Field operators, change of basis and the wavefunction

Units are SI, and nothing here is relativistic, so the course's metric (+, −, −, −) never enters. Let x = (r, σ) and y = (r′, σ′), with ∫dx = Σ_σ ∫d³r and δ(x − y) = δ³(r − r′)δ_σσ′ in m⁻³. Topic 10.01 gave operators cᵢ for an orthonormal set φᵢ with [cᵢ, cⱼ†]₊ = δᵢⱼ, where [A, B]₊ ≡ AB + BA. Assemble ψ(x) = Σᵢ φᵢ(x)cᵢ. Then [ψ(x), ψ†(y)]₊ = Σᵢ φᵢ(x)φᵢ*(y), which completeness makes δ(x − y), and [ψ(x), ψ(y)]₊ = 0 likewise, both derived. Only completeness was used, so every complete orthonormal set gives the same algebra. Dimensions agree: cᵢ is a pure number and φᵢ carries m⁻³⁄², so both sides are in m⁻³. On the vacuum, ψ†(x)|0⟩ = Σᵢ φᵢ*(x)|φᵢ⟩ = |x⟩, a particle at r with spin σ, and in the one-particle limit ⟨0|ψ(x)cᵢ†|0⟩ = φᵢ(x): an orbital is the field's vacuum-to-one-particle amplitude. For hydrogen's 1s state at the nucleus that amplitude is (πa₀³)⁻¹⁄² = 1.47 × 10¹⁵ m⁻³⁄².

Any normalised orbital f gets a creation operator c†(f) = ∫dx f(x)ψ†(x), and for f = φᵢ orthonormality returns cᵢ†. Apply it to a second complete orthonormal set χₖ: aₖ† = ∫dx χₖ(x)Σᵢ φᵢ*(x)cᵢ† = Σᵢ cᵢ†⟨φᵢ|χₖ⟩. That is the matrix Uᵢₖ = ⟨φᵢ|χₖ⟩ that expands the kets, |χₖ⟩ = Σᵢ |φᵢ⟩Uᵢₖ, and the annihilators take its conjugate, aₖ = Σᵢ Uᵢₖ*cᵢ, derived. The algebra survives because U is unitary: [aₖ, aₗ†]₊ = Σᵢ Uᵢₖ*Uᵢₗ = (U†U)ₖₗ = δₖₗ, and Σₖ χₖ(x)aₖ = Σᵢ φᵢ(x)cᵢ, so ψ(x) itself never changed. A change of orbitals is a change of variables on Fock space, a canonical transformation. The orbitals must be orthonormal. Two hydrogen 1s spin-orbitals of the same spin, at H₂'s measured bond length, 0.741 Å = 1.40 a₀, overlap by the quoted S = e^(−R/a₀)(1 + R/a₀ + R²/(3a₀²)) = 0.753, so their operators obey [a_A, a_B†]₊ = 0.753, not 0: they are not independent modes until orthogonalised.

Now rotate only the N occupied orbitals among themselves, with an N × N unitary U. Expanding a₁†a₂†⋯a_N† gives a sum over every way of choosing one cᵢ† from each factor. A term that picks an orbital twice vanishes, since (cᵢ†)² = 0, and every surviving term is a reordering of c₁†⋯c_N†, which costs the sign of the permutation, so the coefficients assemble into a determinant: a₁†⋯a_N†|0⟩ = det U c₁†⋯c_N†|0⟩, derived. With |det U| = 1, the state changes by a phase at most: the orbitals were rearranged, the state was not. The same expansion gives the overlap of any two determinants, det S with Sᵢₖ = ⟨φᵢ|χₖ⟩ over the occupied orbitals. Mixing an occupied orbital with an empty one is different: replacing φ₂ by cos β φ₂ + sin β φ₃ gives det S = cos β, so at β = 20° the new determinant keeps 0.940 of the old amplitude and 0.883 of its probability. That is why Hartree-Fock in topic 10.06 can fix only the occupied subspace, and why canonical and localised orbitals are one state. For bosons the overlap is a permanent instead, with no signs, and the terms that pick an orbital twice survive, so even a rotation among occupied orbitals changes a boson state.

Take |Ψ⟩ = c₁†c₂†|0⟩. The field ψ(x₁) = Σᵢ φᵢ(x₁)cᵢ removes a particle from either orbital, with a sign for moving c₂ past c₁†: ψ(x₁)|Ψ⟩ = φ₁(x₁)c₂†|0⟩ − φ₂(x₁)c₁†|0⟩. A second field reaches the vacuum, so ⟨0|ψ(x₂)ψ(x₁)|Ψ⟩ = φ₁(x₁)φ₂(x₂) − φ₂(x₁)φ₁(x₂), and dividing by √2 gives exactly Quantum Mechanics II's normalised Slater determinant. In general Ψ(x₁, …, x_N) = ⟨0|ψ(x_N)⋯ψ(x₁)|Ψ⟩/√(N!), derived, with inverse |Ψ⟩ = (N!)⁻¹⁄² ∫dx₁⋯dx_N Ψ ψ†(x₁)⋯ψ†(x_N)|0⟩. Antisymmetry is now a theorem: swapping two arguments reorders two anticommuting fields at the cost of −1, and ψ(x)² = 0 makes Ψ vanish when two arguments coincide. Dimensions check, Ψ carrying m^(−3N/2), and so does measurement: the field gives c†(1s↑)c†(2s↑)|0⟩ a spatially antisymmetric wavefunction, which keeps the electrons apart, and helium's 1s2s triplet lies 0.796 eV below the singlet, at 19.820 against 20.616 eV above the ground state.

Without interactions, use the orbitals that diagonalise h = −ħ²∇²/2m + u(r), with u(r) the one-body potential energy, so hφᵢ = εᵢφᵢ and H = Σⱼ εⱼcⱼ†cⱼ, the non-relativistic form of Unit 9's mode sum. The identity [A, BC] = [A, B]₊C − B[A, C]₊ gives [cᵢ, cⱼ†cⱼ] = δᵢⱼcᵢ, so the Heisenberg equation iħ dcᵢ/dt = [cᵢ, H] = εᵢcᵢ has cᵢ(t) = cᵢe^{(−iεᵢt/ħ)}. Then ψ(x, t) = Σᵢ φᵢ(x)e^{(−iεᵢt/ħ)}cᵢ and iħ∂ψ/∂t = hψ(x, t), derived: the field operator obeys exactly the equation a one-particle wavefunction obeys, and since ⟨0|ψ(x, t) = ⟨0|ψ(x)e^(−iHt/ħ), every argument of the recovered Ψ evolves by its own h. Commutators give the same δᵢⱼcᵢ through [A, BC] = [A, B]C + B[A, C], so bosons and fermions share their dynamics and differ only in the algebra. The one-particle limit meets experiment in hydrogen, where εₙ = −13.6 eV/n² gives the 10.2 eV line measured at 121.6 nm. Topic 10.04's interaction adds ∫dy ψ†(y)V(x − y)ψ(y)ψ(x), quoted here, and the equation stops being linear in ψ.

Every calculation keeps finitely many orbitals. With M of them, ψ_M(x) = Σᵢ₌₁ᴹ φᵢ(x)cᵢ still has [cᵢ, cⱼ†]₊ = δᵢⱼ exactly, but the field anticommutator becomes P_M(x, y) = Σᵢ₌₁ᴹ φᵢ(x)φᵢ*(y), the kernel of the projector onto the span. It is idempotent, its trace ∫dx P_M(x, x) is M, and it reproduces a function only if the function lies in the span. For sine orbitals in a box of width L, P_M(L/2, L/2) is 2/L times the number of odd n ≤ M: a peak of height near M/L and width near L/M that becomes δ only as M → ∞. Inside the span, everything above is exact algebra; the error is what the span leaves out, and it differs from correlation. Helium shows the split: the best single hydrogen-like 1s function, e^{(−Z′r/a₀)} with Z′ = 27/16, gives Quantum Mechanics II's −77.49 eV, the complete-basis Hartree-Fock limit of topic 10.08 is −77.87 eV, and the measured energy is −79.0 eV, so 0.38 eV is basis error and about 1.1 eV is correlation, which no basis removes from one determinant.