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Equation 60 · Part 1 · The Entry a Relabeling Cannot Write

Symbol Lambda^dagger

Λ†(HS)=HS⊗IB\Lambda^\dagger(H_S) = H_S \otimes I_B
Λ†\Lambda^\dagger

What this part means

Lambdada^dagger is part of the quantity the equation computes from the expression on the right.

Its job in the formula

Lambdada^dagger is part of the quantity the equation computes from the expression on the right.

The passage around this formula

Return to the genuine coarse-graining set aside earlier: Λ(ρ)\Lambda(\rho) = TrB\mathrm{Tr}_B[ρ\rho] , Hc\mathcal H_c = HS\mathcal H_S , HcH_c = HSH_S . The dual embeds a system observable back into the full space without touching the bath, Λ†(HS)\Lambda^\dagger(H_S) = HSH_S ⊗\otimes IBI_B , so

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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

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