Manuscript X-K-ST@1 · X-K-ST.tex · theorem · 93c39c0f

Theorem: Two-clock complete output

An explicit two-readout, two-route certificate › Complete formula · lines 1522–1585 · proof included or linked

X-K-ST@1#thm:twoclock, X-K-ST@1#eq:twoclock

On the specified input and output boundary, the complete channel is

C_{2} (X) = k, l \in {0, 1} \sum ∣ k ⟩ ⟨ k ∣_{R_{A}} \otimes ∣ k, l, l ⟩ ⟨ k, l, l ∣_{R_{B}} \otimes K_{k l} X K_{k l}^{†} \otimes ω_{A_{1}} .

It is CPTP on every input operator and every reference extension. Both clock inputs are consumed.

For the particular supplied clocks

ρ_{C_{1}} = ∣ + ⟩ ⟨ + ∣, ρ_{C_{2}} = ∣ 0 ⟩ ⟨ 0 ∣,

and arbitrary payload–reference state $ρ_{P E}$ , the nonzero public probabilities are

p_{00} = \frac{1}{2}, p_{10} = \frac{1}{4}, p_{11} = \frac{1}{4}, p_{01} = 0.

The corresponding payload–reference blocks are

\frac{1}{2} ρ_{P E}, \frac{1}{4} (Z_{1} \otimes I_{E}) ρ_{P E} (Z_{1} \otimes I_{E}), \frac{1}{4} ρ_{P E} .

The failure probability is $1/4$ , and the failure output is retained.

Proof. For each fixed $k$ , summing over $l$ gives

l \sum (H^{k})^{†} ∣ l ⟩ ⟨ l ∣ H^{k} = I_{C_{2}},

while $W_{k l}^{†} W_{k l} = I_{P}$ . Hence

k, l \sum K_{k l}^{†} K_{k l} = k \sum ∣ k ⟩ ⟨ k ∣_{C_{1}} \otimes I_{C_{2} P} = I .

The routes return the other payload slot in the independent blank by Theorem thm:route. Copying the already classical $k$ gives the displayed record duplication. This proves (eq:twoclock), including the reference version.

The first clock gives $k = 0, 1$ with probability $1/2$ . If $k = 0$ , the second clock remains $∣ 0 ⟩$ . If $k = 1$ , it becomes $H ∣ 0 ⟩ = ∣ + ⟩$ , giving $l = 0, 1$ equally. Substitute the controlled payload gates.

Displayed formulas in this passage

X-K-ST@1#eq:twoclock · line 1526
$C_{2} (X) = k, l \in {0, 1} \sum ∣ k ⟩ ⟨ k ∣_{R_{A}} \otimes ∣ k, l, l ⟩ ⟨ k, l, l ∣_{R_{B}} \otimes K_{k l} X K_{k l}^{†} \otimes ω_{A_{1}} .$
unlabeled · line 1538
$ρ_{C_{1}} = ∣ + ⟩ ⟨ + ∣, ρ_{C_{2}} = ∣ 0 ⟩ ⟨ 0 ∣,$
unlabeled · line 1544
$p_{00} = \frac{1}{2}, p_{10} = \frac{1}{4}, p_{11} = \frac{1}{4}, p_{01} = 0.$
unlabeled · line 1551
$\frac{1}{2} ρ_{P E}, \frac{1}{4} (Z_{1} \otimes I_{E}) ρ_{P E} (Z_{1} \otimes I_{E}), \frac{1}{4} ρ_{P E} .$
unlabeled · line 1563
$l \sum (H^{k})^{†} ∣ l ⟩ ⟨ l ∣ H^{k} = I_{C_{2}},$
unlabeled · line 1569
$k, l \sum K_{k l}^{†} K_{k l} = k \sum ∣ k ⟩ ⟨ k ∣_{C_{1}} \otimes I_{C_{2} P} = I .$

Theorem: Two-clock complete output

Displayed formulas in this passage

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