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Crooks path and work-distribution relation for a finite protocol

Physical scope

A path-reversal identity for a finite-state protocol whose stage kernels are reversible with respect to their positive canonical laws; it does not extend to arbitrary driven systems without matching support and reversal assumptions.

Statement

Assume the setup of Finite driven protocol and path work, the canonical initial law of Canonical initial preparation for a finite protocol, the stationarity model of Finite bath kernels preserve the stage canonical law, and, additionally, $π_{k} (i) K_{k} (i, j) = π_{k} (j) K_{k} (j, i)$ for every $k, i, j$ . For every forward path with positive probability, define its reverse probability by $P_{R} (ω^{R}) = π_{m} (i_{m}) \prod_{k = 1}^{m} K_{k} (i_{k}, i_{k - 1})$ . Then $P_{R} (ω^{R}) > 0$ and $P_{F} (ω) / P_{R} (ω^{R}) = e^{β (W (ω) - Δ F)}$ . Consequently, at each work value $w$ , $P_{F} (W = w) = e^{β (w - Δ F)} P_{R} (W_{R} = - w)$ . The reverse protocol first applies the stage- $k$ reverse transition at fixed $E_{k}$ , then reverses the quench $E_{k} \to E_{k - 1}$ ; its work is $W_{R} (ω^{R}) = - W (ω)$ .

Facts & Assumptions

Given: The finite protocol with canonical initial and terminal laws, strictly positive $π_{k}$ , stationary stage kernels, and the additional stagewise detailed-balance equations.

[F1]

Finite driven protocol and path work: quench $k$ changes $E_{k - 1}$ to $E_{k}$ before transition $K_{k}$ ; $W$ is work on the system.

[F2]

Finite detailed-balance rate and transition models: the kernel balance equation is $π_{k} (i) K_{k} (i, j) = π_{k} (j) K_{k} (j, i)$ .

[A1]

Canonical initial preparation for a finite protocol: the forward initial law is $π_{0}$ .

[A2]

Finite bath kernels preserve the stage canonical law: each $K_{k}$ preserves $π_{k}$ .

[F3]

Pathwise energy balance for a finite driven protocol: $Δ E = W + Q_{i n}$ with the stated signs.

Proof

Given: [F1], [F2], [A1], [A2], and [F3].

1.1F1F2A1A2

Define the reverse path law by starting with $π_{m} (i_{m})$ and, in reverse order, using $K_{k} (i_{k}, i_{k - 1})$ at fixed energy $E_{k}$ before undoing that quench. By detailed balance, if $K_{k} (i_{k - 1}, i_{k}) > 0$ then $K_{k} (i_{k}, i_{k - 1}) > 0$ , because $π_{k}$ is strictly positive. Thus positive forward paths pair with positive reverse paths.

2.1F1F2F3A1step 1.1algebra

On a paired path, divide the endpoint Gibbs weights and use detailed balance for each transition: $P_{F} (ω) / P_{R} (ω^{R}) = [π_{0} (i_{0}) / π_{m} (i_{m})] \prod_{k = 1}^{m} [π_{k} (i_{k}) / π_{k} (i_{k - 1})]$ . Substituting $π_{k} (i) = e^{- β E_{k} (i)} / Z_{k}$ and collecting terms gives $\log (P_{F} / P_{R}) = β [E_{m} (i_{m}) - E_{0} (i_{0}) - Q_{i n}] + \log (Z_{m} / Z_{0}) = β (W - Δ F)$ by [F3].

3.1F1F3step 2.1algebra∎

The reversed quench at stage $k$ occurs at state $i_{k - 1}$ , so its work term is $E_{k - 1} (i_{k - 1}) - E_{k} (i_{k - 1}) = - [E_{k} (i_{k - 1}) - E_{k - 1} (i_{k - 1})]$ ; summing proves $W_{R} (ω^{R}) = - W (ω)$ . Summing the path ratio over the finite set of paths with $W = w$ proves the stated distribution identity.

Crooks path and work-distribution relation for a finite protocol

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources