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A finite quench has a positive probability of work below delta F

Physical scope

A calculated two-state hypothetical protocol example; it is not an experiment and makes no claim about a measured system.

Example

Take $Ω = {a, b}$ , $β > 0$ , $E_{0} (a) = E_{0} (b) = 0$ , and $E_{1} (a) = 0$ , $E_{1} (b) = ϵ$ for $ϵ > 0$ . Prepare canonically at stage zero and use the identity transition kernel after the quench.

Facts & Assumptions

Given: The finite path model and canonical protocol preparation.

[F1]

Finite driven protocol and path work: work is the energy change at the instantaneous quench and the forward path is weighted by the initial law.

[A1]

Canonical initial preparation for a finite protocol: both initial states have probability $1 / 2$ .

[F2]

Jarzynski equality for a finite protocol with stationary kernels: this model satisfies the exact finite ensemble exponential-work identity.

Verification

Given: $ϵ > 0$ and $β > 0$ .

1.1givenF1algebra

The partition sums are $Z_{0} = 2$ and $Z_{1} = 1 + e^{- β ϵ}$ , so $Δ F = - β^{- 1} \log [(1 + e^{- β ϵ}) / 2]$ . Since $e^{- β ϵ} < 1$ , the ratio lies strictly between $e^{- β ϵ}$ and $1$ , hence $0 < Δ F < ϵ$ .

2.1A1F1step 1.1

The identity kernel leaves the state unchanged, so the two work values are $0$ and $ϵ$ , each with probability $1 / 2$ . Therefore $P_{F} (W < Δ F) = 1 / 2$ .

3.1F2step 1.1step 2.1algebra∎

Directly, $E_{F} e^{- β W} = (1 + e^{- β ϵ}) / 2 = e^{- β Δ F}$ , while the event in step 2.1 has positive probability. Thus the equality is an ensemble statement and does not impose $W \geq Δ F$ on every path.

Depends on

Dependency tree · 2 levels

Sources

C. Jarzynski, Introduction to Nonequilibrium Statistical Physics, lecture notes, Nonequilibrium Work Relations