Equation 61 · The Entry a Relabeling Cannot Write

What does this equation mean?

D_\Lambda = H_f - H_S \otimes I_B = I_S \otimes H_B + H_{SB},

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Inputs and operationsH_f - H_S otimes I_B = I_S otimes H_B + H_SB

Result or conditionD_Lambda

How to read the two sides of this formula. Follow the article passage for the meaning of each quantity.

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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D_\Lambda

Symbol D_Lambda

$D_L$ ambda is part of the quantity the equation computes from the expression on the right.

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H_f

Symbol H_f

$H_f$ is one of the signed contributions combined to compute the quantity on the left.

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H_S

Symbol H_S

$H_S$ is one of the signed contributions combined to compute the quantity on the left.

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I_B

Symbol I_B

$I_B$ is one of the signed contributions combined to compute the quantity on the left.

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I_S

Symbol I_S

$I_S$ is one of the signed contributions combined to compute the quantity on the left.

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H_B

Symbol H_B

$H_B$ is one of the signed contributions combined to compute the quantity on the left.

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H_{SB}

Symbol H_SB

$H_S$ B is one of the signed contributions combined to compute the quantity on the left.

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=

The expressions on both sides represent the same quantity under the stated assumptions.

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addition

Add the term after the plus sign to the term or group before it.

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subtraction

Subtract the following term or group from the preceding one. A leading minus marks a negative quantity.

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subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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How to interpret it

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What the article says around this equation

Return to the genuine coarse-graining set aside earlier: $\Lambda(\rho)$ = $\mathrm{Tr}_B$ [ $\rho$ ] , $\mathcal H_c$ = $\mathcal H_S$ , $H_c$ = $H_S$ . The dual embeds a system observable back into the full space without touching the bath, $\Lambda^\dagger(H_S)$ = $H_S$ $\otimes$ $I_B$ , so $D_\Lambda = H_f - H_S \otimes I_B = I_S \otimes H_B + H_{SB}$ . 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. $\Delta$ $E_\Lambda(\rho)$ = $\mathrm{Tr}$ [ $\rho(I_S \otimes H_B + H_{SB})$ ] for any joint state $\rho$ .

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Sources cited in the article section

[13] Quantum Thermodynamics ↗

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