Manuscript X-I-ST@1 · X-I-ST.tex · theorem · 9745fd40

Theorem: Complete residual-report instrument

Complete destructive instruments and their composition › The complete task-clock instrument · lines 1769–1806 · proof included or linked

X-I-ST@1#thm:instrument, X-I-ST@1#eq:drop

Back to manuscript

Equation (eq:instrument) is CPTP and satisfies

Tr_{R} [(M \otimes id_{E}) (Z)] = Tr_{C} Z

for every operator $Z$ , including all reference extensions.

On a positive branch, the surviving state is the corresponding subnormalized block divided by its trace. No normalized state is assigned to a zero branch. The returned quantum names are those of $A_{fin}$ , not those of $C$ .

Proof. Use the Kraus maps

L_{j} = ∣ j, w_{j} ⟩_{R} \otimes D_{j} .

Orthonormality gives

j \sum L_{j}^{†} L_{j} = j \sum ∣ τ_{j} ⟩ ⟨ τ_{j} ∣ \otimes I = I .

Thus the map is completely positive and trace preserving.

Expand $Z = \sum_{j, l} ∣ τ_{j} ⟩ ⟨ τ_{l} ∣ \otimes Z_{j l}$ . Its $j$ -th output block is $Z_{j j}$ ; summing gives $Tr_{C} Z$ . The calculation is unchanged on adjoining $E$ . The branch and consumption statements follow from the actual domain and codomain.

Displayed formulas in this passage

X-I-ST@1#eq:drop · line 1772
$Tr_{R} [(M \otimes id_{E}) (Z)] = Tr_{C} Z$
unlabeled · line 1789
$L_{j} = ∣ j, w_{j} ⟩_{R} \otimes D_{j} .$
unlabeled · line 1793
$j \sum L_{j}^{†} L_{j} = j \sum ∣ τ_{j} ⟩ ⟨ τ_{j} ∣ \otimes I = I .$

Theorem: Complete residual-report instrument

Displayed formulas in this passage

Nearby in this section