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DΛ=Hf−HS⊗IB=IS⊗HB+HSBD_\Lambda = H_f - H_S \otimes I_B = I_S \otimes H_B + H_{SB}

Why this formula appears here

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 DΛ=Hf−HS⊗IB=IS⊗HB+HSBD_\Lambda = H_f - H_S \otimes I_B = I_S \otimes H_B + H_{SB}. 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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DΛD_\Lambda

Symbol D_Lambda

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

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Published contexts (1)

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DΛ=Hf−HS⊗IB=IS⊗HB+HSB,D_\Lambda = H_f - H_S \otimes I_B = I_S \otimes H_B + H_{SB},

Equation 61 · 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.

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 DΛ=Hf−HS⊗IB=IS⊗HB+HSBD_\Lambda = H_f - H_S \otimes I_B = I_S \otimes H_B + H_{SB}. 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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