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Fixed finite-mode results do not establish a continuum limit

Statement

In Finite-lattice polynomial Hamiltonian data, the regulator is one fixed finite mode count $N$ and one fixed polynomial on $R^{N}$ . The self-adjoint and compact-resolvent conclusions of Self-adjoint unbounded polynomial interaction at fixed regulator concern that fixed Hilbert space and form realization. They assert no continuum or infinite-volume limit. Here “continuum limit” means a limit of a specified family of finite-regulator objects as a cutoff parameter changes, together with an identified comparison space and convergence notion. No such family or convergence structure is part of the two cited items. This scope note does not assert that a continuum limit does not exist.

Facts & Assumptions

Given: the fixed finite-mode data and its selected self-adjoint realization.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): the selected finite-mode realization is constructed under Countable Choice.

[F1]

Finite-lattice polynomial Hamiltonian data: $N$ is a fixed finite integer and the datum defines one $L^{2} (R^{N})$ space and one polynomial differential expression; no limit in $N$ is included.

[F2]

Self-adjoint unbounded polynomial interaction at fixed regulator: the self-adjointness, lower bound, compact resolvent, and ground eigenvalue concern the fixed form realization on that $L^{2} (R^{N})$ .

Proof

technique · direct

Given: [F1] and [F2].

1.1A1F1algebra

The hypotheses in [F1] specify a single integer $N$ , a single coordinate space $R^{N}$ , and a single potential $V$ ; they contain no indexed family with varying cutoff.

2.1A1F1F2step 1.1algebra∎

A convergence assertion for varying cutoffs requires at least a family of objects, a common comparison space or maps between the spaces, and a convergence relation. Neither those data nor a limiting object occur in [F1] or [F2]. Therefore the theorems establish only the stated fixed- $N$ operator properties. The absence of those hypotheses is not a proof of nonexistence of a limit in a separately specified model.

Fixed finite-mode results do not establish a continuum limit

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources