Manuscript XIX-C@1 · XIX-C.tex · proposition · 98b067d9

Proposition: Equal records need not determine a residual source

Records, residual source functionals, and token acquisition › A destructive record is not its residual state · lines 461–479 · proof included or linked

XIX-C@1#prop:residual

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There are two states with identical complete classical record laws but different residual expectations of a fixed observable. Consequently, a residual source functional does not generally factor through the record law.

Proof. Take two qubits and

ρ_{0} = ∣ 0 ⟩ ⟨ 0 ∣_{S} \otimes ∣ 0 ⟩ ⟨ 0 ∣_{B}, ρ_{1} = ∣ 0 ⟩ ⟨ 0 ∣_{S} \otimes ∣ 1 ⟩ ⟨ 1 ∣_{B} .

In both cases the record is $z = 0$ with probability one. For $Z_{B} = ∣ 0 ⟩ ⟨ 0 ∣ - ∣ 1 ⟩ ⟨ 1 ∣$ , the residual expectations are $1$ and $- 1$ . Apply Lemma lem:fibers.

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unlabeled · line 471
$ρ_{0} = ∣ 0 ⟩ ⟨ 0 ∣_{S} \otimes ∣ 0 ⟩ ⟨ 0 ∣_{B}, ρ_{1} = ∣ 0 ⟩ ⟨ 0 ∣_{S} \otimes ∣ 1 ⟩ ⟨ 1 ∣_{B} .$

Proposition: Equal records need not determine a residual source

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