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Free scalar wedge modular geometry — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite two-mode squeeze illustrates but does not construct wedge factorization

Physical scope

A two-oscillator-mode illustration of a reduced state and its entropy. It is not a construction of the continuum wedge algebra's inside/outside factorization, a thermal density matrix for that algebra, or an experimental observation.

Example

Let $H_{R} = H_{L} = ℓ^{2} (N_{0})$ with orthonormal number bases ${∣ n ⟩_{R} : n \in N_{0}}$ and ${∣ n ⟩_{L} : n \in N_{0}}$ . Use inner products linear in the second variable. For a Hilbert space $K$ , write $B (K)$ for its bounded linear operators, and for $u, v \in K$ write $∣ u ⟩ ⟨ v ∣$ for the rank-one operator $w \mapsto u ⟨ v, w ⟩$ . For dimensionless $0 < z < 1$ , define $∣ ψ_{z} ⟩ = \sqrt{1 - z} \sum_{n = 0}^{\infty} z^{n / 2} ∣ n ⟩_{R} \otimes ∣ n ⟩_{L} .$ Define its left-traced operator $ρ_{R}$ by $⟨ m ∣ ρ_{R} ∣ n ⟩ = \sum_{k = 0}^{\infty} ⟨ m, k ∣ ψ_{z} ⟩ \overline{⟨ n, k ∣ ψ_{z} ⟩} .$ This coefficient formula defines the partial trace of this rank-one operator over the left oscillator. The vector is normalized, and $ρ_{R}$ is a positive diagonal operator with eigenvalues $p_{n} = (1 - z) z^{n}$ , whose sum is one. Its entropy in nats, using the natural logarithm, is $S (ρ_{R}) = - \sum_{n = 0}^{\infty} p_{n} \ln p_{n} = - \ln (1 - z) - \frac{z \ln z}{1 - z} .$ This explicitly constructed pair of oscillator modes does not construct a tensor-product factorization of the continuum right-wedge Hilbert space or a density matrix for its local von Neumann algebra.

Example references

Free scalar wedge data and modular-geometric bridge

Diagram: partial trace in the two-mode model

Write $P_{z} = ∣ ψ_{z} ⟩ ⟨ ψ_{z} ∣$ . Set $V_{z} = {c P_{z} : c \in C} \subseteq B (H_{R} \otimes H_{L})$ and $W_{z} = {c ρ_{R} : c \in C} \subseteq B (H_{R})$ . The rank-one reduction used here is the linear map $R_{z} : V_{z} \to W_{z}$ defined by $R_{z} (c P_{z}) = c ρ_{R}$ for $c \in C$ ; its value on $P_{z}$ is specified by the product-basis coefficient formula in the Example. The diagram uses $χ_{z} : C \to V_{z}$ , $χ_{z} (c) = c P_{z}$ , and $r_{z} : C \to W_{z}$ , $r_{z} (c) = c ρ_{R}$ . The verification shows $R_{z} (P_{z}) = ρ_{R}$ , so the diagram commutes. Every arrow is defined only on its stated domain; the reduction arrow acts only on scalar multiples of $P_{z}$ . The TikZ source below specifies that rank-one diagram. Since the library renderer does not execute TikZ, the adjacent array is its rendered algebraic layout. No operation on general trace-class operators or continuum factorization is asserted.

$\begin{array}{ccc} C & \overset{χ_{z}}{\to} & V_{z} \\ ↘_{r_{z}} & ↓_{R_{z}} \\ W_{z} \end{array}$

Facts & Assumptions

Given: $0 < z < 1$ , the two Hilbert spaces, and their number bases.

[F1]

The shared wedge postulate fixes the continuum scalar Fock model and local right-wedge algebra, and explicitly does not assume a tensor-product factorization across that wedge and its complement. Free scalar wedge data and modular-geometric bridge

Verification

Proof technique: Normalize the explicit series, compute the partial trace in the product basis, and evaluate its convergent entropy series.

Given: [F1] and the data in the Example.

1.1givenalgebra

Let $ψ_{z, N} = \sqrt{1 - z} \sum_{n = 0}^{N} z^{n / 2} ∣ n ⟩_{R} \otimes ∣ n ⟩_{L}$ . Orthogonality gives, for $M > N$ , $∥ ψ_{z, M} - ψ_{z, N} ∥^{2} = (1 - z) \sum_{n = N + 1}^{M} z^{n} \leq (1 - z) \sum_{n = N + 1}^{\infty} z^{n} = z^{N + 1} \to 0$ . Thus the partial sums converge in $H_{R} \otimes H_{L}$ to $ψ_{z}$ . Also $(1 - z) \sum_{n = 0}^{N} z^{n} = 1 - z^{N + 1}$ , so continuity of the norm gives $⟨ ψ_{z} ∣ ψ_{z} ⟩ = (1 - z) \sum_{n = 0}^{\infty} z^{n} = 1$ .

2.1givenstep 1.1algebra

The product-basis coefficients are $⟨ m, k ∣ ψ_{z} ⟩ = \sqrt{1 - z} z^{k / 2} δ_{m k}$ . Substitution in the defining partial-trace sum gives $⟨ m ∣ ρ_{R} ∣ n ⟩ = (1 - z) \sum_{k = 0}^{\infty} z^{k} δ_{m k} δ_{n k} = (1 - z) z^{n} δ_{m n}$ . Therefore the eigenvalues are exactly $p_{n} = (1 - z) z^{n}$ , they are positive, and their sum is one by step 1.1.

3.1givenstep 2.1algebra

Put $a_{n} = n z^{n}$ for $n \geq 1$ . Since $a_{n + 1} / a_{n} = z (1 + 1 / n) \to z < 1$ , choose $q$ with $z < q < 1$ ; for all sufficiently large $n$ , $a_{n + 1} \leq q a_{n}$ , so $a_{n} \to 0$ and the tail $\sum a_{n}$ is bounded by a geometric series. Differentiating $\sum_{n = 0}^{N} z^{n} = (1 - z^{N + 1}) / (1 - z)$ gives $\sum_{n = 1}^{N} n z^{n - 1} = (1 - (N + 1) z^{N} + N z^{N + 1}) / (1 - z)^{2}$ . Since $N z^{N} \to 0$ , taking $N \to \infty$ gives $\sum_{n = 0}^{\infty} n z^{n} = z / (1 - z)^{2}$ . Hence $\sum_{n = 0}^{\infty} n p_{n} = z / (1 - z)$ , and $- \sum_{n = 0}^{\infty} p_{n} \ln p_{n} = - \ln (1 - z) - (\ln z) \sum_{n = 0}^{\infty} n p_{n} = - \ln (1 - z) - z \ln z / (1 - z)$ . The series converges because $\sum z^{n}$ and $\sum n z^{n}$ converge for $0 < z < 1$ . In particular, $\sum_{n = 0}^{\infty} p_{n} ∣ \ln p_{n} ∣ = ∣ \ln (1 - z) ∣ + ∣ \ln z ∣ z / (1 - z) < \infty$ .

4.1F1step 1.1step 2.1step 3.1givenconstructalgebra∎

Step 2.1 shows that the coefficient formula in the Example gives $ρ_{R} ∣ n ⟩ = p_{n} ∣ n ⟩$ with $0 < p_{n} < 1$ and $\sum_{n} p_{n} = 1$ ; hence $ρ_{R}$ is bounded and nonzero. Step 1.1 gives $P_{z} \neq 0$ . Every element of $V_{z}$ therefore has a unique form $c P_{z}$ , so $R_{z} (c P_{z}) = c ρ_{R}$ defines a linear map from $V_{z}$ to $W_{z}$ . On $P_{z}$ its matrix coefficients are exactly those of the rank-one coefficient formula in the Example; thus $R_{z} (P_{z}) = ρ_{R}$ and the displayed rank-one diagram commutes. The asserted reduction is confined to this rank-one input. The continuum postulate [F1] instead defines a scalar Fock representation and a right-wedge von Neumann algebra; no map identifying those two constructions is defined or used here. Consequently this example proves the stated oscillator identities but no continuum wedge factorization or density-matrix claim.