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

Theorem: All prefix channels and all message coherences

E2: complete radiation-channel calculation · lines 644–705 · proof included or linked

Em03-E@1#thm:channel, Em03-E@1#eq:N1, Em03-E@1#eq:N2, Em03-E@1#eq:diagonalunits, Em03-E@1#eq:offdiagonalunit

At prefix zero there is no accessible radiation factor. At prefix one,

N_{1} (X) = tr (X) σ_{1} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣_{C_{1}} .

At prefix two,

N_{2} (X) = σ_{O} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣_{C_{1}} \otimes [p (P_{ab} Λ_{η} (X) P_{ab}^{†})^{o} + (1 - p) tr (X) ∣ v ⟩ ⟨ v ∣_{C_{2}}] .

In particular, for $x = 0, 1$ ,

N_{2} (∣ x ⟩ ⟨ x ∣) = N_{2} (∣ 0 ⟩ ⟨ 1 ∣) = σ_{O} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣ \otimes [p (P_{ab} ∣ x ⟩ ⟨ x ∣ P_{ab}^{†})^{o} + (1 - p) ∣ v ⟩ ⟨ v ∣], σ_{O} \otimes \frac{p η}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣ \otimes (P_{ab} ∣ 0 ⟩ ⟨ 1 ∣ P_{ab}^{†})^{o} .

The other off-diagonal matrix unit is the adjoint of (eq:offdiagonalunit).

Proof. Occupation emission leaves the displayed diagonal densities after the daughter states are traced. These operations act on factors independent of the message.

Key preparation gives (eq:keypurification). The inaccessible key purification makes the radiation key diagonal with weight $1/4$ for every $a, b$ . Tracing the still-internal message at prefix one gives (eq:N1).

After ciphertext emission, the successful and unsuccessful source fuel states are orthogonal. Their cross terms therefore vanish in the radiation channel. On success the source message has the common terminal state, leaving its full encrypted density in $C_{2}$ . On failure $C_{2}$ is vacuum, and tracing the remaining source message contributes $tr X$ . Tracing $Z_{e}$ gives $Λ_{η}$ . This proves (eq:N2) on arbitrary operators. Equations (eq:diagonalunits)–(eq:offdiagonalunit) then follow directly.

Displayed formulas in this passage

Em03-E@1#eq:N1 · line 648
$N_{1} (X) = tr (X) σ_{1} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣_{C_{1}} .$
Em03-E@1#eq:N2 · line 656
$N_{2} (X) = σ_{O} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣_{C_{1}} \otimes [p (P_{ab} Λ_{η} (X) P_{ab}^{†})^{o} + (1 - p) tr (X) ∣ v ⟩ ⟨ v ∣_{C_{2}}] .$
Em03-E@1#eq:diagonalunits · line 668
$N_{2} (∣ x ⟩ ⟨ x ∣) = N_{2} (∣ 0 ⟩ ⟨ 1 ∣) = σ_{O} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣ \otimes [p (P_{ab} ∣ x ⟩ ⟨ x ∣ P_{ab}^{†})^{o} + (1 - p) ∣ v ⟩ ⟨ v ∣], σ_{O} \otimes \frac{p η}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣ \otimes (P_{ab} ∣ 0 ⟩ ⟨ 1 ∣ P_{ab}^{†})^{o} .$
Em03-E@1#eq:offdiagonalunit · line 668
$N_{2} (∣ x ⟩ ⟨ x ∣) = N_{2} (∣ 0 ⟩ ⟨ 1 ∣) = σ_{O} \otimes \frac{1}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣ \otimes [p (P_{ab} ∣ x ⟩ ⟨ x ∣ P_{ab}^{†})^{o} + (1 - p) ∣ v ⟩ ⟨ v ∣], σ_{O} \otimes \frac{p η}{4} a, b \sum ∣ a, b ⟩ ⟨ a, b ∣ \otimes (P_{ab} ∣ 0 ⟩ ⟨ 1 ∣ P_{ab}^{†})^{o} .$

Nearby in this section

section: E2: complete radiation-channel calculation