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Jarzynski equality for a finite protocol with stationary kernels

Physical scope

An exact ensemble identity for a finite state space, finite protocol, canonical initial law, and kernels preserving each stage's canonical law; no claim is made for general stochastic processes or for any individual path.

Statement

For the forward path law of Finite driven protocol and path work, under Canonical initial preparation for a finite protocol and Finite bath kernels preserve the stage canonical law, the finite expectation satisfies $E_{F} [e^{- β W}] = Z_{m} / Z_{0} = e^{- β Δ F}$ . This is an ensemble identity; it does not say $W = Δ F$ on each path.

Facts & Assumptions

Given: A nonempty finite state space, finite energies $E_{k}$ , $β > 0$ , a row-stochastic $K_{k}$ , initial law $π_{0}$ , and $π_{k} K_{k} = π_{k}$ at each stage.

[F1]

Finite driven protocol and path work: the path law is $π_{0} (i_{0}) \prod_{k} K_{k} (i_{k - 1}, i_{k})$ , and work at quench $k$ is $E_{k} (i_{k - 1}) - E_{k - 1} (i_{k - 1})$ .

[A1]

Canonical initial preparation for a finite protocol: the initial law is $π_{0} (i) = e^{- β E_{0} (i)} / Z_{0}$ .

[A2]

Finite bath kernels preserve the stage canonical law: $\sum_{i} π_{k} (i) K_{k} (i, j) = π_{k} (j)$ .

[L1]

Canonical sampling assumption for finite and normalized statistical models: the current draft postulate adopts the finite microstate law $P_{β} (i) = e^{- β E (i)} / Z (β)$ for finite nonempty $Ω$ with $0 < Z (β) < \infty$ and $β = (k_{B} T)^{- 1}$ . Only this finite branch is used; the countable and measurable branches are not invoked.

Proof

technique · induction on protocol stages

Given: [F1], [A1], [A2], [L1], with the finite partition sums and canonical laws of the protocol.

1.1F1A1L1givenbasealgebra

For $k \geq 1$ and $j \in Ω$ , define $S_{k} (j)$ by the finite sum over $i_{0}, \dots, i_{k - 1} \in Ω$ of $π_{0} (i_{0}) \prod_{ℓ = 1}^{k} K_{ℓ} (i_{ℓ - 1}, i_{ℓ}) e^{- β W_{k}}$ , with $i_{k} = j$ and $W_{k} = \sum_{ℓ = 1}^{k} [E_{ℓ} (i_{ℓ - 1}) - E_{ℓ - 1} (i_{ℓ - 1})]$ ; put $S_{0} (j) = π_{0} (j)$ . Partitioning this finite sum by the state $i = i_{k - 1}$ gives $S_{k} (j) = \sum_{i} S_{k - 1} (i) e^{- β (E_{k} (i) - E_{k - 1} (i))} K_{k} (i, j)$ . The canonical preparation gives the base $S_{0} (j) = π_{0} (j) = (Z_{0} / Z_{0}) π_{0} (j)$ by [A1] and the finite branch of [L1].

2.1ihstep 1.1F1

For the induction step fix $k \in {1, \dots, m}$ and suppose $S_{k - 1} (i) = (Z_{k - 1} / Z_{0}) π_{k - 1} (i)$ for every $i$ .

3.1F1A2step 1.1step 2.1algebra

Substitute this hypothesis in the recurrence from step 1.1. Since $π_{k - 1} (i) = e^{- β E_{k - 1} (i)} / Z_{k - 1}$ , the previous-stage Boltzmann factor cancels against the quench factor, giving $S_{k} (j) = (Z_{k} / Z_{0}) \sum_{i} π_{k} (i) K_{k} (i, j) = (Z_{k} / Z_{0}) π_{k} (j)$ by [A2]. This proves the induction step at each stage.

4.1F1step 1.1step 3.1algebradischarge-induction∎

The base and induction step establish $S_{k} (j) = (Z_{k} / Z_{0}) π_{k} (j)$ for all stages. At $k = m$ , summing over terminal states and using normalization of $π_{m}$ gives $E_{F} [e^{- β W}] = Z_{m} / Z_{0}$ . By $Δ F = - β^{- 1} \log (Z_{m} / Z_{0})$ , this equals $e^{- β Δ F}$ .

Jarzynski equality for a finite protocol with stationary kernels

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources