Equation 32 · Part 1 · The Entry a Relabeling Cannot Write

Symbol Lambda^dagger

\Lambda^\dagger(H_c) = H_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

…every density matrix are the same operator — the strongest statement follows at once: $\Delta$ $E_\Lambda(\rho)$ = 0 for every fine state $\rho$ if and only if $D_\Lambda$ = 0 as an operator identity, that is, if and only if $\Lambda^\dagger(H_c)$ = $H_f$ exactly. Energy is preserved for every possible input under that one algebraic condition, never as a statistical tendency and never approximately.

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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 article section

[9] Positive Functions on C*-Algebras ↗

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