Equation 14 · Part 4 · The Entry a Relabeling Cannot Write

Symbol Lambda^dagger

\mathrm{Tr}\big[\Lambda(\rho)\,O_c\big] = \mathrm{Tr}\big[\rho\,\Lambda^\dagger(O_c)\big].

\Lambda^\dagger

What this part means

the define.

Its job in the formula

Lambd $a^d$ agger is an input to the expression that computes the quantity on the left.

Full expressionSymbol Lambda^daggerArticle meaning

Where the article explains it

Define $\Lambda^\dagger$ : $\mathcal B(\mathcal H_c)$ $\to$ $\mathcal B(\mathcal H_f)$ by the pairing that must hold for every fine state and every coarse observable $O_c$ , $\mathrm{Tr}\big[\Lambda(\rho)\,O_c\big] = \mathrm{Tr}\big[\rho\,\Lambda^\dagger(O_c)\big]$ .

The passage around this formula

Every channel has a dual. Define $\Lambda^\dagger$ : $\mathcal B(\mathcal H_c)$ $\to$ $\mathcal B(\mathcal H_f)$ by the pairing that must hold for every fine state and every coarse observable $O_c$ , $\mathrm{Tr}\big[\Lambda(\rho)\,O_c\big] = \mathrm{Tr}\big[\rho\,\Lambda^\dagger(O_c)\big]$ . $\Lambda^\dagger$ is positive because $\Lambda$ is completely positive and every $\Lambda(\rho)$ with $\rho$ $\ge$ 0 is itself a…

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the surrounding passage

[9] Positive Functions on C*-Algebras ↗

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