Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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Nearly duplicate Gaussian primitives produce an ill-conditioned overlap matrix

Example

For a,b>0, define normalized s primitives ha(x)=(2a/π)3/4e−a∥x∥2. Their overlap is S(ha,hb)=(2aba+b)3/2. For a=b the two-function overlap matrix has eigenvalues 2 and 0. For distinct exponents a,b it has eigenvalues 1±S(ha,hb), and as b→a its condition number tends to infinity.

Facts & Assumptions

Given: Positive exponents a,b, normalized Gaussian functions as defined in the Example, and the overlap integral in Cartesian Gaussian primitives, contractions, and integral conventions.

[F1]

Gaussian product theorem and finite polynomial shifts: e−a∥x∥2e−b∥x∥2=e−(a+b)∥x∥2.

[F2]

Cartesian Gaussian moments and exponent-parameter derivatives: ∫e−p∥x∥2dx=(π/p)3/2.

Verification

1.1givenF1F2algebra

By [F1]–[F2], the unnormalized overlap is (π/(a+b))3/2 and each norm squared is (π/(2a))3/2 or (π/(2b))3/2. Multiplying by the normalization constants gives S(ha,hb)=(2ab/(a+b))3/2.

2.1givenalgebra∎

The Gram matrix is (1SS1), whose characteristic polynomial is (1−λ)2−S2, so its eigenvalues are 1±S. At a=b, S=1 and the lower eigenvalue is zero. For a≠b, 2ab<a+b, hence 0<S<1; continuity of the displayed formula gives S→1 as b→a, and therefore (1+S)/(1−S)→∞.

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