TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Jarzynski identity for a finite-state driven rate process

Statement

Let $E$ be a finite nonempty state set and $T \geq 0$ . Fix $β > 0$ . If $T > 0$ , for each $x \in E$ let $E_{x} \in C^{1} ([0, T], R)$ be an energy function; if $T = 0$ , let each $E_{x} (0)$ be finite. Let $κ_{x y} : [0, T] \to [0, \infty)$ be bounded continuous transition rates for $x \neq y$ , and let $X_{t}$ be the associated finite-state Markov jump process. Define

$Z_{t} : = \sum_{x \in E} e^{- β E_{x} (t)}, π_{t} (x) : = \frac{e^{- β E_{x} (t)}}{Z_{t}} .$

Assume instantaneous detailed balance:

$π_{t} (x) κ_{x y} (t) = π_{t} (y) κ_{y x} (t) (x \neq y, 0 \leq t \leq T),$

and initial law $P (X_{0} = x) = π_{0} (x)$ . The path work and free-energy difference are

$W_{T} : = {\begin{cases} 0, & T = 0, \\ \int_{0}^{T} E_{X_{t}}^{'} (t) d t, & T > 0, \end{cases} Δ F : = - β^{- 1} \log (Z_{T} / Z_{0}) .$

Then

$E [e^{- β W_{T}}] = \frac{Z_{T}}{Z_{0}} = e^{- β Δ F} .$

If energy is measured in joules, then $β$ has units $J^{- 1}$ , $W_{T}$ and $Δ F$ have units joules, and the exponent is dimensionless. Under a common $C^{1}$ energy shift $E_{x} (t) \mapsto E_{x} (t) + a (t)$ for all $x$ , both $W_{T}$ and $Δ F$ change by $a (T) - a (0)$ , so the identity is unchanged.

Statement references

Finite-state continuous-time Markov jump processes with bounded rates

Finite-state Feynman-Kac evolution for continuous jump rates

The Axiom of Countable Choice ( ${A C}_{ω}$ )

The natural logarithm as the inverse of the exponential function

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$

Facts & Assumptions

Given: A finite nonempty state set, $T \geq 0$ , finite energies at $T = 0$ or $C^{1}$ energies for $T > 0$ , bounded continuous rates, $β > 0$ , instantaneous detailed balance, and initial distribution $π_{0}$ .

[F1]

Finite-state continuous-time Markov jump processes with bounded rates: the process is Markov, has row generator $Q (t)$ with off-diagonal entries $κ_{x y} (t)$ , and has finitely many jumps on the finite interval.

[F2]

Finite-state Feynman-Kac evolution for continuous jump rates: the weighted endpoint masses for potential $V_{x} (t)$ solve the finite-dimensional ODE $G_{y}^{'} = \sum_{x \neq y} G_{x} κ_{x y} - G_{y} λ_{y} - V_{y} G_{y}$ , uniquely for their initial vector.

[F3]

The natural logarithm as the inverse of the exponential function: $\log$ is the inverse of the exponential on $(0, \infty)$ .

[F4]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : for $T > 0$ and $a \in C^{1} ([0, T], R)$ , including one-sided endpoint derivatives, the continuous derivative is integrable and $\int_{0}^{T} a^{'} (t) d t = a (T) - a (0)$ .

[F5]

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ : $\exp (u + v) = \exp (u) \exp (v)$ for all real $u, v$ .

[F6]

The exponential function is smooth and $(\exp)^{'} = \exp$ and The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : differentiating $t \mapsto \exp (- β E_{y} (t))$ gives $- β E_{y}^{'} (t) \exp (- β E_{y} (t))$ .

[F7]

Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ : the integral of a sum of Riemann-integrable functions is the sum of their integrals, including with oriented limits.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice, carried by the path-density construction used in [F2].

Proof

Given: The finite-state driven process and assumptions in the Statement.

1.1F3givenalgebra

If $T = 0$ , then $W_{T} = 0$ by definition and $Z_{T} / Z_{0} = 1$ , so $E [e^{- β W_{T}}] = 1 = Z_{T} / Z_{0}$ ; the definition of $Δ F$ and the inverse relation [F3] give $e^{- β Δ F} = Z_{T} / Z_{0}$ as well. Now suppose $T > 0$ . For each $t$ , $π_{t}$ is a strictly positive probability vector because $E$ is finite and every exponential is positive. Instantaneous detailed balance gives, for each $y$ , $\sum_{x \neq y} π_{t} (x) κ_{x y} (t) = π_{t} (y) \sum_{x \neq y} κ_{y x} (t) = π_{t} (y) λ_{y} (t) .$ Thus the column inflow from the distribution $π_{t}$ equals the generator outflow at $y$ .

1.2F1F2A1

Apply [F2] with potential $V_{y} (t) = β E_{y}^{'} (t)$ and initial law $π_{0}$ . The weighted endpoint masses $G_{y} (t) : = E [e^{- β W_{t}} 1_{{X_{t} = y}}]$ satisfy $G_{y}^{'} (t) = \sum_{x \neq y} G_{x} (t) κ_{x y} (t) - G_{y} (t) λ_{y} (t) - β E_{y}^{'} (t) G_{y} (t), G_{y} (0) = π_{0} (y),$ where $W_{t} : = \int_{0}^{t} E_{X_{s}}^{'} (s) d s$ .

2.1F2F3F6step 1.1step 1.2algebra

Define $C_{y} (t) : = e^{- β E_{y} (t)} / Z_{0} = (Z_{t} / Z_{0}) π_{t} (y)$ . Then $C_{y} (0) = π_{0} (y)$ and [F6] gives $C_{y}^{'} (t) = - β E_{y}^{'} (t) C_{y} (t)$ . By step 1.1, the generator inflow and outflow terms evaluated at $C (t)$ cancel, so $C$ satisfies the same ODE and initial vector as $G$ . Uniqueness in [F2] gives $G_{y} (t) = C_{y} (t)$ for every $y, t$ .

3.1F3step 2.1algebra

Summing step 2.1 over the finite state set at $t = T$ yields $E [e^{- β W_{T}}] = \sum_{y \in E} G_{y} (T) = \frac{1}{Z_{0}} \sum_{y \in E} e^{- β E_{y} (T)} = \frac{Z_{T}}{Z_{0}} .$ By [F3] and the definition of $Δ F$ , $e^{- β Δ F} = Z_{T} / Z_{0}$ , proving the identity.

4.1F3F4F5F7givenstep 3.1algebra∎

For a common shift $a (t)$ , put $δ : = a (T) - a (0)$ . If $T = 0$ , then $δ = 0$ and both $W_{T}$ and $Z_{T} / Z_{0}$ are unchanged, so the claimed invariance holds. Suppose $T > 0$ . Then $Z_{T}^{♯} / Z_{0}^{♯} = e^{- β δ} (Z_{T} / Z_{0})$ . Since $Z_{T} / Z_{0} > 0$ , [F3] and [F5] give $\exp ⁣ (\log (Z_{T} / Z_{0}) - β δ) = e^{- β δ} (Z_{T} / Z_{0}),$ so the inverse relation in [F3] implies $Δ F^{♯} = Δ F + δ$ . By [F7] the shifted path work is $W_{T}^{♯} = W_{T} + \int_{0}^{T} a^{'} (t) d t = W_{T} + δ$ , with the last equality from [F4]. Therefore $W_{T}^{♯} - Δ F^{♯} = W_{T} - Δ F$ .

Remarks

The theorem is a conditional mathematical identity under the stated finite-state rates and instantaneous detailed balance. It does not derive detailed balance or local detailed balance for an arbitrary nonequilibrium reservoir, and it does not establish that a physical apparatus realizes these rates. The exact exponential average can be dominated by rare paths, so the identity alone gives no low-variance or finite-sample guarantee.

Jarzynski identity for a finite-state driven rate process

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources