Two-Nucleon Problem with Pion Theoretical Potential, I

Abstract

Assuming the pion theoretical potential due to the exchange of one pion in the region outside the pion Compton wave length, the deuteron problem is analysed. Since the deuteron has a large radius on account of its small binding energy, the wave function is strongly affected by the outer part of the potential. This outer part , of the pion theoretical potential has the effective coupling constant g 6 2/4rc between the p-wave pion field and a nucleon as the only variable parameter. Thus we can determine the value of the coupling constant from the deuteron problem as 0.065<g.2f4rc<0.090. This value is in good agreement with that inferred from pion reactions.The form of the deuteron wave function with the pion theoretical potential and ,its approximate analytical expression are given.An information is obtained that the tensor potential must be negative inside the pion Compton wave length in order to give the bound deuteron state, which is consistent with the pion theoretical potential in , the same region. § I. IntroductionThe second-order nuclear force has recently attracted much more attention of people than in former years. The triplet P-wave phase shift of low energy proton-proton scattering experiments (E1ab<4.2 Mev) was found to be negativeY This negative phase shift was explained 2 l as due to the characteristic features of the pion theoretical potential at large distances. The main contributing factor to the P-wave phase shift is the repulsive central part of the second-order nuclear force, which is effective when two nucleons are separated by two or more times the pion range fj / p.c (p. : the pion mass) . The pion theory may be expected to be reliable in the region of the outer part of the potential. Therefore the result was regarded as a strong support to the pion theoretical nuclear forces. At the same time, it was found that the quantitative agreement is very good when the coupling constant g 2 /47L in the pseudoscalar pion theory with P-wave coupling is about 0.06---.0.08.Before the above research, many Japanese physicists,2l 3 > 4 l had attacked the problem of nuclear forces dividing the configuration space of nucleons into two regions along the line t)

Chapter 7. On Repulsive Core of Nuclear Forces

“…potential parameters are given in Table 2 · 3. *) 57 A violent cancellation of the repulsion and attraction to give a steep wall as discussed above could be understood by the large values of -Gz and Ga respectively for rc=O in Table 2·3 (the cases OBE-I and VI), and can be seen in Fig~ 2·8 (b). When they have some reasonable values comparable to or less than ten as obtained from the OBE analyses of higher partial waves as will be discussed in §4 ·l/ 5 ) the "supplementary inner" hard core radius rc in (2 · 24) is as large as 2/ M. The low energy parameters and the phase shift by the rc 0 OBE type potential are given in Table 2 At any rate, we could say that the behaviour of a (1S 0 ) requires not only a huge volume of the inner repulsion but also a sufficiently steep gradient of it.…”

“…p-p scattering data extended up to 3 Ge V are all in accord with that Rea CSo) at these energies lie on a curve naturally extrapolated from that at low energies, as shown aJsq in Fig. 2 literally rigid up to at least a few Ge V. *) (2 · 1· ii) The 3 S1 state At low energies, Iwadare, Otsuki, Tamagaki and Watari made an investigation in a way similar to that in the 1 S 0 state, and solved the deuteron problem by assuming the one-pion-exchange tail outside 1/ f.1.. 57 ) However, they could not obtain the minimum value of the hard core radius but only probable values of it. The probable values were 0.4, 0.2 and 0 X 10-13 cm for j 2 / 47C 0.07, 0.08 and 0;09, respectively.…”

A new effective potential for deuteron

“…The OPEP is the potential derived from meson theory in the treatment of the NN system [29], and given by…”

Photodisintegration of the Deuteron at High Energies

“…for the computation of cross sections, and for the wave function of the deuteron we adopted that of 1-0 -T -W. 13 ) In the computation of e. d. we have taken the P wave so that it is shifted by the given phase shife) at r> l/k (k=tnc/fl) and that it continues smoothly and falls to zero at r=O.I/k in r<.l/k. for the computation of cross sections, and for the wave function of the deuteron we adopted that of 1-0 -T -W. 13 ) In the computation of e. d. we have taken the P wave so that it is shifted by the given phase shife) at r> l/k (k=tnc/fl) and that it continues smoothly and falls to zero at r=O.I/k in r<.l/k.…”