Alphabeta Math

1 · Prerequisites

2 · Summary

These examples calculate one-mode bosonic and fermionic weights, a two-site product law, divergence of the unrenormalized trace for countably many identical bosonic modes, and two frozen finite models that separate canonical preparation from dynamical equilibration. Each example states its own model and derives the claimed probabilities or state invariance from its declared prerequisites, including the assigned A-page items and, for the isolated quantum dynamics, the NRQM autonomous-evolution assumption and definition. The NRQM B-leaf examples are not proof premises. The examples are hypothetical calculations, not experimental observations.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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One bosonic and one fermionic mode at a specified temperature

Physical scope

An exact hypothetical canonical calculation for one bosonic oscillator mode and one fermionic two-level mode with beta times excitation energy equal to log 2; it is not a measurement or a thermalization claim.

Example

Assume AC and Countable Choice as required by Finite bosonic and fermionic mode partition and occupation formulas. Choose $T > 0$ kelvins and excitation energy $ε = k_{B} T \log 2$ joules, so $q = e^{- β ε} = 1 / 2$ . Consider separately one bosonic mode with occupations $n = 0, 1, 2, \dots$ and one fermionic mode with occupations $n = 0, 1$ .

Facts & Assumptions

Given: The declared one-mode systems, their positive excitation energy, and the adopted Gibbs preparation.

[F1]

Finite bosonic and fermionic mode partition and occupation formulas: for $q = e^{- β ε}$ , the bosonic partition trace is $(1 - q)^{- 1}$ with mean $q / (1 - q)$ , and the fermionic partition sum is $1 + q$ with excited occupation $q / (1 + q)$ ; these are conditional on its stated quantum preparation and suppliers.

[F2]

SI quantities and unit conventions: $T$ is in kelvins, $k_{B}$ in $J K^{- 1}$ , and $β = (k_{B} T)^{- 1}$ in $J^{- 1}$ .

[A1]

The selected energy is $ε = k_{B} T \log 2 > 0$ joules, and $q = e^{- β ε} = 1 / 2$ .

[A2]

AC and Countable Choice are assumed through [F1] for the quantum Gibbs assignment and the supplied one-mode calculations.

Verification

Given: $q = 1 / 2$ and the finite-mode identities in [F1].

1.1givenF1F2A1algebra

The bosonic trace is $Z_{B} = (1 - 1 / 2)^{- 1} = 2$ , and the probability of occupation $n$ is $2^{- (n + 1)}$ . Its mean occupation is $(1 / 2) / (1 - 1 / 2) = 1$ .

2.1F1F2A1algebra∎

The fermionic partition sum is $Z_{F} = 1 + 1 / 2 = 3 / 2$ , so the ground and excited probabilities are $2 / 3$ and $1 / 3$ , respectively, and the mean occupation is $1 / 3$ . The bosonic and fermionic values belong to distinct declared models.

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Two independent two-level sites give a product thermal law

Physical scope

A calculated two-site finite product model with additive energy and no interaction; it is not an observation or a claim about interacting sites.

Example

Let each site have states $0, 1$ with energies $0, ε$ , and take $T > 0$ with $ε = k_{B} T \log 2$ . The composite has states $(n_{1}, n_{2}) \in {0, 1}^{2}$ and additive energy $E = ε (n_{1} + n_{2})$ .

Facts & Assumptions

Given: Two finite sites, additive energy, common positive temperature, and canonical sampling.

[F1]

Finite statistical microstate model: the composite state set is the Cartesian product and additivity of energy is explicit model data.

[F2]

Finite canonical partition, energy, variance, entropy and free-energy identities: each finite canonical law has probabilities proportional to $e^{- β E}$ .

[F3]

Partition, Gibbs-entropy and free-energy factorization for independent finite systems: under the explicit product and additive-energy assumptions, partition sums factor and the joint canonical law is the product of the two subsystem laws.

[A1]

Set $β ε = \log 2$ , hence the single-site excited-state weight is $q = 1 / 2$ .

Verification

Given: The model data and $q = 1 / 2$ .

1.1F1F2F3A1algebra

Each site has $Z_{1} = 1 + q = 3 / 2$ and probabilities $(2 / 3, 1 / 3)$ . By [F3], $Z_{12} = Z_{1}^{2} = 9 / 4$ and the joint probabilities of $(00, 01, 10, 11)$ are $(4 / 9, 2 / 9, 2 / 9, 1 / 9)$ .

2.1F2F3step 1.1algebra∎

The expected total occupation is the sum of the two marginal means, $1 / 3 + 1 / 3 = 2 / 3$ , so the mean energy is $2 ε / 3$ . This conclusion follows for the declared noninteracting product model only.

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Identical bosonic modes do not have a finite infinite-mode Gibbs trace

Physical scope

A countable family of independent identical bosonic modes at fixed positive excitation energy; the divergent positive trace rules out this particular normalized Gibbs density and says nothing about finite mode sets or renormalized models.

Example

Assume AC and Countable Choice as required by Finite bosonic and fermionic mode partition and occupation formulas and the trace-class suppliers. Take modes indexed by $j \in N$ , each with excitation energy $ε > 0$ joules, and let $q = e^{- β ε} \in (0, 1)$ . Let $F$ be the set of finite-support occupation sequences $n = (n_{j})_{j \geq 1}$ with $n_{j} \in N_{0}$ . Enumerate $F$ by increasing $R (n) = \sum_{j : n_{j} > 0} (j + n_{j})$ and lexicographic order of the finite list of nonzero pairs $(j, n_{j})$ within each level. Each level is finite because $j, n_{j} \leq R (n)$ on its support, and every configuration occurs; skipping empty levels gives a bijection $N \to F$ . Define $ℓ^{2} (F) : = {c : F \to C : \sum_{n \in F} ∣ c (n) ∣^{2} < \infty}, ⟨ c, d ⟩ : = \sum_{n \in F} c (n) \overline{d (n)} .$ The vector $∣ n ⟩$ is the indicator of ${n}$ ; these canonical occupation vectors form an orthonormal basis. Relabeling by the displayed enumeration is an isometry with $ℓ^{2} (N, C)$ , whose completeness and dense finite-support subspace are supplied by Completeness and linear domain of the diagonal sequence model; the coordinate pairing makes the displayed indicator vectors orthonormal, and their finite linear span is exactly that dense subspace, so they form an orthonormal basis; the definition of square-summable families and their inner product is Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ . Define the bounded positive diagonal Gibbs weight $K_{β} ∣ n ⟩ = q^{\sum_{j} n_{j}} ∣ n ⟩$ . On the subspace with $n_{j} = 0$ for every $j > M$ , the corresponding finite-mode partition sum is the trace of that restriction.

Facts & Assumptions

Given: The independent identical-mode Fock model, positive $β ε$ , and the canonical trace convention.

[F1]

Finite bosonic and fermionic mode partition and occupation formulas: for each finite collection of independent bosonic modes the Gibbs trace is the product of the one-mode traces $(1 - q)^{- 1}$ .

[F2]

Canonical sampling assumption for finite and normalized statistical models: the A-page countable canonical law is adopted only for a specified countable model with a positive finite partition sum.

[F3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ), Trace class operator, and Trace of a positive operator is the sum of its eigenvalues: under Countable Choice, a positive trace-class operator has finite trace equal to the sum of its nonnegative eigenvalues with multiplicity.

[F4]

Countable canonical model: the nonnegative countable partition sum is the supremum of its finite subsums, so a finite normalizer bounds the sum over every subset of the listed states.

[F5]

SI quantities and unit conventions: $ε$ is in joules and $β$ in $J^{- 1}$ , so $β ε$ is dimensionless.

[F6]

Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ defines $ℓ^{2} (F)$ and its inner product. The explicit enumeration identifies it isometrically with $ℓ^{2} (N, C)$ , which is complete with a dense finite-support subspace by Completeness and linear domain of the diagonal sequence model. The indicator vectors are orthonormal by the coordinate pairing, and their finite linear span is that dense subspace, so they form an orthonormal basis.

[A1]

The basis indices are all finite-support occupation sequences with physical energy $E_{n} = ε \sum_{j} n_{j}$ joules, so their Gibbs eigenvalues are $e^{- β E_{n}} = q^{\sum_{j} n_{j}}$ ; every finite-mode subset is a subset of this full basis.

[A2]

AC and Countable Choice are assumed through the finite-mode Gibbs result and the trace-class definitions/theorems cited above.

Verification

Given: $0 < q < 1$ and the finite-mode subspaces described above.

1.1F1A1F5algebra

The trace contribution from states supported on the first $M$ modes is the finite product $Z_{M} = (1 - q)^{- M}$ by [F1]. Since $c = (1 - q)^{- 1} > 1$ , $Z_{M} = c^{M}$ is unbounded: writing $c = 1 + r$ with $r > 0$ , induction gives $(1 + r)^{M} \geq 1 + M r$ . The unit convention [F5] makes $β ε$ dimensionless.

2.1A1F1F2F3F4F6step 1.1algebra∎

If $K_{β}$ were trace class, [F3] would make its positive eigenvalue sum finite and equal to its trace. For each $M$ , the configurations with $n_{j} = 0$ for $j > M$ form a subset of the enumerated states [A1]; [F4] therefore makes their finite-mode partition sum $Z_{M}$ no larger than the full normalizer. [F1] identifies that subset sum as $Z_{M}$ , and step 1.1 makes $Z_{M}$ unbounded, a contradiction. Hence $K_{β}$ is not trace class and the countable partition sum has no finite value. The preparation condition [F2] fails, so this model has no normalized countable canonical law of the stated form.

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An identity dynamics leaves a noncanonical finite preparation unchanged

Physical scope

A two-state finite model with explicitly frozen identity dynamics; it shows that a canonical preparation rule alone supplies no relaxation dynamics and is not an empirical statement.

Example

Let $Ω = {a, b}$ with energies $E_{a} = 0$ , $E_{b} = ε > 0$ joules, and fix $T > 0$ kelvins. The separately adopted canonical law has probabilities proportional to $(1, e^{- β ε})$ . For the dynamical comparison, use a dimensionless model-time parameter $τ \geq 0$ , stipulate the identity transition map $Φ_{τ} (a) = a$ , $Φ_{τ} (b) = b$ for all $τ$ , and start with the point mass $δ_{a}$ .

Facts & Assumptions

Given: A two-state energy model, positive temperature, the canonical preparation rule, and the explicitly stipulated identity dynamics.

[F1]

Finite statistical microstate model: the two microstates and their fixed finite energies define an admissible finite statistical model.

[F2]

Canonical sampling assumption for finite and normalized statistical models: the canonical probabilities are the normalized weights $e^{- β E_{x}}$ ; this is an adopted preparation rule and includes no time-evolution equation.

[A1]

The comparison dynamics is the deterministic identity map $Φ_{τ}$ for every dimensionless model time $τ \geq 0$ , and the initial law for that comparison is $δ_{a}$ .

Verification

Given: $ε > 0$ , $T > 0$ , and the stipulated dynamics.

1.1F1F2algebra

Put $q = e^{- β ε}$ , so $0 < q < 1$ . The canonical law is $P_{β} (a) = 1 / (1 + q)$ and $P_{β} (b) = q / (1 + q)$ ; in particular it is not $δ_{a}$ .

2.1A1step 1.1algebra∎

The identity map preserves the initial point mass at every model time: $(Φ_{τ})_{*} δ_{a} = δ_{a}$ . Hence the time-evolved law does not converge to the canonical law, whose probability of $b$ is strictly positive. This is a counterexample to deriving equilibration from the sampling rule alone.

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Two isolated two-level sites do not relax an energy eigenstate to Gibbs form

Physical scope

A hypothetical closed two-site quantum model initialized in any pure energy eigenstate, including a superposition within a degenerate eigenspace; unitary evolution preserves that state and does not produce its finite-temperature Gibbs mixture. This is not an experiment.

Example

Assume AC and Countable Choice for the canonical mode preparation. Let $H_{1} = H_{2} = C^{2}$ with orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ , and define the factor number maps by $N_{r} ∣ 0 ⟩ = 0$ , $N_{r} ∣ 1 ⟩ = ∣ 1 ⟩$ . The composite Hilbert space is the finite Hilbert tensor product $H_{1} \hat{\otimes} H_{2} ≅ C^{4}$ from Hilbert tensor construction: inner product, completion and product bases, with orthonormal product basis $∣ n_{1} n_{2} ⟩$ , $n_{r} \in {0, 1}$ . On this basis define the site-number operators $N_{1} ∣ n_{1} n_{2} ⟩ = n_{1} ∣ n_{1} n_{2} ⟩$ and $N_{2} ∣ n_{1} n_{2} ⟩ = n_{2} ∣ n_{1} n_{2} ⟩$ . For $T > 0$ kelvins set $ε = k_{B} T \log 2$ joules and define the noninteracting total Hamiltonian $H = ε (N_{1} + N_{2}), H ∣ n_{1} n_{2} ⟩ = ε (n_{1} + n_{2}) ∣ n_{1} n_{2} ⟩ .$ The quantum Gibbs preparation at $T$ is used for the thermal comparison, and the closed-system unitary evolution is generated by $H$ with $U_{t} = e^{- i t H / ℏ}$ , where $t$ is in seconds and $ℏ > 0$ is in $J s$ . Start the dynamics in $ρ_{0} = ∣ ψ ⟩ ⟨ ψ ∣$ , where $∥ ψ ∥ = 1$ and $H ψ = λ ψ$ for $λ \in {0, ε, 2 ε}$ . Here $∣ ψ ⟩ ⟨ ψ ∣$ means the rank-one map $v \mapsto ⟨ v, ψ ⟩ ψ$ , with inner product linear in the first argument. This includes every unit vector in the degenerate eigenspace $span {∣ 01 ⟩, ∣ 10 ⟩}$ , as well as the ground and top eigenvectors. The preparation stays constant for every time and does not converge to the Gibbs density. The choice $ψ = ∣ 00 ⟩$ gives the explicit ground-state comparison below.

Facts & Assumptions

Given: The finite product energy model, the assigned positive temperature, the unitary evolution postulate, and the stated initial state.

[F1]

The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature: under AC and Countable Choice, a declared self-adjoint semibounded energy generator with a positive finite Gibbs trace is assigned the density $ρ_{β} = e^{- β H} / Z$ .

[F2]

Finite bosonic and fermionic mode partition and occupation formulas: for a finite tensor product of independent fermionic two-level factors with additive total energy, the trace is the product of the one-mode traces $1 + q$ , and the joint occupation-basis law is the product of the one-mode Gibbs weights.

[F3]

Autonomous dynamics postulate P4: self-adjoint generators and unitary evolution and Assigned energy observable, unitary flow and the two pictures of autonomous evolution: closed-system evolution is $ρ (t) = U_{t} ρ_{0} U_{t}^{*}$ with $U_{t} = e^{- i t H / ℏ}$ .

[A1]

Each site has $q = 1 / 2$ , and the initial vector $ψ$ is any unit energy eigenvector of $H$ ; $∣ 00 ⟩$ is the ground-state special case.

Verification

Given: $q = 1 / 2$ , the product model, a unit energy eigenvector, and the closed-system evolution.

1.1F1F2A1algebra

The product trace in [F2] is $Z = (1 + q)^{2} = 9 / 4$ . By the Gibbs assignment [F1] and the product-basis energies from the Statement, $P (n_{1}, n_{2}) = q^{n_{1} + n_{2}} / Z$ ; at $q = 1 / 2$ , the probabilities on $(00, 01, 10, 11)$ are $(4 / 9, 2 / 9, 2 / 9, 1 / 9)$ . In particular, the Gibbs probability of $∣ 11 ⟩$ is $1 / 9 > 0$ .

2.1F1F3A1step 1.1algebra

For any stated eigenvector, the diagonal spectral evolution in [F3] gives $U_{t} ψ = e^{- i t λ / ℏ} ψ$ , including every vector in the degenerate middle eigenspace. The phase has modulus one, so $ρ (t) = ∣ ψ ⟩ ⟨ ψ ∣$ for every $t$ . The Gibbs density is diagonal with the four strictly positive weights of step 1.1, hence has rank four, whereas $ρ (t)$ has rank one. They are distinct; the constant orbit cannot converge to the Gibbs density in any matrix norm.

3.1F3A1step 1.1algebra∎

In the special case $ψ = ∣ 00 ⟩$ , $H ∣ 00 ⟩ = 0$ , $U_{t} ∣ 00 ⟩ = ∣ 00 ⟩$ for every $t$ ; thus $ρ (t) = ∣ 00 ⟩ ⟨ 00 ∣$ . Its probability for the projector $∣ 11 ⟩ ⟨ 11 ∣$ is zero at every time, unlike the Gibbs value $1 / 9$ . The state therefore never reaches this Gibbs density under the declared unitary evolution.

Finite Statistical Bridge — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

One bosonic and one fermionic mode at a specified temperature

Example

Facts & Assumptions

Verification

Two independent two-level sites give a product thermal law

Example

Facts & Assumptions

Verification

Identical bosonic modes do not have a finite infinite-mode Gibbs trace

Example

Facts & Assumptions

Verification

An identity dynamics leaves a noncanonical finite preparation unchanged

Example

Facts & Assumptions

Verification

Two isolated two-level sites do not relax an energy eigenstate to Gibbs form

Example

Facts & Assumptions

Verification

Sources