Manuscript VI@1 · VI.tex · proposition · b5b60afb

Proposition: Exact external-response comparison

Bare dressing, solution coefficients, and external response › The compactly switched minimal scalar vertex · lines 1298–1339 · proof included or linked

VI@1#prop:external-comparison, VI@1#eq:external-duhamel, VI@1#eq:Szero, VI@1#eq:external-smearing

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For $P ϕ_{0} = 0$ ,

ϕ_{1, g} [h, ϕ_{0}] = S_{0} (g h) ϕ_{0},

where

(S_{0} (a) ϕ_{0}) (t) = \int_{0}^{t} \frac{sin (( t - s ) Ω _{m} )}{Ω _{m}} L_{a} ϕ_{0} (s) d s .

For $u_{f} = - G_{A} f$ , its smearing is

\int f ϕ_{1, g} d^{4} x = \int (Q_{g h} u_{f}) ϕ_{0} d^{4} x .

Proof. Equation (eq:QL) gives $Q_{g h} ϕ_{0} = L_{g h} ϕ_{0}$ . Both sides of (eq:external-duhamel) have zero Cauchy data at $t = 0$ and solve $P ψ = - L_{g h} ϕ_{0}$ . Klein–Gordon uniqueness proves equality.

For the smeared identity, use transposition $G_{R}^{*} = G_{A}$ , then formal self-adjointness of $Q_{g h}$ :

- \int f G_{R} Q_{g h} ϕ_{0} = - \int (G_{A} f) Q_{g h} ϕ_{0} = \int (Q_{g h} u_{f}) ϕ_{0} .

All vertex integrations involve compact coefficients.

Displayed formulas in this passage

VI@1#eq:external-duhamel · line 1301
$ϕ_{1, g} [h, ϕ_{0}] = S_{0} (g h) ϕ_{0},$
VI@1#eq:Szero · line 1306
$(S_{0} (a) ϕ_{0}) (t) = \int_{0}^{t} \frac{sin (( t - s ) Ω _{m} )}{Ω _{m}} L_{a} ϕ_{0} (s) d s .$
VI@1#eq:external-smearing · line 1314
$\int f ϕ_{1, g} d^{4} x = \int (Q_{g h} u_{f}) ϕ_{0} d^{4} x .$
unlabeled · line 1331
$- \int f G_{R} Q_{g h} ϕ_{0} = - \int (G_{A} f) Q_{g h} ϕ_{0} = \int (Q_{g h} u_{f}) ϕ_{0} .$

Proposition: Exact external-response comparison

Displayed formulas in this passage

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