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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.

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

Facts & Assumptions

Example

Verification

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