On Dineutron and Deuteron Binding Energies

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1. Introduction

The current mainstream particle theory, the standard model SM [1] , can say nothing about binding energies of two nucleon nuclei. In its place, the scalar strong interaction hadron theory SSI has been proposed [2] [3] . 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 ^{16}Be decay. Some related considerations are given in Section 6.

2. Triplet Deuteron Binding Energy

In SSI [3] (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, *R _{a}* is in the “hidden” relative space

When the two nucleons in Figure 1 are far apart or *R _{nn}* is large, they do not interact with each other. When they are sufficiently close to each other, they form a triplet deuteron

${q}_{u}=2e/3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{q}_{uu}=4e/3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{q}_{d}=-e/3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{q}_{dd}=-2e/3$ (1)

Consider the total electrostatic energy *E _{em}*

${E}_{em2H1}=\frac{{q}_{uu}{q}_{dd}+{q}_{d}{q}_{u}}{{R}_{nn}}+\frac{{q}_{uu}{q}_{u}+{q}_{d}{q}_{dd}}{\sqrt{{R}_{nn}^{2}+{R}_{a}^{2}}}=\frac{10{e}^{2}}{9}\frac{1}{{R}_{nn}}\left[-1+\frac{1}{\sqrt{1+{R}_{a}^{2}/{R}_{nn}^{2}}}\right]\le 0$ (2)

For large neutron-proton separation *R _{nn}*,

Figure 1. Schematic configuration of the spin
$\uparrow \uparrow $ component of a triplet deuteron ^{2}*H*_{1}. *X* denotes the coordinate and *S* the spin* *of the diquarks and quarks. The subscripts refer to the quark content. *S _{uu} *=

Let the diquark *uu *be fixed at *X _{uu} *in Figure 1. Then the

${R}_{nn}\ge 2{R}_{a},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{R}_{a}}{{R}_{nn}}\le 0.5$ (3)

With *R _{a} *» 3.23 fm from the caption of Figure 1,

${E}_{em2H1}\ge -2.0653\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{MeV}$ (4)

The lower limit of *R _{nn}* = 2

The present treatment is phenomenological and *R _{nn}* = 2

The above *R _{a}*/

$\frac{proton\text{\hspace{0.17em}}charge\text{\hspace{0.17em}}radius\text{\hspace{0.17em}}}{deuteron\text{\hspace{0.17em}}charge\text{\hspace{0.17em}}radius}=\frac{0.843\text{\hspace{0.17em}}fm}{2.128\text{\hspace{0.17em}}fm}=0.395<0.5\text{\hspace{0.17em}}$ (5)

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 *R _{a}* ≈ 3.23 fm in the “hidden” relative space

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 *E _{em}*

$\begin{array}{c}{E}_{em2H0}=\frac{{q}_{uu}{q}_{u}+{q}_{d}{q}_{dd}}{{R}_{nn}}+\frac{{q}_{uu}{q}_{dd}+{q}_{d}{q}_{u}}{\sqrt{{R}_{nn}^{2}+{R}_{a}^{2}}}=\frac{10{e}^{2}}{9}\frac{1}{{R}_{nn}}\left[1-\frac{1}{\sqrt{1+{R}_{a}^{2}/{R}_{nn}^{2}}}\right]\\ =-{E}_{em2H1}\ge 0\end{array}$ (6)

For large neutron-proton separation *R _{pn}*,

4. Singlet Dineutron Binding Energy

Perform the interchange *u
$\leftrightarrow $ d* for the proton *p
$\uparrow $ * in Figure 1. This turns *p
$\uparrow $ * into *n
$\uparrow $ * which is the same as the* n
$\uparrow $ * on the right half of Figure 1. By Pauli’s principle, these two neutrons *n
$\uparrow $ * and *n
$\uparrow $ * cannot coexist. However, if the spin of this left neutron is switched to
$\downarrow $ , then the aggregate *n
$\downarrow $ * and *n
$\uparrow $ *, assigned to the singlet dineutron ^{2}*n*_{0}, can exist. The configuration is depicted in Figure 3.

Analogous to (6), the total electrostatic energy *E _{em}*

${E}_{em2n0}=\frac{{q}_{u}{q}_{u}+{q}_{dd}{q}_{dd}}{{R}_{nn}}+\frac{{q}_{u}{q}_{dd}+{q}_{dd}{q}_{u}}{\sqrt{{R}_{nn}^{2}+{R}_{a}^{2}}}=\frac{8{e}^{2}}{9}\frac{1}{{R}_{nn}}\left[1-\frac{1}{\sqrt{1+{R}_{a}^{2}/{R}_{nn}^{2}}}\right]\ge 0$ (7)

which is simply 0.8 × *E _{em}*

Figure 2. Schematic configuration of a singlet deuteron ^{2}*H*_{0}* *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 ^{2}*n*_{0}. The right neutron configuration is the same as that in Figure 1. Switching the spin directions in *n
$\uparrow $ * on the right leads to the *n
$\downarrow $ * configuration on the left.

5. Triplet Dineutron Binding Energy and Decay

In Section 3, the right neutron configuration in Figure 1 has been inverted to become the right configuration in Figure 2. Perform the same inversion to the right neutron in Figure 3 yields Figure 4.

Analogous to (2), the total electrostatic energy *E _{em}*

$\begin{array}{c}{E}_{em2n1}=\frac{{q}_{u}{q}_{dd}+{q}_{dd}{q}_{u}}{{R}_{nn}}+\frac{{q}_{u}{q}_{u}+{q}_{dd}{q}_{dd}}{\sqrt{{R}_{nn}^{2}+{R}_{a}^{2}}}\\ =\frac{8{e}^{2}}{9}\frac{1}{{R}_{nn}}\left[-1+\frac{1}{\sqrt{1+{R}_{a}^{2}/{R}_{nn}^{2}}}\right]=-{E}_{em2n1}\le 0\end{array}$ (8)

Comparison with (2) together with the lower limit of (4) yields the triplet dineutron ^{2}*n*_{1} binding energy *E _{B}*

^{2}*n*_{1} is thus a stable nucleus, like ^{2}*H*_{1}, but its binding energy 1.78 MeV is weaker than 2.2245 MeV for ^{2}*H*_{1}. Energetically, it can decay into deuteron via neutron beta decay,

${}^{2}{n}_{1}\to {\text{\hspace{0.17em}}}^{2}{H}_{1}+{e}^{-}+{\stackrel{\xaf}{\nu}}_{e}$ (9)

where *n*_{e}* *denotes the electron neutrino. Since each of the neutrons in ^{2}*n*_{1} 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 ^{16}Be decay [6] . It is assigned to ^{2}*n*_{1} here but needs not decay according to (9). Instead, it can decay into two neutrons when excess energy is available. The predicted ^{2}*n*_{1}* *binding energy 1.78 MeV has not been measured.

Figure 4. Schematic configuration of the spin
$\downarrow \downarrow $ component of a triplet dineutron ^{2}*n*_{1}. 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 *E _{B}*

Thus, in the two nucleon system, only the triplet deuteron ^{2}*H*_{1} and the triplet dineutron* *^{2}*n*_{1} are bound electromagnetically. The former is 20% stronger than the latter. The remaining four combinations, ^{2}*H*_{0}, ^{2}*n*_{0} 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.

References

[1] Burgess, C. and Moore, G. (2007) The Standard Model, A Primer. Cambridge University Press, Cambridge.

https://doi.org/10.1017/CBO9780511819698

[2] Hoh, F.C. (1994) International Journal of Theoretical Physics, 33, 2325-2349.

https://doi.org/10.1007/BF00673960

[3] Hoh, F.C. (2019) Scalar Strong Interaction Hadron Theory II. Nova Science Publishers, New York.

[4] Hoh, F.C. (2021) Journal of Modern Physics, 12, 139-160.

[5] Hoh, F.C. (2020) Journal of Modern Physics, 11, 967-975.

https://doi.org/10.4236/jmp.2020.117060

[6] Spyrou, A., Kohley, Z., Baumann, T., Bazin, D., et al. (2012) Physical Review Letters, 108, Article ID: 102501.