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Coulomb screening must accumulate every discarded-block bound

Example

Suppose a finite screening queue has three candidate input-basis Coulomb quartets (which may be singleton-contraction primitive quartets), each with a verified weighted energy bound $w_{j} = 0.4 τ$ , and the total discarded-energy budget is $τ > 0$ . Choose the permitted screening execution that skips a candidate whenever its cumulative bound fits the budget. With exact accumulation, the first two blocks are skipped and contribute a cumulative bound $0.8 τ$ . The third would raise the cumulative bound to $1.2 τ$ , so it must be evaluated or left unscreened. The reported discarded bound is $0.8 τ$ ; none of the skipped integrals is claimed to be zero. This is also an execution of the outward-rounded rule when these sums are represented exactly, for example $τ = 5$ and $w_{j} = 2$ in a format containing $2, 4, 5, 6$ . Otherwise outward accumulation may conservatively prevent a skip or report a larger bound.

Facts & Assumptions

Given: A total screening tolerance $τ > 0$ , three finite candidate records with verified weighted Cauchy–Schwarz bounds $0.4 τ$ , exact accumulation of these bounds, and the permitted policy that skips each candidate when the budget guard allows it. For an execution of Finite primitive and contracted Gaussian integral evaluation with certified screening, assume the displayed sums are represented exactly in its format.

[F1]

Positive Coulomb bilinear form, Cauchy–Schwarz, and screening bound: each weighted discarded contribution is bounded by its certified diagonal Cauchy–Schwarz product, and total error is bounded by the sum of those terms.

[F2]

Soundness and accumulated error bounds for the finite Gaussian integral evaluator: every successful reported discarded-energy bound dominates the absolute contribution of all skipped terms.

[F3]

Finite primitive and contracted Gaussian integral evaluation with certified screening: a candidate is skipped only when adding its certified weighted bound keeps the running total at most the requested tolerance.

Verification

1.1givenF1F3algebra

The first bound is $0.4 τ \leq τ$ , and the cumulative sum after two is $0.8 τ \leq τ$ , so [F3] permits both skips. The selected policy therefore skips both and, with the assumed exact accumulation, records $0.8 τ$ .

2.1givenF1F2F3step 1.1algebra∎

Adding the third gives $0.8 τ + 0.4 τ = \frac{6}{5} τ > τ$ . By [F3], the screening guard therefore rejects skipping that third block and requires its evaluation or retention. By [F1]–[F2], the reported $0.8 τ$ bounds only the first two omitted contributions and does not assert exact sparsity.

Depends on

Used by

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Sources

Trygve Helgaker, Molecular Integral Evaluation, 10th Sostrup Summer School lecture notes (2008) (standard reference, not scraped)
L. E. McMurchie and E. R. Davidson, One and Two Electron Integrals over Cartesian Gaussian Functions, UCRL-13791 (1976) (standard reference, not scraped)