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Finite-regulator Gibbs predictions

Physical scope

Trace-normalized bounded-observable expectations at T>0 for one fixed finite regulator. No infinite-volume or continuum trace, empirical data, or unbounded-observable formula is asserted.

Statement

Assume Countable Choice and the finite-regulator coercivity hypotheses of Finite-regulator Gibbs trace and bounded expectations. Use the model state of Adopted finite-regulator equilibrium state and the inverse-energy parameters of Finite-regulator temperature and inverse-energy parameter. Let $α_{j} : = \sqrt{a_{j} c}$ , $j = 1, \dots, N$ , for the dimensionless kinetic coefficients and coercivity constant in the model, and let $C_{0} \geq 0$ be its potential lower-bound shift. Then

$0 < Z_{T} = \sum_{k \geq 0} e^{- \hat{β} λ_{k}} \leq e^{\hat{β} C_{0}} \prod_{j = 1}^{N} \frac{e^{- \hat{β} α_{j}}}{1 - e^{- 2 \hat{β} α_{j}}} < \infty,$

where $λ_{k}$ are the eigenvalues of $h$ . The adopted state $ρ_{T}$ is a positive trace-class operator with trace one. For every bounded linear operator $B : H_{N} \to H_{N}$ ,

$Tr (ρ_{T} B) = Z_{T}^{- 1} \sum_{k \geq 0} e^{- \hat{β} λ_{k}} ⟨ B ψ_{k}, ψ_{k} ⟩,$

and the series is absolutely convergent. If $E_{a}$ is an effect in a supplied normalized finite instrument of Physical interpretation of a supplied constructed quantum model, this trace is its adopted canonical model probability. These are conditional finite-regulator predictions, not measured outcomes or a continuum-limit claim.

Facts & Assumptions

Given: Countable Choice; $N \geq 1$ ; dimensionless $a_{j}, c > 0$ and $C_{0} \geq 0$ ; the fixed coercive real polynomial; $E_{*} > 0$ joules; $T > 0$ ; $β_{p h y s} = (k_{B} T)^{- 1}$ ; and the selected form realization $h$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable family of nonempty sets has a choice function. The QFT-IR01 Gibbs-trace proof assumes this principle.

[A2]

Adopted finite-regulator equilibrium state: using its supplied complete eigenbasis and eigenvalues $(ψ_{k}, λ_{k})$ , the postulate defines $e^{- γ h}$ by $G_{γ} f = \sum_{k} e^{- γ λ_{k}} ⟨ f, ψ_{k} ⟩ ψ_{k}$ . It defines $e^{- β_{p h y s} H_{p h y s}} f = \sum_{k} e^{- β_{p h y s} E_{*} λ_{k}} ⟨ f, ψ_{k} ⟩ ψ_{k}$ ; since $\hat{β} = E_{*} β_{p h y s}$ , this is the same operator $G_{\hat{β}} = e^{- \hat{β} h}$ . Thus $ρ_{T} = G_{\hat{β}} / Z_{T}$ .

[A3]

Physical interpretation of a supplied constructed quantum model: a supplied normalized finite instrument assigns model probability $Tr (ρ E_{a})$ to its outcome effect $E_{a}$ , but only after the density operator and instrument are actually constructed.

[F1]

Finite-regulator Gibbs trace and bounded expectations: for dimensionless $γ > 0$ , the supplier defines $ρ_{γ}$ as the operator-norm limit of finite-rank truncations and gives its action $ρ_{γ} f = \sum_{k} e^{- γ λ_{k}} ⟨ f, ψ_{k} ⟩ ψ_{k}$ . It proves that this operator is positive trace class, that its trace is positive and finite with the oscillator-product upper bound, and that for bounded $B$ the trace $Tr (ρ_{γ} B)$ is the absolutely convergent eigenbasis sum.

[F2]

Finite-regulator temperature and inverse-energy parameter: $β_{p h y s} = 1 / (k_{B} T)$ has unit $J^{- 1}$ and $\hat{β} = E_{*} β_{p h y s} > 0$ is dimensionless.

Proof

technique · direct

1.1A2F1F2givenalgebra

By [F2], $\hat{β} = E_{*} β_{p h y s} > 0$ is dimensionless. For each $f \in H_{N}$ , the series defining $e^{- β_{p h y s} H_{p h y s}} f$ has coefficients $e^{- β_{p h y s} E_{*} λ_{k}} ⟨ f, ψ_{k} ⟩$ , while the series defining $e^{- \hat{β} h} f$ has coefficients $e^{- \hat{β} λ_{k}} ⟨ f, ψ_{k} ⟩$ . These coefficients are equal term by term, and both series converge in $H_{N}$ ; hence $e^{- β_{p h y s} H_{p h y s}} = e^{- \hat{β} h} = G_{\hat{β}}$ as operators. The trace supplied by [F1] is therefore $Z_{T} = \sum_{k} e^{- \hat{β} λ_{k}}$ .

2.1A1A2F1step 1.1algebra

Under Countable Choice [A1], apply [F1] at $\hat{β} > 0$ . It gives $Z_{T} > 0$ , the stated finite upper bound, and trace-class positivity of $G_{\hat{β}} = e^{- \hat{β} h}$ . Dividing by $Z_{T}$ yields $Tr (ρ_{T}) = 1$ and preserves positivity and trace class.

3.1A1A2A3F1step 2.1algebra∎

Under the same Countable Choice assumption [A1], for bounded $B$ , [F1] gives the absolutely convergent sum for $Tr (G_{\hat{β}} B)$ . Since $ρ_{T} = Z_{T}^{- 1} G_{\hat{β}}$ , linearity of the trace and $Z_{T} > 0$ give the displayed normalized formula. If $B = E_{a}$ is an effect in the instrument of [A3], that postulate supplies its physical model probability interpretation; the finite trace calculation supplies its value.

Finite-regulator Gibbs predictions

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources