Abstract
With the nuclear forces as derived in the first part of this paper the deuteron problem is integrated. In §7, natural units are introduced, viz., the range of the nuclear forces 1κ=μc as the unit of length and the quantity E0=μ2c2M as the unit of energy. With the adopted value of the meson mass, μ=177 electron masses, we have 1κ=2.185×1013 cm and E0=8.68 Mev. The quantum states to be expected are discussed (§8); each state is characterized by the total angular momentum J, the total spin S=0 or 1, and the parity. For some states, the orbital momentum L is defined uniquely by these quantum numbers (in this case, L=J); for others, we have a linear combination of wave functions with two different values of L, viz., J+1 and J1. The angular coordinates are eliminated from the Schrödinger equation (§9) and the radial wave equations obtained. The order of the states is discussed qualitatively (§10) and it is made plausible that the ground state is a combination of a S3 and a D13 state, both in the neutral and in the symmetrical theory. The next higher triplet state is probably P13. Then the wave equation is solved numerically for the S1 state (§11), the position of this state being taken from experiments on the scattering of slow neutrons by protons. The method of the numerical solution is described, and a table constructed giving the interaction constant a=(2Mμ)f2c as a function of the cut-off distance x0=κr0; a is found to depend only slightly on x0. The next section (§12) deals with the numerical integration for the triplet (ground) state.