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Gibbs jump commuting relative entropy production

Statement

Use the finite connected Gibbs jump model of Connected Gibbs jump generators satisfy detailed balance. For a diagonal density $ρ_{t} = \sum_{a} p_{a} (t) ∣ a ⟩ ⟨ a ∣$ , the populations satisfy $p_{a}^{'} = \sum_{b \neq a} (p_{b} w_{b a} - p_{a} w_{a b})$ . Define the dimensionless classical relative entropy

$K (p ∣ π) : = \sum_{a \in E} p_{a} \log \frac{p_{a}}{π_{a}}, 0 \log (0 / π_{a}) : = 0,$

where $\log$ is the natural logarithm. For every $t > 0$ , $p_{a} (t) > 0$ for all $a$ , and

$\frac{d}{d t} K (p_{t} ∣ π) = - \frac{1}{2} \sum_{\begin{array}{c} a, b \in E \\ a \neq b \end{array}} π_{a} w_{a b} (u_{a} - u_{b}) (\log u_{a} - \log u_{b}) \leq 0, u_{a} : = p_{a} / π_{a} .$

Equality in this derivative holds exactly when $p_{t} = π$ . For any initial probability vector, including one with zero entries, $t \mapsto K (p_{t} ∣ π)$ is continuous at $t = 0$ and nonincreasing on $[0, \infty)$ . This is a finite commuting-state statement; it asserts no entropy-production theorem for noncommuting densities.

Facts & Assumptions

Given: The finite state set $E$ , positive Gibbs weights $π_{a}$ , connected rates $w_{a b}$ , and the population equation in Connected Gibbs jump generators satisfy detailed balance.

[F1]

Connected Gibbs jump generators satisfy detailed balance: $π_{a} w_{a b} = π_{b} w_{b a}$ , the positive-rate graph is connected, and diagonal populations obey $p_{a}^{'} = \sum_{b \neq a} (p_{b} w_{b a} - p_{a} w_{a b})$ .

Proof

1.1givenF1chooseconstructalgebra

The population equation in [F1] is $p_{t}^{'} = p_{t} W$ for row probability vectors, so $p_{t} = p_{0} e^{t W}$ . Let $λ > \max_{a} r_{a}$ and $P : = I + W / λ$ . Then $P$ is a finite stochastic matrix with $P_{a a} > 0$ , and the matrix exponential satisfies

e^{t W} = e^{- λ t} \sum_{n = 0}^{\infty} \frac{(λ t)^{n}}{n!} P^{n} .

For any initial probability vector, choose $a$ with $p_{a} (0) > 0$ . For every $b$ , graph connectivity gives a finite positive-rate path from $a$ to $b$ ; the corresponding power $P^{n}$ has $(P^{n})_{a b} > 0$ . Its coefficient in the series is positive for $t > 0$ , so $p_{b} (t) > 0$ .

2.1givenF1step 1.1algebra

For $t > 0$ , put $u_{a} = p_{a} (t) / π_{a} > 0$ and, for $a \neq b$ , put $c_{a b} : = π_{a} w_{a b} = c_{b a}$ . Differentiating $K$ and using $\sum_{a} p_{a}^{'} = 0$ gives

\frac{d}{d t} K (p_{t} ∣ π) = \sum_{a} p_{a}^{'} \log u_{a} = \sum_{a \neq b} c_{a b} (u_{b} - u_{a}) \log u_{a} .

Pairing the $(a, b)$ and $(b, a)$ terms yields the displayed symmetric formula. Since $\log$ is increasing, each $(u_{a} - u_{b}) (\log u_{a} - \log u_{b}) \geq 0$ .

3.1givenF1step 2.1algebra∎

Equality in step 2.1 forces $u_{a} = u_{b}$ on every positive-rate edge. Connectedness makes all $u_{a}$ equal, and $\sum_{a} p_{a} = \sum_{a} π_{a} = 1$ then gives $u_{a} = 1$ , so $p_{t} = π$ . Conversely $p_{t} = π$ makes every term zero. Finally $x \log (x / π_{a}) \to 0$ as $x ↓ 0$ , so $K (p_{t} ∣ π)$ is continuous at $t = 0$ ; applying the derivative inequality on $[ε, t]$ and taking $ε ↓ 0$ proves nonincrease from any initial probability vector.

Depends on

Connected Gibbs jump generators satisfy detailed balance

Gibbs jump commuting relative entropy production

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources