Equation 12 · Part 1 · The Entry a Relabeling Cannot Write

Symbol Lambda^dagger

\Lambda^\dagger: \mathcal B(\mathcal H_c) \to \mathcal B(\mathcal H_f)

\Lambda^\dagger

What this part means

the define.

Its job in the formula

Lambd $a^d$ agger is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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 $O_c$ ,

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 $O_c$ ,

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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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See this notation across published equations →

Sources cited in the article section

[9] Positive Functions on C*-Algebras ↗

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