LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Factorization of finite additive canonical densities

Statement

Use the finite additive spectral data of Additive spectral Hamiltonian and normalized ensembles. For every $N \geq 1$ , $Z_{N} = Z_{1}^{N}$ . For each product label $j = (j_{1}, \dots, j_{N})$ , the diagonal entry of $γ_{N}$ is $\prod_{a = 1}^{N} p_{j_{a}}$ and its off-diagonal entries are zero. For each integer $k$ with $1 \leq k \leq N$ , write $x, x^{'} \in {1, \dots, q}^{k}$ and $y \in {1, \dots, q}^{N - k}$ , and define the first- $k$ reduced matrix $R_{N, k}$ by $(R_{N, k})_{x, x^{'}} : = \sum_{y \in {1, \dots, q}^{N - k}} (γ_{N})_{(x, y), (x^{'}, y)} .$ When $k = N$ , the tail-label set consists of the single empty tuple. Then $R_{N, k} = γ_{k}$ , where $γ_{k}$ is the diagonal matrix on $V_{k} = (C^{q})^{\otimes k}$ with entries $(γ_{k})_{x, x} = \prod_{a = 1}^{k} p_{x_{a}}$ , and the canonical state has expected additive energy $N E$ .

Facts & Assumptions

Given: A finite alphabet ${1, \dots, q}$ , real levels $e_{j}$ , finite $β \in R$ , and the canonical matrices and partition sums of Additive spectral Hamiltonian and normalized ensembles.

[F1]

Additive spectral Hamiltonian and normalized ensembles: the energy of a product label is $\sum_{a} e_{j_{a}}$ and its canonical weight is the exponential of $- β$ times that energy.

[F2]

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ : for real $x, y$ , $\exp (x + y) = \exp (x) \exp (y)$ .

[F3]

The real exponential function and the number $e$ by a power series: $e^{x}$ denotes the real exponential $\exp (x)$ defined by its power series.

Proof

1.1givenF1F2F3algebra

By [F1] and the notation in [F3], each label weight uses the real exponential $e^{x} = \exp (x)$ . Apply [F2] repeatedly and use distributivity of finite sums to obtain $Z_{N} = \sum_{j_{1}, \dots, j_{N}} \prod_{a = 1}^{N} e^{- β e_{j_{a}}} = \prod_{a = 1}^{N} \sum_{j_{a} = 1}^{q} e^{- β e_{j_{a}}} = Z_{1}^{N} .$ For each product label, division by this positive denominator makes its canonical diagonal entry $\prod_{a} p_{j_{a}}$ , while off-diagonal entries vanish. This is the matrix description of $γ_{N}$ in the Statement.

2.1givenF1step 1.1algebra∎

For first-block labels $x, x^{'}$ , the matrix entry $(γ_{N})_{(x, y), (x^{'}, y)}$ vanishes when $x \neq x^{'}$ , since $γ_{N}$ is diagonal in the product basis. For $x = x^{'}$ , its value is $\prod_{a = 1}^{k} p_{x_{a}} \prod_{a = k + 1}^{N} p_{y_{a - k}}$ . Summing over the tail labels in the definition of $R_{N, k}$ gives $(R_{N, k})_{x, x} = \prod_{a = 1}^{k} p_{x_{a}}$ , because each tail factor has sum $\sum_{j = 1}^{q} p_{j} = 1$ . Hence $R_{N, k} = γ_{k}$ . Finally the canonical state's mean energy is $\sum_{j_{1}, \dots, j_{N}} (\prod_{a} p_{j_{a}}) \sum_{a} e_{j_{a}} = \sum_{a = 1}^{N} \sum_{j_{a}} p_{j_{a}} e_{j_{a}} = N E,$ because every other factor sum is one.

Factorization of finite additive canonical densities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources