One-body operators: the sum of ⟨i|f|j⟩ c†ᵢ cⱼ

SI units throughout; x stands for position and spin together, and nothing here is relativistic, so Unit 9's (+, −, −, −) metric never enters. Order a set of orthonormal spin-orbitals once and for all, and write a determinant as an ordered product of creators on |0⟩, with topic 10.01's anticommutators quoted. Act with c†ᵢ cⱼ. If orbital j is empty, cⱼ reaches |0⟩ and gives zero. If j stands in slot k, cⱼ anticommutes past k − 1 creators, a sign (−1)^(k−1), and removes c†ⱼ; c†ᵢ then needs k − 1 swaps to reach slot k, the same sign again. The signs cancel: c†ᵢ cⱼ writes i into j's slot and leaves every other orbital where it was. If i is another occupied orbital, c†ᵢ appears twice and the result vanishes, the Pauli principle; if i = j the state comes back, so c†ⱼ cⱼ = nⱼ counts. With three orbitals, c†₃ c₁ c†₁ c†₂|0⟩ = c†₃ c†₂|0⟩ = −c†₂ c†₃|0⟩: the minus sign appears only when the result is put back into the standard order.

Now apply F⁽ᴺ⁾ = Σₙ f(n) to the Slater determinant of the same orbitals, (1/√N!) det[φ(xₙ)]. Each term of the determinant has one orbital per coordinate and f(n) acts on one factor, so F⁽ᴺ⁾ returns a sum of N determinants, the k-th with its column-k orbital φⱼ replaced by fφⱼ. Insert completeness, fφⱼ = Σᵢ φᵢ ⟨i|f|j⟩. A column equal to another occupied orbital kills its determinant; otherwise orbital i stands in j's slot with the rest untouched. That is, term for term and sign for sign, what Σᵢⱼ ⟨i|f|j⟩ c†ᵢ cⱼ produced. Determinants of a complete orthonormal set span every N-particle space, so equal action on all of them makes the operators equal: derived. Three checks. c†ᵢ cⱼ is dimensionless, so F has the unit of f. For N = 1, ⟨0|cₖ F c†ₗ|0⟩ = ⟨k|f|l⟩, the one-particle operator itself. And F|0⟩ = 0: an empty box has no kinetic energy. Nothing in F mentions N, so one operator serves every N and conserves it, each term holding one creator and one annihilator.

Substituting ψ(x) = Σᵢ φᵢ(x) cᵢ from topic 10.02 folds the double sum into F = ∫ dx ψ†(x) f ψ(x); since ⟨i|f|j⟩ and c†ᵢ cⱼ change by inverse unitary matrices, F is the same in every basis. Kinetic energy is T = ∫ dx ψ†(−ħ²∇²/2m)ψ, diagonal in plane waves: T = Σ (ħ²k²/2m) c†(k, σ) c(k, σ). A potential U(r) gives V = ∫ d³r U(r) n(r), where n(r) = Σ_σ ψ†_σ(r) ψ_σ(r) is the second-quantised Σₙ δ³(r − rₙ), in m⁻³. In a box of volume Ω, V = (1/Ω) Σ Ũ(q) c†(k + q, σ) c(k, σ), with Ũ the Fourier transform of U: the potential scatters one electron by q at a time. On a determinant only diagonal terms survive, so ⟨Φ|F|Φ⟩ = Σ over occupied i of ⟨i|f|i⟩. Helium's effective-charge determinant, Z_eff = 27/16, has ⟨T⟩ = 2(27/16)² × 13.606 eV = 77.5 eV. The virial theorem, quoted, turns the measured total energy, −79.0 eV, into ⟨T⟩ = 79.0 eV: this determinant misses 1.5 eV of kinetic energy.

Spin is one-body too. A spin-orbital is a spatial orbital times ↑ or ↓, with single-particle elements ⟨a↑|s_z|a↑⟩ = ħ/2, ⟨a↓|s_z|a↓⟩ = −ħ/2 and ⟨a↑|s₊|a′↓⟩ = ħ⟨a|a′⟩, so S_z = (ħ/2) Σₐ (nₐ↑ − nₐ↓) and S₊ = ħ Σₐ c†ₐ↑ cₐ↓, derived. Put two electrons in one spatial orbital, Φ = c†ₐ↑ c†ₐ↓|0⟩. Only the a term of S₊ acts: cₐ↓ anticommutes past c†ₐ↑ and removes c†ₐ↓, leaving −c†ₐ↑ c†ₐ↑|0⟩ = 0. So S₊Φ = 0 and S_zΦ = 0, and S² = S₋S₊ + S_z² + ħS_z gives S²Φ = 0: a singlet, which written out is Quantum Mechanics II's φ(r₁)φ(r₂)(↑↓ − ↓↑)/√2, recovered as a check. The same cancellation makes every closed shell a singlet. Experiment agrees: helium's 1s² ground state is ¹S₀ with no triplet partner, and its lowest triplet level, 1s2s ³S₁, lies 19.82 eV higher. Different orbitals behave differently: S₊ c†ᵤ↑ c†ᵥ↓|0⟩ = ħ c†ᵤ↑ c†ᵥ↑|0⟩, whose squared norm ħ²(1 − |⟨u|v⟩|²) vanishes only when v is u itself, up to a phase.

Every term of F moves at most one electron, so its matrix elements between determinants of one orthonormal set take one of three forms, fixed by how many spin-orbitals differ, derived. None: ⟨Φ|F|Φ⟩ = Σ over occupied i of ⟨i|f|i⟩. One, a replaced by r in a's slot: ⟨r|f|a⟩. Two or more: zero, however large the single-particle elements are. The electric-dipole operator D = −e Σₙ rₙ is one-body and blind to spin, so between single determinants a photon moves one electron, keeps its spin, and needs a non-zero one-electron element. Helium bears this out. Its strongest absorption line from the ground state is at 58.4 nm, 21.22 eV: 1s² → 1s2p ¹P, one orbital changed. A one-photon jump to 2s2p changes two orbitals and has no amplitude between determinants, yet Madden and Codling found it in 1963, a resonance near 60.1 eV in helium's synchrotron absorption, because neither state is really a single determinant. Any intensity in a two-electron jump measures correlation, the part of the wavefunction one determinant leaves out.

From the anticommutators, [c†ᵢ cⱼ, c†ₖ cₗ] = δⱼₖ c†ᵢ cₗ − δᵢₗ c†ₖ cⱼ, derived in two lines: the rule |i⟩⟨j| obeys. So the commutator of two one-body operators is the one-body operator of [f, g], and second quantisation preserves commutators. The components of S obey [Sₓ, S_y] = iħS_z for every N with no further work, and [T, X] for X = Σₙ xₙ is the one-body operator of [p²/2m, x] = −iħp/m. Products differ. In FG, writing cⱼ c†ₖ = δⱼₖ − c†ₖ cⱼ splits Σ fᵢⱼ gₖₗ c†ᵢ cⱼ c†ₖ cₗ into a one-body piece with matrix fg and a remainder Σ fᵢⱼ gₖₗ c†ᵢ c†ₖ cₗ cⱼ that moves two electrons at once, the second-quantised sum of f(n)g(m) over n ≠ m. S² = S⋅S is the case that matters: its pair part, 2 Σ sₙ⋅sₘ over n < m, is two-body, which is why a determinant can be an eigenstate of S_z and not of S². Operators acting on pairs are topic 10.04's.