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Ensembles And Limits — Examples

1 · Prerequisites

2 · Summary

These examples calculate one countable partition sum and three finite probability models. Each example checks definitions and conditional results on the A page; no measurement or physical preparation is asserted.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

A lower energy bound does not ensure a countable partition function

Facts & Assumptions

Given: As stated in the item.

[F1]

Countable canonical model: $Z$ is the extended nonnegative sum and a canonical probability law is defined only when $0 < Z < \infty$ .

[A1]

The state set is $N$ and every state has zero energy and unit weight.

Example

Take $I = N$ , $E_{n} = 0$ , and $g_{n} = 1$ for every $n$ . The energy list is bounded below, but $Z (β) = + \infty$ for every real $β$ , so this model has no normalized canonical law.

Verification

Given: The countable model in [F1] with the data in [A1].

1.1F1A1algebra

Every summand is $g_{n} e^{- β E_{n}} = 1$ , so the first $N$ terms sum to $N$ for every $N$ ; their supremum is infinite.

2.1F1step 1.1∎

Thus $Z (β) = + \infty$ , which violates the finite-normalization condition in [F1]; no normalized canonical probabilities $p_{n}$ are defined.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

Canonical mean and variance for a two-level system

Facts & Assumptions

Given: As stated in the item.

[F1]

Countable canonical model: the finite partition sum and normalized state weights are defined as stated there.

[F2]

Summable canonical differentiation: a finite energy list has the required local summable moment envelope on every compact parameter interval (each of finitely many terms is bounded there), so $- \partial_{β} \log Z$ is the mean and $\partial_{β}^{2} \log Z$ is the variance.

[A1]

$E_{0} = 0$ , $E_{1} = ε > 0$ , $g_{0} = g_{1} = 1$ , and $β > 0$ .

Example

Let the state set be ${0, 1}$ , with energies $E_{0} = 0$ , $E_{1} = ε > 0$ , unit weights, and $β > 0$ . Writing $q = e^{- β ε}$ , the excited-state probability is $q / (1 + q)$ , the mean energy is $ε q / (1 + q)$ , and the energy variance is $ε^{2} q / (1 + q)^{2} > 0$ .

Verification

Given: The finite canonical model [F1] with data [A1].

1.1F1A1algebra

Direct summation gives $Z = 1 + e^{- β ε} = 1 + q$ , hence $p_{1} = q / (1 + q)$ and $p_{0} = 1 / (1 + q)$ .

2.1F1F2A1step 1.1algebra∎

The finite weighted sum gives $\overset{ˉ}{E} = ε p_{1} = ε q / (1 + q)$ . The second centered moment is $p_{0} (0 - \overset{ˉ}{E})^{2} + p_{1} (ε - \overset{ˉ}{E})^{2} = ε^{2} q / (1 + q)^{2}$ , which is positive because $q \in (0, 1)$ . The values also agree with the finite-list derivative identities of [F2].

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

A perfectly correlated uniform pair has zero conditional entropy

Facts & Assumptions

Given: As stated in the item.

[F1]

Mutual information of finite random variables: $I (M; R) = H (M) + H (R) - H (M, R) = H (M) - H (M ∣ R)$ .

[F2]

Conditional entropy of finite memory given side information: conditional entropy is computed from the conditional laws on positive-probability register values.

[A1]

The pair law assigns probability $1 / 2$ to $(0, 0)$ and $(1, 1)$ and zero to the other two pairs.

Example

Let $M, R \in {0, 1}$ have joint probabilities $p (0, 0) = p (1, 1) = 1 / 2$ and $p (0, 1) = p (1, 0) = 0$ . Then $H (M) = H (R) = \log 2$ , $H (M, R) = \log 2$ , $H (M ∣ R) = 0$ , and $I (M; R) = \log 2$ .

Verification

Given: The finite joint probability law in [A1].

1.1A1algebra

Both marginals are uniform, so their entropies are $- 2 (1 / 2) \log (1 / 2) = \log 2$ . The joint law has two positive entries of size $1 / 2$ , so $H (M, R) = \log 2$ .

2.1F1F2A1step 1.1∎

Conditional on either value of $R$ , $M$ is determined, so its conditional law has one probability $1$ and one $0$ ; therefore $H (M ∣ R) = 0$ . Substituting these values into [F1] gives $I (M; R) = \log 2$ .

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

A finite canonical law differs from a narrow-shell microcanonical law

Facts & Assumptions

Given: As stated in the item.

[F1]

Countable canonical model: finite canonical probabilities are normalized exponential weights.

[F2]

Summable canonical differentiation: if each compact interval $K \subset J$ admits a summable nonnegative majorant $a_{i}$ with $g_{i} (1 + ∣ E_{i} ∣^{2}) e^{- β E_{i}} \leq a_{i}$ for all $i$ and $β \in K$ , then $Z$ is positive, finite and $C^{2}$ on $J$ , and the canonical law is normalized.

[F3]

The natural logarithm as the inverse of the exponential function identifies the exponential as a bijection from $R$ onto $(0, \infty)$ ; The exponential function is strictly increasing supplies its monotonicity.

[A1]

The state set has energies $0$ and $ε > 0$ , the weights are one, $β > 0$ , and $0 < Δ < ε$ . The finite-shell convention is the uniform law on states with energies in $[0, Δ)$ when the shell is nonempty.

Example

Let a two-state model have energies $0$ and $ε > 0$ , with unit state weights and $β > 0$ . Adopt the finite-shell convention that the microcanonical law is uniform over states whose energy lies in $[0, Δ)$ when the shell is nonempty. For $0 < Δ < ε$ , that shell contains only the ground state, whereas the canonical law assigns the excited state probability $e^{- β ε} / (1 + e^{- β ε}) > 0$ . Thus the two finite-ensemble laws differ.

Verification

Given: The finite model and shell convention in [A1].

1.1F1F2F3A1

Take $J = (0, \infty)$ . For any nonempty compact interval $K \subset J$ , set $a_{0} = 1$ and $a_{1} = (1 + ε^{2}) e^{- ε \min K}$ . These two finite nonnegative numbers are summable and bound the ground and excited terms in [F2], by [F3]. Thus the hypotheses of [F2] hold on $J$ . The canonical partition sum is $1 + e^{- β ε}$ ; its excited probability is $e^{- β ε} / (1 + e^{- β ε})$ , strictly positive because the exponential is positive.

2.1A1step 1.1∎

The shell condition $0 \leq E_{i} < Δ$ includes the state of energy zero and excludes the state of energy $ε$ , since $Δ < ε$ . Uniform weighting of this one-state shell gives excited probability zero, which differs from step 1.1.

Ensembles And Limits — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

A lower energy bound does not ensure a countable partition function

Facts & Assumptions

Example

Verification

Canonical mean and variance for a two-level system

Facts & Assumptions

Example

Verification

A perfectly correlated uniform pair has zero conditional entropy

Facts & Assumptions

Example

Verification

A finite canonical law differs from a narrow-shell microcanonical law

Facts & Assumptions

Example

Verification

Sources