Equation 19 · Part 1 · The Entry a Relabeling Cannot Write

Symbol Lambda^dagger

\Lambda^\dagger(I_c) = I_f

\Lambda^\dagger

What this part means

the define.

Its job in the formula

Lambd $a^d$ agger is part of the quantity the equation computes from the expression on the right.

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

$\Lambda^\dagger$ is positive because $\Lambda$ is completely positive and every $\Lambda(\rho)$ with $\rho$ $\ge$ 0 is itself a valid state; it is unital, $\Lambda^\dagger(I_c)$ = $I_f$ , because $\Lambda$ preserves trace for every input. This duality is the same construction Stinespring used to show that any completely positive map admits a dilation to a larger, unitary description [ 9 ] , and it lets a coarse Hamiltonian be pulled back onto the fine space without the fine space’s own structure ever being disturbed.

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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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