Manuscript Hawk04-B@2 · Hawk04-B.tex · proposition · ba3b0713

Proposition: Energy and selected entropy do not locate the code

The corrected destination ledger · lines 2170–2199 · proof included or linked

Hawk04-B@2#prop:diagnostics

A source can remain in one fixed zero-energy vacuum while an inaccessible memory retains the entire diary. Also, two complete isometries can give the same zero radiation entropy for every pure code input but have radiation recovery errors $0$ and $a_{d}$ , respectively.

Proof. Let $B = S \otimes M$ , choose $H_{S} ∣ vac ⟩ = 0$ , and set

V ∣ ψ ⟩ = ∣ 0 ⟩_{R} ∣ vac ⟩_{S} ∣ ψ ⟩_{M} .

The source energy is identically zero, the complete residual recovers exactly, and radiation is constant. Its recovery error is $a_{d}$ by Lemma lem:unitary-mixture. The residual is not logical-input independent.

For the entropy comparison, take

V_{good} ∣ ψ ⟩ = ∣ ψ ⟩_{R} ∣ 0 ⟩_{B}, V_{bad} ∣ ψ ⟩ = ∣ 0 ⟩_{R} ∣ ψ ⟩_{B} .

For pure inputs, both radiation entropies are zero. Their recovery errors are $0$ and $a_{d}$ .

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$V ∣ ψ ⟩ = ∣ 0 ⟩_{R} ∣ vac ⟩_{S} ∣ ψ ⟩_{M} .$
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$V_{good} ∣ ψ ⟩ = ∣ ψ ⟩_{R} ∣ 0 ⟩_{B}, V_{bad} ∣ ψ ⟩ = ∣ 0 ⟩_{R} ∣ ψ ⟩_{B} .$

Proposition: Energy and selected entropy do not locate the code

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