Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Canonical weights uniquely maximize finite-state entropy at their mean energy

Physical scope

Finite probability distributions on the declared microstate set at a fixed positive temperature; the result is a conditional variational property of the model and does not derive the physical canonical-preparation postulate.

Statement

Let $(Ω, ϵ)$ be a finite nonempty microstate model, let $T > 0$ , and let $q_{x} = e^{- β E_{x}} / Z$ be its canonical probabilities at $β = (k_{B} T)^{- 1}$ . Set $\overset{ˉ}{E} = \sum_{x} q_{x} E_{x}$ . Among all probability vectors $r = (r_{x})_{x \in Ω}$ with $\sum_{x} r_{x} = 1$ and $\sum_{x} r_{x} E_{x} = \overset{ˉ}{E}$ , the dimensionless Shannon entropy $h (r) : = - \sum_{x : r_{x} > 0} r_{x} \log r_{x}$ is uniquely maximized by $r = q$ . The corresponding physical Gibbs entropy is $S (r) : = k_{B} h (r)$ in $J K^{- 1}$ and has the same unique maximizer. This is a finite-model variational theorem; it does not establish that a physical sample is prepared canonically.

Facts & Assumptions

Given: A finite nonempty state set with fixed energies $E_{x}$ in joules, $T > 0$ , $β = (k_{B} T)^{- 1}$ , the canonical law $q$ , and a competing probability vector $r$ with the same mean energy.

[F1]

Finite statistical microstate model: $Ω$ is finite and nonempty and $E_{x}$ is fixed.

[F2]

Canonical sampling assumption for finite and normalized statistical models: $q_{x} = e^{- β E_{x}} / Z$ with $q_{x} > 0$ and $\sum_{x} q_{x} = 1$ .

[F3]

Finite canonical partition, energy, variance, entropy and free-energy identities: $\overset{ˉ}{E} = \sum_{x} q_{x} E_{x}$ , $\log q_{x} = - β E_{x} - \log Z$ , and the physical entropy satisfies $S (q) = k_{B} (\log Z + β \overset{ˉ}{E})$ . Dividing by the positive $k_{B}$ gives $h (q) = \log Z + β \overset{ˉ}{E}$ .

[F4]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t and The natural logarithm as the inverse of the exponential function: $\log t = \int_{1}^{t} d u / u$ for $t > 0$ .

[A1]

The competitor has $r_{x} \geq 0$ , $\sum_{x} r_{x} = 1$ , and $\sum_{x} r_{x} E_{x} = \overset{ˉ}{E}$ .

Proof

Given: The finite model, its canonical weights, and a competing probability vector with the same mean.

1.1F2F4A1algebra

For every $t > 0$ , [F4] gives $\log t \leq t - 1$ , with equality only at $t = 1$ : for $t > 1$ , integrate $1 / u < 1$ on $(1, t)$ ; for $0 < t < 1$ , integrate $1 / u > 1$ on $(t, 1)$ and reverse the endpoints. For every $x$ with $r_{x} > 0$ , apply this to $t = q_{x} / r_{x}$ to obtain $r_{x} \log (r_{x} / q_{x}) \geq r_{x} - q_{x}$ . Summing over the support of $r$ gives $D : = \sum_{x : r_{x} > 0} r_{x} \log (r_{x} / q_{x}) \geq 1 - \sum_{x : r_{x} > 0} q_{x} \geq 0$ . Equality forces full support and $r_{x} = q_{x}$ for every state.

2.1F1F2F3A1step 1.1algebra∎

Using [F3] and the fixed-mean condition [A1], $D = \sum_{x : r_{x} > 0} r_{x} \log r_{x} + β \overset{ˉ}{E} + \log Z$ . Therefore $h (r) = β \overset{ˉ}{E} + \log Z - D \leq β \overset{ˉ}{E} + \log Z = h (q)$ , so $S (r) = k_{B} h (r) \leq k_{B} h (q) = S (q)$ . Since $k_{B} > 0$ , equality holds only when $D = 0$ , which by step 1.1 is equivalent to $r = q$ . This proves the unique maximum of both entropies.

Canonical weights uniquely maximize finite-state entropy at their mean energy

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources