The current mainstream particle theory, the standard model SM  , can say nothing about binding energies of two nucleon nuclei. In its place, the scalar strong interaction hadron theory SSI has been proposed   . The purpose of this paper is to provide estimates of the binding energies in the two nucleon system based upon SSI.
In Section 2, the triplet deuteron binding energy is estimated and compared to data. Section 3 shows that the singlet deuteron is not a bound state. The same conclusion is reached for the singlet dineutron in Section 4. In Section 5, the triplet dineutron binding energy is estimated and its decay mode is shown. This dineutron is assigned to the observed one in 16Be decay. Some related considerations are given in Section 6.
2. Triplet Deuteron Binding Energy
In SSI  (Ch 9), the ground state of a proton consists of a diquark uu and a quark d.
Consider such a free proton and change its u quarks into d quarks and vice versa. The result is a neutron with the same spin configuration. Put these two nucleons next to each other to form a deuteron, as is depicted in Figure 1.
The configuration in Figure 1 is fixed by the: following two conditions: 1) the total spin in a two nucleon system can only be 1 with the nucleon spins parallel or 0 with the spins antiparallel and 2) both nucleons are on equal footing and the configuration remains symmetric with respect to them after an interchange or a space inversion.
In SSI, Ra is in the “hidden” relative space x between the diquark and the quark  (A1) and is determined by strong interquark forces. It is hence not observable. On the other hand, the internucleon distance Rnn in the laboratory space is observable.
When the two nucleons in Figure 1 are far apart or Rnn is large, they do not interact with each other. When they are sufficiently close to each other, they form a triplet deuteron 2H1. Let q denote quark charge, then
Consider the total electrostatic energy Eem2H1 of the triplet deuteron in Figure 1. There are four contributions, between uu and dd, uu and u, d and dd, and d and u so that
For large neutron-proton separation Rnn, Eem2H1 ≈ 0 but < 0. Let now the neutron move closer to the proton, Rnn decreases and Eem2H1 becomes more negative. However, Rnn cannot be too small because the neutron will then experience the short range strong interaction force from the proton which prevents the merger of these two nucleons.
Figure 1. Schematic configuration of the spin component of a triplet deuteron 2H1. X denotes the coordinate and S the spin of the diquarks and quarks. The subscripts refer to the quark content. Suu = Sdd = 1 and Su = Sd = −1/2 with signs given by the arrow for + and for −. Xuu = xI and Xd = xII in  (10.1.1) or  (A2). Ra ≈ 3.23 fm  (2.5) is the mean value of the diquark-quark distance. The neutron configuration is obtained from the proton configuration via the interchange u d. Rnn is distance between the both nucleons.
Let the diquark uu be fixed at Xuu in Figure 1. Then the d quark can lie anywhere on the surface of a sphere with radius Ra centered at Xuu to form the same proton. This scenario can be taken over by the neutron in Figure 1 via the interchange u d. The same neutron is obtained when the u quark is Ra away from Xdd. Since the proton and the neutron cannot occupy the same space,
With Ra » 3.23 fm from the caption of Figure 1, e2 = 4π/137, (2) and (3) yield the estimate
The lower limit of Rnn = 2Ra in (3) leads to the triplet deuteron binding energy EB2H1 = -Eem2H1 = 2.0653 MeV which is 7.7% smaller than the measured 2.2245 MeV.
The present treatment is phenomenological and Rnn = 2Ra used above is not an exact criterion. If Rnn = 1.846Ra were used instead, the predicted binding energy EB2H1 will agree with the measured one. Similarly, the diquark-quark distance is a continuous variable represented by an average value Ra here. If Ra ≈ 3.23 fm 3 fm, then prediction and data will likewise agree.
The above Ra/Rnn ratios are also in qualitative agreement with the ratio
Note here that the charge radius arises from long range electromagnetic interactions and is an observable in the laboratory space. It is not to be confused with the unobservable short range strong interaction interquark radius Ra ≈ 3.23 fm in the “hidden” relative space x between quarks.
3. Singlet Deuteron Binding Energy
The neutron spin in Figure 1 has been chosen to be parallel to that of the proton spin. But it can also be chosen to be antiparallel to the proton spin. This can be achieved by inverting the neutron configuration in Figure 1 upside down or, equivalently, by rotating this configuration 180˚. This is carried out in Figure 2.
Analogous to (2), there are also four contributions, now between uu and u, d and dd, uu and dd, and d and u so that the total electrostatic energy Eem2H1 in (2) is changed to Eem2H0,
For large neutron-proton separation Rpn, Eem2H0 ≈ 0 but now > 0. Let now the neutron move closer to the proton, Rnn decreases and Eem2H0 becomes more positive. Using the lower limit in (3), the singlet deuteron binding energy EB2H0 = −Eem2H0 = −2.0653 MeV is negative so that there is no stable spin 0 deuteron, in agreement with observation.
4. Singlet Dineutron Binding Energy
Perform the interchange u d for the proton p in Figure 1. This turns p into n which is the same as the n on the right half of Figure 1. By Pauli’s principle, these two neutrons n and n cannot coexist. However, if the spin of this left neutron is switched to , then the aggregate n and n , assigned to the singlet dineutron 2n0, can exist. The configuration is depicted in Figure 3.
Analogous to (6), the total electrostatic energy Eem2n0 for 2n0 reads
which is simply 0.8 × Eem2H0 = −1.652 MeV for the lower limit of (3). Analogous to the singlet deuteron 2H0 case in Section 3, the singlet dineutron 2n0 binding energy EB2n0 = −Eem2n0 = −1.652 MeV is similarly negative so there is no stable spin 0 dineutron.
Figure 2. Schematic configuration of a singlet deuteron 2H0 with spin 0. The proton configuration is the same as that in Figure 1. The neutron configuration has been inverted upside down relative to that in Figure 1, indicated by the subscript inv.
Figure 3. Schematic configuration of a singlet dineutron 2n0. The right neutron configuration is the same as that in Figure 1. Switching the spin directions in n on the right leads to the n configuration on the left.
5. Triplet Dineutron Binding Energy and Decay
Analogous to (2), the total electrostatic energy Eem2n1 for 2n1 reads
Comparison with (2) together with the lower limit of (4) yields the triplet dineutron 2n1 binding energy EB2n1 = −Eem2n1 = 0.8 × EB2H1 = 1.652 MeV. This value is expected to be 7.7 % smaller than its actual value by analogy to the triplet deuteron case mentioned beneath (4). Correcting for this discrepancy, the binding energy EB2n1 becomes 1.077 × EB2n1 = 1.78 MeV.
2n1 is thus a stable nucleus, like 2H1, but its binding energy 1.78 MeV is weaker than 2.2245 MeV for 2H1. Energetically, it can decay into deuteron via neutron beta decay,
where ne denotes the electron neutrino. Since each of the neutrons in 2n1 can decay separately, the decay time is expected to be half of the neutron decay time or 440 sec.
While the deuteron can be the nucleus of an atom, the deuterium, the neutral dineutron cannot form an atom. The decay (9) implies that there are no free dineutrons in nature, just like neutrons.
Dineutron has been first observed some years ago in 16Be decay  . It is assigned to 2n1 here but needs not decay according to (9). Instead, it can decay into two neutrons when excess energy is available. The predicted 2n1 binding energy 1.78 MeV has not been measured.
Figure 4. Schematic configuration of the spin component of a triplet dineutron 2n1. The right neutron configuration is the same as that in Figure 2. The left neutron configuration is the same as that in Figure 3.
6. Related Considerations
The third member in the two nucleon system, the diproton pp, can be treated in an entirely analogous fashion. The results are that the binding energies of a singlet diproton EB2p0 as well as that for a triplet diproton EB2p1 are the classical value −e2/Rnn for large Rnn. For the lower limit in (3), Rnn = 2Ra, EB2p1 = −19.26 MeV and EB2p0 = −14.1 MeV. Thus, pp is not bound, as expected.
Thus, in the two nucleon system, only the triplet deuteron 2H1 and the triplet dineutron 2n1 are bound electromagnetically. The former is 20% stronger than the latter. The remaining four combinations, 2H0, 2n0 and the triplet and singlet diprotons are not electromagnetically bound states. This is reflected in the composition of nuclei which consist mostly of pn pairs, to an less degree nn pairs but no pp pair.
The above results show that the binding energies of the deuteron and dineutron are of long range, electromagnetic nature arising from the quark charges. Short range, strong interaction forces between the nucleons set the limits of these binding energies but do not contribute to them.
For heavier nuclei, the above electrostatic confinement is insufficient and strong interaction forces, akin to those that led to the confinement characterized by Ra in Section 2, need be included. To accommodate the additional quarks, the two dimensional Figures 1-4 here have to be extended to three dimensions.