Manuscript Em03-B@2 · Em03-B-v2.tex · theorem · c1a0977d

Theorem: Exact coherent extraction and sharp tail

B4 extraction: explicit swaps and control certificates › The exact finite-domain interaction · lines 1201–1241 · proof included or linked

Em03-B@2#thm:extraction, Em03-B@2#eq:extractionchannel, Em03-B@2#eq:sharpK

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Tracing out $c$ after $U_{N, K}$ , with initial register $∣ 0 ⟩_{Q}$ , induces

E_{N, K} (X) = J_{N} V_{N} P_{N} X P_{N} V_{N}^{†} J_{N}^{†} + n = N + 1 \sum^{K} ⟨ n ∣ X ∣ n ⟩ ∣ N ⟩ ⟨ N ∣_{Q} + n > K \sum ⟨ n ∣ X ∣ n ⟩ ∣ 0 ⟩ ⟨ 0 ∣_{Q} .

For every normal memory–reference state,

d_{tr} ((E_{N, K} \otimes id) (ρ), (B_{N} \otimes id) (ρ)) = ρ Pr (N_{c} > K) .

For the ideal target packet this error is $q_{h}^{K + 1}$ .

Proof. For $0 \leq n \leq N$ , the output memory is the same vacuum. Consequently all low-block coherences survive in $Q$ . For $N < n \leq K$ , memory state $n$ labels the environment and the register is $N$ . For $n > K$ , the untouched input has memory state $n$ and register zero. Distinct such memory states are orthogonal, proving (eq:extractionchannel).

The channel difference on a referenced input is exactly

(∣ 0 ⟩ ⟨ 0 ∣_{Q} - ∣ N ⟩ ⟨ N ∣_{Q}) \otimes H, H = n > K \sum ⟨ n ∣ ρ ∣ n ⟩ ⪰ 0.

Because $N \geq 1$ , the two register states are orthogonal. Its half trace norm is $tr H$ .

Displayed formulas in this passage

Em03-B@2#eq:extractionchannel · line 1205
$E_{N, K} (X) = J_{N} V_{N} P_{N} X P_{N} V_{N}^{†} J_{N}^{†} + n = N + 1 \sum^{K} ⟨ n ∣ X ∣ n ⟩ ∣ N ⟩ ⟨ N ∣_{Q} + n > K \sum ⟨ n ∣ X ∣ n ⟩ ∣ 0 ⟩ ⟨ 0 ∣_{Q} .$
Em03-B@2#eq:sharpK · line 1215
$d_{tr} ((E_{N, K} \otimes id) (ρ), (B_{N} \otimes id) (ρ)) = ρ Pr (N_{c} > K) .$
unlabeled · line 1235
$(∣ 0 ⟩ ⟨ 0 ∣_{Q} - ∣ N ⟩ ⟨ N ∣_{Q}) \otimes H, H = n > K \sum ⟨ n ∣ ρ ∣ n ⟩ ⪰ 0.$

Theorem: Exact coherent extraction and sharp tail

Displayed formulas in this passage

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