From the Pauli field to occupation numbers on any basis

Work in SI units; four-vectors appear only when recalling Unit 9, with the metric (+, −, −, −). Unit 9's Dirac field holds bₚₛ, which removes an electron, and d†ₚₛ, which creates a positron, so a Hamiltonian built from it has pieces b†d† that make a pair and db that remove one, each linking states at least 2mc² = 1.022 MeV apart. Atoms come nowhere near: ionising hydrogen costs 13.6 eV, 1.33 × 10⁻⁵ of the threshold, and even lead's 1s electrons, bound by 88.0 keV, reach only 0.086. Energy conservation forbids real pairs, and we drop the virtual ones too, the no-pair approximation. What remains commutes with N = Σ b†ₚₛ bₚₛ, so Fock space splits into sectors of fixed electron number, and a neutral atom of nuclear charge Z lives entirely in the sector N = Z. The neglect has a measured price, quoted rather than derived: vacuum polarisation, the leading virtual-pair effect, lowers hydrogen's 2S level by about 27 MHz, 2.6% of the measured 1057.8 MHz Lamb shift.

With the d's gone only positive-energy modes remain, and Unit 6's Foldy-Wouthuysen rotation separates their two large components from two small ones, smaller by about |p|/2mc. That reduction is quoted here, not repeated: the electron field becomes a two-component Pauli field. In a box of volume V with periodic walls the momenta are discrete, bₚₛ is dimensionless with Unit 9's anticommutator, and ψₛ(r) = (1/√V) Σₚ e^(ip⋅r/ħ) bₚₛ carries m^(−3/2), so ψ†ψ is an electron density in m⁻³ whose integral is N. The free-field limit is exact: each mode has energy √(p²c² + m²c⁴) = mc² + p²/2m − p⁴/(8m³c²) + …, the constant mc²N set aside and the next term Unit 6's kinetic correction. With a potential, Unit 6's spin-orbit and Darwin terms enter as one-body operators. The expansion parameter for a 1s electron is Zα: 0.0073 in hydrogen but 0.60 in lead, where two components stop being adequate although N is still conserved.

First quantisation labels particles: particle 1 in orbital i₁, particle 2 in i₂, and so on. For identical particles symmetrisation erases those labels, and what survives is how many particles sit in each orbital. Choose any orthonormal one-particle set φ₁, φ₂, … (spin-orbitals, for electrons) and fix its order once. The ket |n₁, n₂, …⟩ has nᵢ particles in φᵢ, with nᵢ = 0, 1, 2, … for bosons and 0 or 1 for fermions, and N = n₁ + n₂ + …, while the vacuum |0⟩ has every nᵢ = 0; Fock space is the sum of all the fixed-N sectors. Counting tests the bookkeeping: M orbitals hold C(M, N) fermion states of N particles but C(M + N − 1, N) boson states. Carbon's 2p² configuration has M = 6 spin-orbitals and N = 2, so 15 fermion states, exactly the 9 + 5 + 1 sublevels of the measured ³P, ¹D and ¹S terms, the last two lying 1.26 eV and 2.68 eV above the ground level. Bosons would have given 21 states, and distinguishable particles 36.

Write QM-II's symmetrised state with the 1/√N! a Slater determinant carries: |i₁ … i_N⟩_S = (1/√N!) Σ_P P|i₁⟩|i₂⟩ … |i_N⟩, summed over all N! permutations P of the particles. Its squared norm counts the permutations that leave the list unchanged, n₁! n₂! …, so the normalised ket is |n₁, n₂, …⟩ = |i₁ … i_N⟩_S / √(n₁! n₂! …). Define a†ₖ as adding k to the list, |i₁ … i_N⟩_S → |k i₁ … i_N⟩_S. Only nₖ! changes, to (nₖ + 1)!, so a†ₖ|…, nₖ, …⟩ = √(nₖ + 1)|…, nₖ + 1, …⟩. The adjoint lowers with √nₖ, a†ₖ aₖ counts nₖ, and aₖ a†ₗ − a†ₗ aₖ = δₖₗ, Unit 9's algebra for scalar modes. For occupations (2, 1, 0), a†₁ gives √3|3, 1, 0⟩: a third boson joins a doubly occupied orbital with three times the probability of joining an empty one. The factor is measured: two photons meeting at a 50:50 beam splitter leave together, each exit with probability |√2/2|² = 1/2, the Hong-Ou-Mandel effect.

Antisymmetrise instead: |i₁ … i_N⟩_A = (1/√N!) Σ_P (−1)^P P|i₁⟩ … |i_N⟩ is QM-II's Slater determinant, normalised when the orbitals differ and zero when one repeats. Let |n₁, n₂, …⟩ be the determinant with its occupied orbitals listed in the fixed order, and define c†ₖ as adding k as the top row. To restore the order, row k must be swapped past every occupied orbital before it, each swap of two rows flips a determinant's sign, and if k is already occupied two rows coincide. Hence c†ₖ|…, nₖ, …⟩ = (−1)^{(n₁ + … + nₖ₋₁)}(1 − nₖ)|…, nₖ + 1, …⟩, and its adjoint cₖ carries the same sign. Take orbital k before orbital l: c†ₗ sees k occupied only if c†ₖ acted first, so the two orders differ by −1 and c†ₖ c†ₗ = −c†ₗ c†ₖ, with (c†ₖ)² = 0. The same count gives cₖ c†ₗ = −c†ₗ cₖ for k ≠ l, since only one of the two operators sees the other's orbital, and with cₖ c†ₖ + c†ₖ cₖ = (1 − nₖ) + nₖ = 1 the algebra is complete. For example, c†₂|1, 0, 1, 1⟩ = −|1, 1, 1, 1⟩, since one occupied orbital precedes orbital 2.

Neither derivation asked what the orbitals are, only that they are orthonormal and ordered. So cᵢ c†ⱼ + c†ⱼ cᵢ = δᵢⱼ holds for plane waves, where it is Unit 9's algebra for bₚₛ, for hydrogenic orbitals, and for the Hartree-Fock orbitals this unit will compute. Orthonormality matters. A determinant is linear in each row, so creating a general one-particle state g gives c†(g) = Σₖ ⟨k|g⟩ c†ₖ, and the anticommutator of c(f) with c†(g) is the overlap ⟨f|g⟩. For two hydrogen 1s orbitals 1.4 a₀ apart that overlap is 0.753, not 0, which is why such orbitals are orthogonalised before use; topic 10.02 develops the change of basis. The order is a convention: renumbering the orbitals multiplies each ket by a fixed ±1, which changes the signs of individual matrix elements but no probability or energy. Building QM-II's determinant row by row gives c†₁ c†₂ … c†_N|0⟩ = |1 2 … N⟩_A, and N = 1 returns c†ᵢ|0⟩ = |φᵢ⟩, the one-particle limit, with no sign left to track.