2021
DOI: 10.1016/bs.aiq.2021.05.010
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High-precision Hy-CI and E-Hy-CI studies of atomic and molecular properties

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Cited by 7 publications
(3 citation statements)
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“…Note: All energies in this work are variational; all calculations were carried out in quadruple-double-with exponent (QDE) precision. 1 For a detailed discussion of both E-Hy-CI and Hy-CI, including how they compare with the other methods in this table and elsewhere in this research, see the authors' review article [14]. 2 6 here represents n orbs = 7, 6, 5, 4, 3, 2 and 1, respectively, for l = 0−6.…”
Section: Tablementioning
confidence: 99%
See 1 more Smart Citation
“…Note: All energies in this work are variational; all calculations were carried out in quadruple-double-with exponent (QDE) precision. 1 For a detailed discussion of both E-Hy-CI and Hy-CI, including how they compare with the other methods in this table and elsewhere in this research, see the authors' review article [14]. 2 6 here represents n orbs = 7, 6, 5, 4, 3, 2 and 1, respectively, for l = 0−6.…”
Section: Tablementioning
confidence: 99%
“…For papers discussing the integrals arising in E-Hy-CI, see [8,[10][11][12][13]; for a review of both Hy-CI and E-Hy-CI, see [14].…”
Section: Statementioning
confidence: 99%
“…The other approach to the complexity mitigation involves expressing the N -electron Ψ in terms of a (in general infinite) sum of products of quantities that depend on fewer than N coordinates { x k }. Although many variants of this tensor decomposition are possible (including those based upon geminals ), the vast majority of electronic structure calculations employs its most extreme form normalΨ ( x 1 , ... , x N ) = prefix∑ scriptJ C scriptJ k = 1 N ψ j k ( x k ) where scriptJ = { j 1 , ... , j N } , that carries it all the way down to the orthonormal one-electron eigenfunctions (commonly known as spinorbitals) {ψ j } ≡ {ψ j ( x 1 )} of the ŝ z (σ 1 ) operator. This representation of Ψ, which is just the FCI (full configuration interaction) expansion written in terms of the Hartree products rather than the Slater determinants, is exact provided that {ψ j } is a complete set .…”
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confidence: 99%