Starting from deuteron, the binding energy, or the p-n interaction is 2.2 MeV.

From triton, 3H, the total binding energy is 8.5 MeV, in which, there are only 3 interactions, two p-n and one n-n. Assume the p-n interaction does not change, the n-n interaction is 4.1 MeV.

The total binding energy of 3He is 7.7 MeV. The p-p interaction is 3.3 MeV.

Notices that we neglected the 3-body force in 3H and 3He. And it is strange that the n-n and p-p interaction is stronger then p-n interaction.

In 4He, the total binding energy becomes 28.3 MeV. I try to decompose the energy in term of 2-body, 3-body, and 4-body interaction.

If we only assume 2-body interaction, the interaction strength from n-p, n-n, and p-p are insufficient. One way to look is the 1-particle separation energy.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n).
The proton separation energy is 19.8 MeV = 2(p-n) + (p-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p).

There is no solution for above 3 equations. Thus, only consider 2-body interaction is not enough.

The neutron separation energy is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

Assuming the 2-body terms are the same in 2H, 3H, and 3He, the (n-p-p) and (n-n-p) is 4.03 MeV, which is strange again, as the Coulomb repulsion should make the (n-p-p) interaction smaller then the (n-n-p) interaction. The (n-n-p-p) interaction is -4.03 MeV.

Lets also add 3-body force in 3H and 3He.

The neutron separation energy of 3H is 6.3 MeV = (p-n) + (n-n) + (n-n-p)
The toal energy of 3H is 8.5 MeV = 2(p-n) + (n-n) + (n-n-p)
The toal energy of 3He is 7.7 MeV = 2(p-n) + (p-p) + (n-p-p)
The neutron separation energy of 4He is 20.6 MeV = 2(p-n) + (n-n) + 2(n-n-p) + (n-p-p)
The proton separation energy is 4He 19.8 MeV = 2(p-n) +(p-p) + 2(n-p-p) + (n-n-p).
The total energy of 4He is 28.3 MeV = 4(p-n) + (n-n) + (p-p) + 2(n-n-p) + 2(n-p-p) + (n-n-p-p).

We have 6 equations, with 6 unknown [ (p-n) , (n-n) , (p-p) , (n-n-p), (n-p-p), and (n-n-p-p)]. Notice that the equation from proton separation energy of 3He is automatically satisfied. The solution is

(p-n) = 2.2 MeV
(p-p) = – 4.7 – (n-n) MeV
(n-n-p)  = 4.1 – (n-n) MeV
(n-p-p) = 8 + (n-n) MeV
(n-n-p-p) = 0 MeV

It is interesting that there is redundant equation. But still, the (p-n) interaction is 2.2 MeV, and 4-body (n-n-p-p) becomes 0 MeV. Also the (n-p-p) is more bound than (n-n-p) by 3.9 + 2(n-n) MeV. If (n-p-p) should be more unbound, than (n-n) must be negative and smaller than -1.95 MeV.

Since the interaction strength has to be on the s-orbit (mainly), by considering 2H, 3H, 3He, and 4He exhausted all possible equations (I think). We need other way to anchor either (n-n), (p-p), (n-p-p), and (n-n-p) interactions.

Use the Coulomb interaction, the Coulomb interaction should add -1.44 MeV on the NN pair (assuming the separation is a 1 fm). Lets assume the (n-n) – (p-p) = 1.44 MeV

(n-n) = -1.63 MeV
(p-p) = – 3.07 MeV
(n-n-p)  = 5.73 MeV
(n-p-p) = 6.37  MeV

The (p-p) is more unbound than (n-n) as expected, but the (n-p-p) is more bound than (n-n-p) by 0.64 MeV. This is surprising! We can also see that, the 3-body interaction play an important role in nuclear interaction.

According to this analysis, the main contribution of the binding energies of 3H and 3He are the 3-body force.

In 3H:  (n-n) + 2(n-p) + (n-n-p) = -1.6 + 4.4 + 5.7 = 8.5 MeV
In 3He: (p-p) + 2(n-p) + (n-p-p) = -3.1 + 4.4 + 6.4 = 7.7 MeV

Worked on the algebra, when ever the difference  (n-n) – (p-p)  > 0.8 MeV, the (n-p-p) will be more bound that (n-n-p). Thus, the average protons separation should be more than 1.8 fm. I plot the interactions energies with the change of Coulomb energy below.

The (n-n) and (p-p) are isoscalar pair, where tensor force is zero. While the (n-p) quasi-deuteron is isovector pair. Thus, the difference between (n-n) and (n-p) reflect the tensor force in s-orbit, which is 3.8 MeV. In s-orbit, there is no spin-orbital interaction, therefore, we can regard the tensor force is 3.8 MeV for (n-p) isovector pair.

Following this method, may be, we can explore the NN interaction in more complex system, say the p-shell nuclei. need an automatic method. I wonder the above analysis agreed present interaction theory or not. If not, why?