**Speaker**

Marviw Schulz, Karlsruhe Institute of Technology

**Abstract**

The question of the maximal ionization of atoms, how many electrons an atom can bind, is a long–standing open problem in Mathematical Physics, listed, for example, as Problem 10D in Barry Simon’s \emph{Fifteen Problems in Mathematical Physics} To the best of our knowledge, there is no experimental evidence that a single neutral atom (in the vacuum) can bind more than one additional electron. Thus the conjecture on the maximal allowed number of electrons an atom with nuclear charge \(Z\) can bind is \(N_c(Z) = Z+1\). However, even the much weaker conjecture, \(N_c(Z) = Z+K\) for some fixed \(K<\infty\) is not known to be true, except in some simplified models. Although there has been no proof of the above conjecture, even in its weaker form, there has been some progress in several ways. Firstly, based on some ideas of Benguria, Lieb proved in 1984 that the maximal number of electrons an atom of charge \(Z\) can bind is \(N_c<2Z+1\). This result holds irrespective of the statistics of the particles and extends to molecules. Secondly, 33 years later Nam extended the \emph{Benguria-Lieb-Method} and improved the result to \({N_c<1.22Z+3.0 \ Z^{1/3}}\). Nam's result holds for atoms and needs the Fermi statistics. We prove that \(N_c< c_1(s) Z + c_2(s) Z^{1/3}\) for any \(s \in (0,3], Z\geq 2\) for some functions \(c_{1} ,c_2\) depending on \(s\) with \begin{equation*} \begin{split} c_1(2) &= (2(\sqrt{2}-1))^{-1} \leq 1.2072\\ c_1(3) &= \frac{2}{3}\frac{\sqrt[3]{1+\sqrt{2}}}{(1+\sqrt{2})^{2/3}-1} \leq 1.1185 \end{split} \end{equation*} and \(c_2(2)<3,c_2(3)\leq 4.3\). The fermionic symmetry is used, hence our result holds only for fermions. Indeed, Benguria and Lieb showed in that an equivalent result for bosons cannot hold since \(\lim_{Z\to \infty} N_c/Z \approx 1.21\) for bosons. Since we know \(c_1(s)\) for any \(s\in(0,3]\) explicitly we can also formulate a new result in the bosonic case. Joint work with Dirk Hundertmark and Nikolaos Pattakos.