Manuscript III-B@1 · III-B-v1.tex · theorem · d2c98f62

Theorem: Clock instrument

Finite Enter experiments and the clock instrument › Complete destructive reading · lines 1546–1582 · proof included or linked

III-B@1#thm:instrument, III-B@1#eq:clock

Back to manuscript

The complete destructive instrument is

M_{C} (Z) = k = 0 \sum^{3} ∣ k ⟩ ⟨ k ∣_{R} \otimes D_{k} Z D_{k}^{†} = k = 0 \sum^{3} ∣ k ⟩ ⟨ k ∣_{R} \otimes Tr_{C} [(P_{k} \otimes I_{X}) Z (P_{k} \otimes I_{X})] .

It is completely positive and trace preserving. For a state $ρ_{C X E}$ , its $k$ -block is

σ_{X E, k} = (⟨ τ_{k} ∣ \otimes I_{X E}) ρ_{C X E} (∣ τ_{k} ⟩ \otimes I_{X E}) .

All identities hold with every untouched reference $E$ . The clock names are not returned as quantum inputs.

Proof. Apply the unitary sending $∣ τ_{k} ⟩$ to computational basis $∣ k ⟩$ , then perform destructive Enter on both clock qubits. Its recorded Kraus operators are $L_{k} = ∣ k ⟩_{R} \otimes D_{k}$ , and

k \sum L_{k}^{†} L_{k} = k \sum P_{k} \otimes I_{X} = I_{C X} .

This proves the channel assertion. The second expression follows from the rank-one projectors. Tensoring the Kraus operators with $I_{E}$ proves the complete reference identity. Their codomain contains $X$ , not $C$ .

Displayed formulas in this passage

III-B@1#eq:clock · line 1549
$M_{C} (Z) = k = 0 \sum^{3} ∣ k ⟩ ⟨ k ∣_{R} \otimes D_{k} Z D_{k}^{†} = k = 0 \sum^{3} ∣ k ⟩ ⟨ k ∣_{R} \otimes Tr_{C} [(P_{k} \otimes I_{X}) Z (P_{k} \otimes I_{X})] .$
unlabeled · line 1559
$σ_{X E, k} = (⟨ τ_{k} ∣ \otimes I_{X E}) ρ_{C X E} (∣ τ_{k} ⟩ \otimes I_{X E}) .$
unlabeled · line 1574
$k \sum L_{k}^{†} L_{k} = k \sum P_{k} \otimes I_{X} = I_{C X} .$

Theorem: Clock instrument

Displayed formulas in this passage

Nearby in this section