Manuscript Em03-E@1 · Em03-E-v1.tex · theorem · c4b07e6a

Theorem: Complete complementary output

The full complementary channel and leakage constants · lines 973–1022 · proof included or linked

Em03-E@1#thm:complement, Em03-E@1#eq:fullcomplement

Up to insertion of fixed factors, the prefix-two complementary channel is

N_{2}^{c} (X) = δ_{A} \otimes \frac{1}{4} a, b \sum ∣ a, b; a, b ⟩ ⟨ a, b; a, b ∣_{K C} \otimes [p ∣ 0 ⟩ ⟨ 0 ∣_{F_{2}} \otimes ∣ 0 ⟩ ⟨ 0 ∣_{Q} \otimes C_{η} (X) \oplus (1 - p) ∣ 1 ⟩ ⟨ 1 ∣_{F_{2}} \otimes W_{ab} X W_{ab}^{†}] .

The direct-sum notation refers to the orthogonal fuel sectors. The key purification $C$ and dephasing environment are part of this output.

At prefix one the complementary channel, apart from independent occupation-source and unused-cell factors, is

X ⟼ \frac{1}{4} a, b \sum ∣ a, b; a, b ⟩ ⟨ a, b; a, b ∣_{K C} \otimes W_{ab} X W_{ab}^{†} .

It is reversible when all its factors are accessible.

Proof. Tracing the emitted key eliminates different $a, b$ cross terms but retains both the source key and its purifying copy. Tracing the ciphertext eliminates success–failure cross terms because occupied and vacuum radiation states are orthogonal. On success, tracing the emitted qubit leaves $C_{η} (X)$ ; on failure, the source qubit and its dephasing environment retain the isometric state $W_{ab} X W_{ab}^{†}$ . Occupation registers contribute the independent diagonal daughter densities. This proves the complete formula.

At prefix one the encrypted qubit has not been emitted. Reading the retained key, undoing $P_{ab}$ , and reversing the known dephasing unitary on $Q Z_{e}$ recovers the input.

Displayed formulas in this passage

Em03-E@1#eq:fullcomplement · line 977
$N_{2}^{c} (X) = δ_{A} \otimes \frac{1}{4} a, b \sum ∣ a, b; a, b ⟩ ⟨ a, b; a, b ∣_{K C} \otimes [p ∣ 0 ⟩ ⟨ 0 ∣_{F_{2}} \otimes ∣ 0 ⟩ ⟨ 0 ∣_{Q} \otimes C_{η} (X) \oplus (1 - p) ∣ 1 ⟩ ⟨ 1 ∣_{F_{2}} \otimes W_{ab} X W_{ab}^{†}] .$
unlabeled · line 998
$X ⟼ \frac{1}{4} a, b \sum ∣ a, b; a, b ⟩ ⟨ a, b; a, b ∣_{K C} \otimes W_{ab} X W_{ab}^{†} .$

Theorem: Complete complementary output

Displayed formulas in this passage

Nearby in this section