Manuscript En10-B@1 · En10-B.tex · proposition · a1105bad

Proposition: Opposite charge does not generally invert graph evolution

Derived channel and complete acquisition instrument › Why this does not yet derive a physical positron channel · lines 539–567 · proof included or linked

En10-B@1#prop:charge

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There are finite charge-phase graph generators for which changing the charge sign does not give the inverse propagator. Consequently the antiqubit inverse response cannot be justified solely by assigning opposite-charge labels.

Proof. Consider the permissible two-site family

H_{q} = J_{*} (e^{i q α} ∣ 0 ⟩ ⟨ 1 ∣ + e^{- i q α} ∣ 1 ⟩ ⟨ 0 ∣), J_{*} > 0.

It is self-adjoint. Its sum with $H_{- q}$ is $2 J_{*} cos (q α) X$ , which need not vanish. At $α = 0$ , both charge signs give $J_{*} X$ . For $0 < J_{*} T /ℏ < π /2$ ,

e^{- i T H_{- q} /ℏ} = e^{- i T J_{*} X /ℏ} \neq = e^{+ i T J_{*} X /ℏ} .

Hence charge-sign reversal is not, in this family, full Hamiltonian inversion.

Displayed formulas in this passage

unlabeled · line 549
$H_{q} = J_{*} (e^{i q α} ∣ 0 ⟩ ⟨ 1 ∣ + e^{- i q α} ∣ 1 ⟩ ⟨ 0 ∣), J_{*} > 0.$
unlabeled · line 560
$e^{- i T H_{- q} /ℏ} = e^{- i T J_{*} X /ℏ} \neq = e^{+ i T J_{*} X /ℏ} .$

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section: Why this does not yet derive a physical positron channel