ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

Depends on

Dependency tree · 2 levels

Sources

David Tong, Statistical Physics, §3.5, canonical and grand-canonical occupation sums