Manuscript XVI-J@1 · XVI-J.tex · lemma · 6c1f306d

Lemma: Time ordering and inverse identity

Ambient propagator and the punctured numerical square › Definition and normalization · lines 445–479 · proof included or linked

XVI-J@1#lem:DF

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The spatial Fourier branches of $d$ are

d (t, p) = {e^{- i ∣ p ∣ t} / (2∣ p ∣), e^{i ∣ p ∣ t} / (2∣ p ∣), t > 0, t < 0.

The distribution fixed by (eq:DF) is symmetric under $z \mapsto - z$ and satisfies

□ d = - i δ, D_{F}^{T} = D_{F}, K D_{F} = i id .

Proof. For $t < 0$ ,

\frac{e ^{- i ∣ p ∣ t}}{2∣ p ∣} + i \frac{sin ( t ∣ p ∣ )}{∣ p ∣} = \frac{e ^{i ∣ p ∣ t}}{2∣ p ∣} .

For $t > 0$ , the advanced summand vanishes. The temporal derivative of the two branches has jump $- i$ . Hence the wave operator gives $- i δ$ . Using $P^{2} = id$ proves the tensor inverse identity. The two branches also prove symmetry.

Displayed formulas in this passage

unlabeled · line 448
$d (t, p) = {e^{- i ∣ p ∣ t} / (2∣ p ∣), e^{i ∣ p ∣ t} / (2∣ p ∣), t > 0, t < 0.$
unlabeled · line 458
$□ d = - i δ, D_{F}^{T} = D_{F}, K D_{F} = i id .$
unlabeled · line 469
$\frac{e ^{- i ∣ p ∣ t}}{2∣ p ∣} + i \frac{sin ( t ∣ p ∣ )}{∣ p ∣} = \frac{e ^{i ∣ p ∣ t}}{2∣ p ∣} .$

Lemma: Time ordering and inverse identity

Displayed formulas in this passage

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