Let H(→ρ, →r) represent the (M-1)-body Hamiltonian that results from fixing the center of mass →R of an M-body system, where →r is the relative coordinate of a particular pair of particles, and →ρ represents the M-2 remaining internal coordinates. With Eₜₕᵣ the lowest continuum threshold associated with H, the number of bound states of the system is the number of negative eigenvalues of H-Eₜₕᵣ. A simple lower bound on H was derived by Hahn and Spruch through the use of an adiabatic-like approximation in which the (M-1)-body problem is attacked by considering first an (M-2)-body problem and then a one-body problem. With Eₐ₀(r) the lowest energy of the system for →R and →r fixed, one finds $H({{→}}{{ρ}}, {{→}}{r}){-}{E}ₜₕᵣ>~{^}{1}({{→}}{{ρ}}){H}⁽¹⁾({{→}}{r}), where {H}⁽¹⁾{≡}{T}_{{{→}}{r}}+{E}ₐ₀(r){-}{E}ₜₕᵣ{≡}{T}_{{{→}}{r}}+{V}⁽¹⁾(r).$ ${H}⁽¹⁾$ is a one-body Hamiltonian, ${T}_{{{→}}{r}}$ is the kinetic energy operator for the relative motion of the particular pair, and ${^}{1}({{→}}{{ρ}})$ is the unit operator in the space of quadratically integrable functions of ${{→}}{{ρ}}$. The adiabatic potential ${E}ₐ₀({ ̇ \.{}{}}{r})$ has been tabulated for a number of systems, primarily atomic and molecular. A necessary condition for the existence of a bound state of $H$ is that the lowest eigenvalue of ${H}⁽¹⁾$ be negative.The method is formulated fairly generally and discussed in some detail for the case of Coulomb interactions. It is shown that neither He+${e}⁺$ nor ${α}+p+{e}^{{-}}$ can form bound states. We also find lower bounds of -0.068 and -0.065 eV on the energy of the ground state and the first excited state, respectively, of the H+${e}⁺$ system; presumably there is no bound state of this system, but we are unable to prove it. The system H+${e}^{{-}}$ is also considered.
No takes yet. Share an insight, caveat, or question.
Gertler et al. (1968) studied this question.
Synapse has enriched 5 closely related papers on similar clinical questions. Consider them for comparative context: