Manuscript X-G@1 · X-G.tex · section · 01566e14

Bare versus completed observables

The one-loop coefficient and its graph inventory › Bare versus completed observables · lines 1317–1352

X-G@1#eq:fullcoefficient

For an actual variational theory, write

V = κ V_{1} + κ^{2} V_{2} + O (κ^{3}),

and let the bare relational insertion be

O = F_{f} + κ O_{1} + κ^{2} O_{2} + O (κ^{3}), F_{f} = \int f ϕ d^{4} x .

At first order,

O_{1} = ϕ (c_{1} [h_{cl} + h]), c_{1} [u] = \partial_{μ} (f X_{1}^{μ} [u]) .

The scalar completion $- ϕ (L_{1, g} [u] E_{m}^{A} f)$ is generated by the scalar response. It must not also be inserted as an independent bare term and then regenerated by the same scalar retarded map.

With

R_{V} (O) = n \geq 0 \sum \frac{1}{n !} R_{n} (V^{\otimes n}; O),

the coefficient at order $κ^{2}$ is

O_{2} + R_{1} (V_{1}; O_{1}) + R_{1} (V_{2}; F_{f}) + \frac{1}{2} R_{2} (V_{1}, V_{1}; F_{f}) .

This identity is combinatorial. It does not identify a density-switched tree coefficient with X-F's source-switched one.

Displayed formulas in this passage

X-G@1#eq:fullcoefficient · line 1345
$O_{2} + R_{1} (V_{1}; O_{1}) + R_{1} (V_{2}; F_{f}) + \frac{1}{2} R_{2} (V_{1}, V_{1}; F_{f}) .$
unlabeled · line 1320
$V = κ V_{1} + κ^{2} V_{2} + O (κ^{3}),$
unlabeled · line 1324
$O = F_{f} + κ O_{1} + κ^{2} O_{2} + O (κ^{3}), F_{f} = \int f ϕ d^{4} x .$
unlabeled · line 1329
$O_{1} = ϕ (c_{1} [h_{cl} + h]), c_{1} [u] = \partial_{μ} (f X_{1}^{μ} [u]) .$
unlabeled · line 1340
$R_{V} (O) = n \geq 0 \sum \frac{1}{n !} R_{n} (V^{\otimes n}; O),$