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

Equation 14 · Evolutionary Physics

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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

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 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…

Meanings in this article

• $\Lambda$: completely positive and every $\Lambda(\rho)$ with $\rho$ $\ge$ 0 is itself a valid state.
• $\Lambda^\dagger$: the define.

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