MATHPH In the literature: open

Excess electron number of atoms bounded independently of nuclear charge

In plain words

Show that there is one fixed number C such that no atom, however heavy, can bind more than Z + C electrons. Bind means the extra electrons stay attached instead of flying off to infinity.

Precise statement

For $H_{N,Z} = \sum_{i=1}^N (-\hbar^2/(2m) \Delta_i - Z e^2/|x_i|) + \sum_{i<j} e^2/|x_i - x_j|$ on antisymmetric $L^2((R^3 x \{\mathrm{up},\mathrm{down}\})^N)$ with a static point nucleus, let $E(N,Z)$ be its ground-state energy and $N_c(Z)$ the largest N with $E(N,Z) < E(N-1,Z)$. Prove or disprove $N_c(Z) \le Z + C$ with C independent of Z. Answer: a proof with explicit or existential C, or a sequence $Z_k$ with $N_c(Z_k) - Z_k$ unbounded.

What would settle it

A proof that bounds $N_{c}(Z) - Z$ uniformly in $Z$, for example by controlling the screened nuclear potential felt by the outermost electrons in the many-body ground state.

Status in the literature

Best known: $N_{c} < 2Z + 1$ (Lieb 1984), $N_{c} < 1.22 Z + 3 Z^{1/3}$ (Nam 2012), $N_{c}/Z \to 1$ (Lieb, Sigal, Simon, Thirring 1988) and $N_{c} - Z \le C Z^{5/7}$ for large Z (Fefferman, Seco 1990; Seco, Sigal, Solovej 1990); the bounded-excess statement is proved in Hartree-Fock theory (Solovej, Annals 2003) and in Thomas-Fermi-Dirac-von Weizsacker theory (Frank, Nam, Van Den Bosch, CPAM 2018), not for the Schrodinger Hamiltonian.