Manuscript XIII-A@1 · XIII-A.tex · proposition · f56e91c7

Proposition: The bγ scalar

A3: the coefficientwise restriction kernel › The independent auxiliary smearing · lines 1001–1043 · proof included or linked

XIII-A@1#prop:commutator, XIII-A@1#eq:commutator, XIII-A@1#eq:block-evaluation

Back to manuscript

Put $χ = E_{0} h$ . Then

[L_{q}, G_{h}]_{⋆_{H}} = i ℏ C (q, h) 1, C (q, h) = \int q^{μν} \partial_{μ} \partial_{ν} χ d^{4} x .

The formula holds for every admissible replacement $H^{'}$ with the same causal part.

Proof. There is at most one contraction between linear functionals. Their commutator is therefore exactly

i ℏ ⟨ u_{q}, Δ_{bγ} v_{h} ⟩ 1 .

In covariant flat identification, $v_{h}$ is $η h$ . Applying only the prescribed block gives

Δ_{bγ} v_{h} = - D P (η χ) = - \partial^{ν} χ .

Indeed $D$ already contains trace reversal, so $P^{2} = id$ leaves the divergence of $η χ$ with the displayed minus sign.

Using (eq:uq),

⟨ u_{q}, - \partial χ ⟩ = \int q^{μν} \partial_{μ} \partial_{ν} χ d^{4} x - \frac{1}{2} \int (η_{μν} q^{μν}) □ χ d^{4} x .

The causal solution satisfies $□ χ = 0$ , proving (eq:commutator). The calculation keeps the coefficient $- 1$ in $Δ_{bγ}$ throughout. Finally, admissible changes leave the graded antisymmetric part unchanged.

Displayed formulas in this passage

XIII-A@1#eq:commutator · line 1004
$[L_{q}, G_{h}]_{⋆_{H}} = i ℏ C (q, h) 1, C (q, h) = \int q^{μν} \partial_{μ} \partial_{ν} χ d^{4} x .$
XIII-A@1#eq:block-evaluation · line 1024
$Δ_{bγ} v_{h} = - D P (η χ) = - \partial^{ν} χ .$
unlabeled · line 1034
$⟨ u_{q}, - \partial χ ⟩ = \int q^{μν} \partial_{μ} \partial_{ν} χ d^{4} x - \frac{1}{2} \int (η_{μν} q^{μν}) □ χ d^{4} x .$
unlabeled · line 1019
$i ℏ ⟨ u_{q}, Δ_{bγ} v_{h} ⟩ 1 .$

Proposition: The bγ scalar

Displayed formulas in this passage

Nearby in this section