# Why are the dineutron and diproton unbound?

It is known that there are no diproton or dineutron nuclei.

Does this mean that two protons or neutrons are not actually attracted to each other? Even if the attraction was weak, wouldn't it cause bound states anyway?

• This seemed like two separate questions, and one of the questions was a duplicate of this one: physics.stackexchange.com/q78107 Therefore I've taken the liberty of editing out that part to focus on the other part. Hope that's OK with you.
– user4552
Nov 18 '14 at 21:44
• My question was quantitative, and now you made it qualitative. I would like a plot U(r) for neutrons and protons. Nov 18 '14 at 21:56
• OK, sorry if I messed up what you wanted. However, I think that would be better if it was separate from the question of why the dineutron and diproton are unbound. How about asking your quantitative question about $U(r)$ as a separate question? But actually the answer is not going to be well defined, for the reasons described in my answer to the qualitative question.
– user4552
Nov 18 '14 at 21:58
• If you want to salvage your old text for a new question, you can look at the edit history by clicking on "edited x minutes ago."
– user4552
Nov 18 '14 at 22:04
• @BenCrowell sure I agree, this is why I didn't roll back your edits, since they are ok. If there is no general question about nonexistence of nn and pp on SE yet, then it should be here of course. Nov 18 '14 at 23:46

Great question!

My answer would be that in order to get a bound state, we need to have a potential that is deeper than the kinetic energy the two particles have. We have a better chance of getting a potential of the right depth to bind two nucleons if:

(1) They don't repel charge-wise. Compared to nucleon interaction the Coulomb force isn't that strong, but it's still worth considering. This makes two protons look a little less attractive as a bound system.

(2) The two nucleons being aligned in spin gives some extra binding. This is a property of the nuclear force, in which there is a term

${S}_{12}(\hat{r},\hat{r}) = (\sigma_{1}\cdot\hat{r})(\sigma_{2}\cdot\hat{r})-3(\sigma_{1}\cdot\sigma_{2})$.

The two nucleons in a zero orbital angular momentum state (state of lowest energy) can only align in spin if they are antialigned in isospin, by Pauli exclusion.

The second point is the most important one: nucleons antialigned in isospin can be aligned in spin in an orbital angular-momentum zero state (S-state), by Pauli. This alignment of spins gives them the extra binding they need to form a bound state, because of the dot-product in spin in in the NN interaction, as in the term I mentioned. A proton and a neutron are antialigned in isospin. This means they can align in spin in an S-state, which gives them the "extra" bit of binding energy and lets them stay bound.

• Welcome to Physics.SE! This is a great first answer. Another way to put it might be that if the deuteron had any spin-singlet excited states, then we might look for diproton or dineutron "mirror nuclei" with similar energies; however the deuteron has no bound excitations.
– rob
Jun 3 '15 at 13:20
• I remember reading that the dineutron and diproton are unbound by only ~.1 MeV, a tiny amount (though I can't find a reference for this at the moment). So if the strong force were just a little bit stronger, they would exist. Jun 3 '15 at 17:03
• So you are saying that the proton and a neutron are attracting each other, but neutrons cannot attract each other strong enough, and protons can't attract each other strong enough either? So there is an extra attraction between protons and neutrons? Apr 1 '18 at 12:05
• @ÁrpádSzendrei yes, there's the extra attraction due to aligned-spin state being allowed, unlike the $pp$ or $nn$ cases, where it's forbidden. Apr 1 '18 at 18:01
• @WillLevine: I remember reading that the dineutron and diproton are unbound by only ~.1 MeV, a tiny amount (though I can't find a reference for this at the moment). So if the strong force were just a little bit stronger, they would exist. Re the dineutron, the number may actually be zero, per the 2012 experimental results by Thoennessen et al. But the diproton has to be much more unbound, because the Coulomb repulsion is on the order of 1 MeV.
– user4552
Feb 18 '19 at 17:46

The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest the strong nuclear force could successfully fight against. Experiments in 2012 give evidence that the dineutron may be weakly bound, or that it may be a resonance state that is close enough to bound to create the same kind of strong correlations in a detector that you would get from a dineutron. So it appears that the strong nuclear force is not quite strong enough, but this is not even clear experimentally.

If the dineutron isn't bound, the diproton is guaranteed not to be bound. The nuclear interaction is the same as in the dineutron, by isospin symmetry, but in addition there is an electrical repulsion.

• So you are saying that the Neutrons are only held by the strong force of the Protons in a nucleus (that hass protons and neutrons)? So the Protons have a stronger strong force then the Neutrons? Because based on what you say, even the Protons' strong force would not be enough to go against the momentum (because of uncertainty principle). Apr 1 '18 at 11:58
• @ÁrpádSzendrei, it is my understanding that the more nucleons you have the stronger the overall binding force. I'm not sure about deuterium, though. Apr 3 '18 at 3:17
• @ÁrpádSzendrei, this answer explains why deuterium is stable. Apr 3 '18 at 3:24
• @ÁrpádSzendrei: So you are saying that the Neutrons are only held by the strong force of the Protons in a nucleus (that hass protons and neutrons)? No. So the Protons have a stronger strong force then the Neutrons? No, isospin symmetry says that it's the same, not stronger. This is not an argument about protons and neutrons, it's simply a general argument about why systems of nucleons need not be bound, even though the nuclear force is attractive. Various systems can be stable or metastable against particle emission, e.g., you can have alpha decay.
– user4552
Jun 26 '18 at 1:42