Can one bind three electrons with a single proton?

Of course not for an ideal H– – atom. But with the help of an intense homogeneous magnetic field B, the question deserves to be reconsidered. It is known (see, e.g. Baumgartner et al. in Commun Math Phys 212(3):703–724, 2000; Brummelhuis and Duclos in J Math Phys 47:032103, 2006) that as B → ∞ and in the clamped nucleus approximation, this ion is described by a one-dimensional Hamiltonian Ni=1−2i−Z(xi)+1ijN(xi−xj)actinginL2(R3)(1) where N = 3, Z = 1 is the charge of the nucleus, and δ stands for the well known “delta” point interaction. We present an extension of the “skeleton method” (Cornean et al. in Few-Body Syst 38(2–4):125–131, 2006; Proc Symp Pure Math AMS 77:657–672, 2008) to the case of three degree of freedom. This is a tool, that we learn from Rosenthal (J Chem Phys 35(5):2474–2483, 1971) for the case N = 2, which reduces the spectral analysis of (1) to determining the kernel a system of linear integral operators acting on the supports of the delta interactions. As an application of this method we present numerical results which indicates that (1) has a bound state for Z = 1 and N = 3.