Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Half-flux ground-state crossing on the ring

Physical scope

Exact level comparison in the ideal spinless ring model with periodic wavefunctions. It does not include spin, finite-width effects, disorder, or an experimental measurement.

Example

In the ideal spinless ring model, at dimensionless flux α=qΦ/(2πℏ)=1/2 the ground energy is twofold degenerate at the Fourier labels n=0 and n=1. At α=0 the ground state is unique at n=0. The scale κ=ℏ2/(2mR2) is in joules and α is dimensionless.

Facts & Assumptions

Given: the ideal ring Hamiltonian, κ=ℏ2/(2mR2)>0, and its dimensionless flux α.

[F1]

Its complete Fourier eigenbasis has energies En=κ(n−α)2 for n∈Z (Aharonov–Bohm ring spectrum and flux periodicity, Spectrum and gauge periodicity of a periodic flux operator).

Verification

1.1givenA1F1algebra

At α=1/2, En=κ(n−1/2)2. The nearest integers to 1/2 are exactly 0 and 1, and both have squared distance 1/4; every other integer has distance at least 3/2. Therefore E0=E1=κ/4 is the ground energy and no other Fourier label attains it. The complete basis in [F1] makes the ground eigenspace exactly two-dimensional, conditional on the ideal ring assumptions.

2.1A1F1algebra∎

At α=0, En=κn2≥0, with equality exactly at the integer n=0. The eigenbasis in [F1] then gives a one-dimensional ground eigenspace, spanned by e0(ϕ)=(2π)−1/2, again conditional on the ideal ring assumptions.

Depends on

Dependency tree · 2 levels

Sources