Manuscript X-C@3 · X-C-v3.tex · section · 2260f5f8

Two diagnostics

Two diagnostics · lines 1503–1544

Take $ψ = ψ (t)$ , with smooth support away from zero, and $g = g_{0} (t)$ near the relevant causal compact. Spatial cutoffs outside that compact make the backgrounds and switch admissible without affecting the evaluation. Define

Ψ = \int_{0}^{t} σ_{ε} ψ d s, Ψ_{2} = \int_{0}^{t} (σ_{ε} ψ)^{2} d s, Γ = \int_{0}^{t} g_{0} ψ d s, Γ_{2} = \int_{0}^{t} (g_{0} ψ)^{2} d s .

On the smooth free solution $u = t$ ,

X_{1}^{0} = - Ψ, X_{2}^{0} = Ψ_{2}, ρ_{1, g} = - Γ, ρ_{2, g} = Γ_{2},

and the spatial coordinate coefficients vanish in the observation region.

The bare second dressing gives $Ψ σ_{ε} ψ - Ψ_{2}$ ; the mixed term gives $- Ψ g_{0} ψ$ ; the iteration gives $Γ_{2}$ . Consequently

⟨ e_{2}, t ⟩ = \int f (Γ_{2} - Ψ_{2} + Ψ ψ (σ_{ε} - g_{0})) d^{4} x .

The first coefficient similarly gives $\int f (Ψ - Γ) d^{4} x$ . Both numbers vanish if $g_{0} ψ = σ_{ε} ψ$ throughout the contributing past interval.

These are classical evaluations of the fixed representatives, not elements of $A_{0}$ , states on the Wick algebra, or quantum expectation values. The calculation checks signs and the retained iteration; it is not an extension or Ward proof.

Displayed formulas in this passage

unlabeled · line 1510
$Ψ = \int_{0}^{t} σ_{ε} ψ d s, Ψ_{2} = \int_{0}^{t} (σ_{ε} ψ)^{2} d s, Γ = \int_{0}^{t} g_{0} ψ d s, Γ_{2} = \int_{0}^{t} (g_{0} ψ)^{2} d s .$
unlabeled · line 1517
$X_{1}^{0} = - Ψ, X_{2}^{0} = Ψ_{2}, ρ_{1, g} = - Γ, ρ_{2, g} = Γ_{2},$
unlabeled · line 1528
$⟨ e_{2}, t ⟩ = \int f (Γ_{2} - Ψ_{2} + Ψ ψ (σ_{ε} - g_{0})) d^{4} x .$

Nearby in this section

section: Two diagnostics