ISSN:
1089-7690
Source:
AIP Digital Archive
Topics:
Physics
,
Chemistry and Pharmacology
Notes:
Given a homonuclear diatomic molecule AB with an even number of electrons N in the state ψ. What is the state of atom A in the molecule? Using the lemma that 〈ψ||HA ||ψ〉/〈ψ||ψ〉≥E0A where HA is the Hamiltonian for atom A only, having NA =N/2 electrons, and E0A is the ground state energy of A, the following general definition is suggested and quantitatively tested for the hydrogen atom in the hydrogen molecule. Define N-electron functions ψA and ψB by ||ψ||2= 1/2 [||ψA||2+||ψB||2], ψA =ψ[1+F]1/2, and ψB =ψ[1−F]1/2, with ψA, ψB, and ψ all normalized and F a function symmetric with respect to electron interchange. Then ρ=ρA+ρB and F can be determined to minimize the promotion energy 〈ψA||HˆA||ψA〉−E0A. Calculations are performed for the H2 ground state and b3∑+u excited state, as functions R, employing accurate wave functions including correlation. For the ground state at the equilibrium internuclear distance, the promotion energy is found to be 0.0447 a.u. or 28 kcal/mol. The electron density for the atom in the molecule is examined, and extensions to heteronuclear and polyatomic molecules briefly discussed.
Type of Medium:
Electronic Resource
URL:
http://dx.doi.org/10.1063/1.450467
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