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Explicit diagonal Fock domains and occupation trace — Examples

1 · Prerequisites

2 · Summary

Worked examples compute one-mode Bose/Fermi corrections, low-temperature oscillator and spin entropy, canonical two-mode correlations, ideal Fermi degeneracy pressure, and grand-condensate fluctuations. They retain each model's assumptions and do not introduce empirical observations.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Quantum occupation fluctuations and the classical approximation boundary

Physical scope

A one-mode exact comparison of Bose, Fermi, and first-order classical occupations at q=1/2, with no empirical interpretation.

Example

Take one dimensionless mode with fugacity $q = 1 / 2$ . The first-order dilute term for the mean occupation is $q = 1 / 2$ . As an explicit comparison, define the Poisson law by $P_{q} (n) = e^{- q} q^{n} / n!$ for $n \in N_{0}$ ; its mean and variance are both $q$ . The exact Bose law is geometric and the exact Fermi law is Bernoulli. This declared calculation illustrates the distinct occupation statistics and why a small-error conclusion needs a small-fugacity condition; the Poisson comparison is not asserted to follow from the linked theorem's mean and log-partition bounds.

Verification

Given: ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $q = 1 / 2$ , and the exact occupation and variance formulas of Controlled dilute Bose/Fermi occupation limit and Explicit diagonal Fock domains and occupation trace.

[F1] Controlled dilute Bose/Fermi occupation limit, Statement: Bose and Fermi mode means are $q / (1 - q)$ and $q / (1 + q)$ , with explicit errors relative to $q$ .

[F2] Explicit diagonal Fock domains and occupation trace, Statement: the one-mode Bose and Fermi variances are $q / (1 - q)^{2}$ and $q / (1 + q)^{2}$ .

1.1givenF1algebra

Substitution of $q = 1 / 2$ gives ${\overset{ˉ}{n}}_{B} = 1$ , ${\overset{ˉ}{n}}_{F} = 1 / 3$ , while the first-order value is $1 / 2$ . Thus the Bose error is $1 / 2$ and the Fermi error magnitude is $1 / 6$ .

2.1givenF2algebra∎

For the comparison law, $E N = e^{- q} \sum_{n \geq 1} n q^{n} / n! = q$ and $E [N (N - 1)] = e^{- q} \sum_{n \geq 2} n (n - 1) q^{n} / n! = q^{2}$ , so $Var (N) = q^{2} + q - q^{2} = q = 1 / 2$ . The exact variances from [F2] are $q / (1 - q)^{2} = 2$ for bosons and $q / (1 + q)^{2} = 2 / 9$ for fermions. These are finite exact model values; since $q$ is not near zero, no small-error claim follows from this example.

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Zero-point energy and ground-degeneracy entropy

Physical scope

A direct low-temperature comparison of the one-dimensional harmonic oscillator and the zero-field two-state spin model covered by the linked theorem. It records a model result, not a universal third-law statement or observation.

Example

For an oscillator with gap $ϵ = ℏ ω > 0$ , the $T ↓ 0$ canonical state approaches the nondegenerate ground level, with energy $ϵ / 2$ and entropy zero. For the separate spin Hamiltonian $H_{s} = - h σ_{z}$ , choosing $h = 0$ leaves two exactly degenerate states equally occupied at every temperature, with entropy $k_{B} \log 2$ . These model calculations show why ground degeneracy must be included when stating a low-temperature entropy limit.

Verification

Given: ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $ϵ > 0$ , the oscillator and spin models in Exact quantum oscillator and spin ensembles with their adopted canonical Gibbs preparations, and $k_{B} > 0$ .

[F1] Exact quantum oscillator and spin ensembles, Statement: oscillator energy tends to $ϵ / 2$ and entropy to zero; the spin entropy is $k_{B} \log 2$ when $h = 0$ .

1.1givenF1algebra

Since $e^{- β ϵ} \to 0$ as $T ↓ 0$ , the oscillator formulas give $\overset{ˉ}{n} \to 0$ , $U = ϵ (1 / 2 + \overset{ˉ}{n}) \to ϵ / 2$ , and the entropy function in [F1] tends zero.

2.1givenF1algebra∎

With $h = 0$ , the spin Gibbs weights are both $1 / 2$ , so $S = - k_{B} (2) (1 / 2) \log (1 / 2) = k_{B} \log 2$ independently of $T$ . Both conclusions follow from these Hamiltonians together with the adopted canonical Gibbs preparations.

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A fixed-number constraint removes independent occupations

Physical scope

An exact two-mode, one-particle canonical calculation compared with the distinct grand-canonical independent-mode preparation; no ensemble-equivalence claim is made.

Example

Consider two bosonic modes with energies $ϵ_{1}, ϵ_{2} \in R$ in J and inverse temperature $β > 0$ in J $^{- 1}$ . Represent occupation outcomes by $(n_{1}, n_{2}) \in N_{0}^{2}$ and define $N_{i} (n_{1}, n_{2}) : = n_{i}$ . Under a canonical preparation restricted to total number $N = N_{1} + N_{2} = 1$ , put $w_{i} = e^{- β ϵ_{i}}$ and $p_{i} = w_{i} / (w_{1} + w_{2})$ ; then the occupations satisfy $N_{1} + N_{2} = 1$ . For a grand-canonical comparison on these same two modes, choose chemical potential $μ < \min (ϵ_{1}, ϵ_{2})$ in J and put $q_{i} = e^{- β (ϵ_{i} - μ)} \in (0, 1)$ . The grand law has independent mode occupations, by factorization of its two geometric weights. This is the same product-law mechanism as in the linked periodic ideal model under its trace hypotheses.

Verification

Given: ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $β > 0$ , two finite energies, and the adopted canonical and grand preparations.

[A1] Quantum statistics and model equilibrium preparation: fixed-number canonical preparation is the Gibbs state restricted to the selected number sector; grand preparation is a separate trace state.

[F1] Finite spectral periodic Bose/Fermi gas, Statement: occupations are independent in the diagonal grand state under its trace hypotheses.

1.1givenA1algebra

The one-particle sector has exactly the two basis states $∣ 1, 0 ⟩$ and $∣ 0, 1 ⟩$ , with weights $w_{1}, w_{2}$ . Normalization gives probabilities $p_{1}, p_{2}$ with $p_{1} + p_{2} = 1$ , so $E N_{i} = p_{i}$ and $N_{1} N_{2} = 0$ in every canonical state outcome.

2.1step 1.1F1algebra∎

For the same two modes, the grand weights are $q_{1}^{n_{1}} q_{2}^{n_{2}}$ , so $Ξ = (1 - q_{1})^{- 1} (1 - q_{2})^{- 1}$ and the normalized joint law is $\prod_{i = 1}^{2} (1 - q_{i}) q_{i}^{n_{i}}$ on $N_{0}^{2}$ ; it is independent. The linked periodic theorem [F1] gives the same product law for its own modes under its trace hypotheses. By step 1.1, the fixed-number covariance is $Cov (N_{1}, N_{2}) = E (N_{1} N_{2}) - E N_{1} E N_{2} = - p_{1} p_{2} < 0$ . Thus the total-number constraint changes the joint law in this two-mode model, without implying any general finite-volume ensemble-equivalence failure.

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Fermi sea energy and ideal degeneracy pressure

Physical scope

A three-dimensional ideal Fermi sea at fixed density, calculating its Fermi wave number, mean energy per particle, and grand thermodynamic pressure from the linked zero-temperature theorem. It is not a material equation of state or measured pressure.

Example

For degeneracy $g = 2$ and number density $ρ > 0$ in the declared homogeneous ideal model, the zero-temperature occupied region is the ball $∣ k ∣ < k_{F}$ , where $k_{F} = (3 π^{2} ρ)^{1 / 3}$ . The mean kinetic energy per particle is $3 E_{F} / 5$ and the ideal grand thermodynamic pressure is $2 ρ E_{F} / 5$ , with $E_{F} = ℏ^{2} k_{F}^{2} / (2 m)$ . These values are model predictions under the ideal-gas assumptions.

Verification

Given: ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $g = 2$ , $ρ > 0$ , $α = ℏ^{2} / (2 m)$ , and the zero-temperature formulas of Ideal Fermi continuum and zero-temperature endpoint, with its SI convention: $m > 0$ is the mass parameter in kg, $ℏ > 0$ is the reduced Planck constant parameter in J s, $α$ has units J m $^{2}$ , $ρ$ is number per volume in m $^{- 3}$ , $k$ and $k_{F}$ have units m $^{- 1}$ , $E_{F}$ and $u / ρ$ have units J, $u$ has units J m $^{- 3}$ , and $p$ has units Pa $=$ J m $^{- 3}$ . The internal multiplicity $g = 2$ is dimensionless.

[F1] Ideal Fermi continuum and zero-temperature endpoint, Statement: $k_{F} = (6 π^{2} ρ / g)^{1 / 3}$ , $u = (3 / 5) ρ E_{F}$ , and $p = 2 u / 3$ .

1.1givenF1algebra

Setting $g = 2$ in [F1] gives $k_{F} = (3 π^{2} ρ)^{1 / 3}$ and $E_{F} = α k_{F}^{2} = ℏ^{2} k_{F}^{2} / (2 m)$ . Dividing the energy density $u$ by the number density $ρ$ gives mean energy per particle $3 E_{F} / 5$ .

2.1step 1.1F1algebra∎

The pressure identity in [F1] yields $p = (2 / 3) (3 / 5) ρ E_{F} = 2 ρ E_{F} / 5$ . The result applies to the ideal continuum model and supplies no material-specific or empirical pressure.

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Grand condensate variance remains macroscopic

Physical scope

The grand-ensemble ground-mode variance and tail at a fixed positive excess density above the ideal critical density, as a direct instance of the linked theorem. It makes no canonical or interacting-gas claim.

Example

In the three-dimensional ideal Bose grand ensemble at fixed density $ρ > ρ_{c}$ , put $ρ_{0} = ρ - ρ_{c} > 0$ . The ground-mode mean is asymptotic to $ρ_{0} V$ , but ${Var}_{L} (N_{0} / V) \to ρ_{0}^{2}$ , and for every $x > 0$ in m $^{- 3}$ its density tail satisfies $P_{L} (N_{0} / V \geq x) \to e^{- x / ρ_{0}}$ . Thus the condensate density does not concentrate at its mean limit in this grand ensemble.

Verification

Given: ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $V = L^{3}$ in m $^{3}$ , $ρ_{0} > 0$ in m $^{- 3}$ , $x > 0$ in m $^{- 3}$ , and the grand geometric ground-mode law in Dimension-restricted fixed-mean-density Bose condensation.

[F1] Dimension-restricted fixed-mean-density Bose condensation, Statement: $E_{L} N_{0} / V \to ρ_{0}$ , ${Var}_{L} (N_{0} / V) \to ρ_{0}^{2}$ , and $P_{L} (N_{0} / V \geq x) \to e^{- x / ρ_{0}}$ .

1.1givenF1algebra

The linked theorem gives $E_{L} N_{0} = ρ_{0} V + o (V)$ . Substitution into the geometric variance ${\overset{ˉ}{n}}_{0, L} (1 + {\overset{ˉ}{n}}_{0, L})$ and division by $V^{2}$ gives limit $ρ_{0}^{2}$ .

2.1givenF1algebra∎

At $x = 2 ρ_{0}$ , $P_{L} (N_{0} / V \geq 2 ρ_{0}) \to e^{- 2} > 0$ . Every such event differs from the limiting mean density $ρ_{0}$ by at least $ρ_{0}$ , so convergence in probability to $ρ_{0}$ is impossible. This conclusion is specific to the grand preparation; a fixed-number state has $N_{0}$ constrained by total number.

Explicit diagonal Fock domains and occupation trace — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Quantum occupation fluctuations and the classical approximation boundary

Example

Verification

Zero-point energy and ground-degeneracy entropy

Example

Verification

A fixed-number constraint removes independent occupations

Example

Verification

Fermi sea energy and ideal degeneracy pressure

Example

Verification

Grand condensate variance remains macroscopic

Example

Verification

Sources