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ΔEΛ(ρ)=Tr[ρ(IS⊗HB+HSB)]\Delta E_\Lambda(\rho) = \mathrm{Tr}[\rho(I_S \otimes H_B + H_{SB})]

Why this formula appears here

an exact operator identity, not an approximation valid in some limit: the ledger for a system-only description of a system-plus-bath is precisely the bath’s own energy plus the coupling energy between the two, nothing more and nothing less. Δ\Delta EΛ(ρ)E_\Lambda(\rho) = Tr\mathrm{Tr}[ρ(IS⊗HB+HSB)\rho(I_S \otimes H_B + H_{SB})] for any joint state ρ\rho .

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ΔEΛ\Delta E_\Lambda

Symbol Δ E_Lambda

Δ ELE_Lambda is part of the quantity the equation computes from the expression on the right.

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ρ\rho

Symbol ρ

ρ is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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Published contexts (1)

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ΔEΛ(ρ)=Tr[ρ(IS⊗HB+HSB)]\Delta E_\Lambda(\rho) = \mathrm{Tr}[\rho(I_S \otimes H_B + H_{SB})]

Equation 62 · Evolutionary Physics

The Entry a Relabeling Cannot Write

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

an exact operator identity, not an approximation valid in some limit: the ledger for a system-only description of a system-plus-bath is precisely the bath’s own energy plus the coupling energy between the two, nothing more and nothing less. Δ\Delta EΛ(ρ)E_\Lambda(\rho) = Tr\mathrm{Tr}[ρ(IS⊗HB+HSB)\rho(I_S \otimes H_B + H_{SB})] for any joint state ρ\rho .

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