Real analyticity away from the nucleus of pseudorelativistic Hartree–Fock orbitals

Abstract

We prove that the Hartree-Fock orbitals of pseudorelativistic atoms, that is, atoms where the kinetic energy of the electrons is given by the pseudorelativistic operator √ − + 1 − 1, are real analytic away from the origin. As a consequence, the quantum mechanical ground state of such atoms is never a Hartree-Fock state.Our proof is inspired by the classical proof of analyticity by nested balls of Morrey and Nirenberg. However, the technique has to be adapted to take care of the nonlocal pseudodifferential operator, the singularity of the potential at the origin, and the nonlinear terms in the equation.

Comment on ‘Solutions to quasi-relativistic multi-configurative Hartree–Fock equations in quantum chemistry’, by C. Argaez and M. Melgaard

“…So the corresponding statement 5 in Proposition 9.1 is in fact not proved in the paper. Fortunately, the real-analyticity was recently proved 2 by Dall'Acqua, Fournais, Østergaard Sørensen and Stockmeyer in [5], and Proposition 9.1 is finally correct.…”

New Bounds on the Maximum Ionization of Atoms

“…In fact, due to a technical gap the original proof of Lieb in [10] works properly only when Z α < 1/2. However, it is posible to fill this gap to obtain the bound up to Z α < 2/π [5]. On the other hand, to our knowledge, no result about the asymptotic behavior of N c /Z is available for the pseudo-relativistic model, although within pseudo-relativistic Hartree-Fock theory it was recently shown by Dall'Acqua and Solovej (2010) [4] that N c ≤ Z + const.…”

Real Analyticity of Solutions to Schrödinger Equations Involving a Fractional Laplacian and Other Fourier Multipliers

“…For a proof, see [1,Lemma B.1]. For p, q ∈ [1, ∞], denote by · Bp,q the operator norm on bounded operators from L p (R n ) to L q (R n ).…”

“…The proof in [1, Lemma C.2] (for n = 3, m > 0, s = 1/2) uses the explicit expression for the Green's function for n = 3 [4, (IX.30)]. We give a modified proof, which holds for all n ≥ 1, and m > 0, s ∈ (0, 1), as well as s = 1/2, m = 0, referring to [1] for further details. We start with m > 0, s ∈ (0, 1).…”