(k, σ)|F⟩, of momentum ħq and energy ε(k + q) − ε(k) = (ħ²/m)(k⋅q + q²/2), with |k| ≤ k_F ≤ |k + q|. The top takes k on the Fermi surface along q: (ħ²/m)(k_F q + q²/2). For q ≤ 2k_F two points of the Fermi surface lie q apart, so pairs reach zero energy; beyond, the floor is (ħ²/m)(q²/2 − k_F q). Derived. For q ≪ k_F the band is 0 to ħv_F q, v_F = 1.07 × 10⁶ m s⁻¹ in sodium; for q ≫ k_F it is centred on the free recoil ħ²q²/2m with half-width ħv_F q, a free electron Doppler-broadened by its Fermi motion. The missing gap is measured: pairs of every small energy make the electronic specific heat linear in T, γ = 1.38 mJ mol⁻¹ K⁻² in sodium against 1.09 here (quoted). Above the band at small q sits the plasmon, ħωₚ = 6.05 eV at rₛ = 3.93, 5.7 eV measured (quoted). As q → 0 every pair costs nothing, so it is no single pair but a coherent sum the Coulomb interaction builds.

Unit 9's bₚₛ, the electron above the gap, corresponds to cₖσ outside the sphere, and its dₚₛ, the positron, to the hole operator bₖσ: same letter, different jobs. The algebra matches, (εₖ − μ)c†c = (μ − εₖ)b†b + constant against −Eₚdd† = Eₚd†d − Eₚ, and so does the removal rule: reversed momentum and spin projection, charge +e. Three things differ. Dirac's sea is bottomless and its constant −Σ Eₚ infinite; the Fermi sea holds N electrons and its constant is −(2/5)Nε_F. Dirac's sea charge is removed by hand; here the ionic background cancels it. Dirac holes cost |E| ≥ mc², so a pair needs 1.022 MeV; a Fermi-sea hole costs 0 to ε_F, a pair as little as you like. One rule survives with its reason inverted: one photon makes no pair in either. A Dirac pair needs E ≥ √((2mc²)² + (ħcq)²), above the photon line ħcq; jellium pairs reach only ħv_F q(1 + q/2k_F), far below it, since c/v_F = 281 in sodium. And the Dirac vacuum looks the same in every frame, while a Fermi sea seen from a moving frame is a displaced sphere carrying current: it fixes the metal's rest frame.