Attraction between neutrons

Is there any simple explanation as to why two neutrons won’t bond together in an atomic “nucleus” (I’m not sure that would be the word). Certainly a proton and a neutron will and I thought the strong (color) force was independent of nucleon type.

Wikipedia says if the strong force were 2% stronger, then Helium-2 would be stable. I assume that means if it were just a bit stronger, it would overcome the electrostatic repulsion, but what is it that is pushing two neutrons apart and keeping them from bonding?

(And yes I’m aware of neutron stars. I’m thinking of just two neutrons in otherwise empty space.)

That depends on what you mean by “simple”. The simplest explanation that I know of is based on a concept called isospin, which is analogous to quantum mechanical spin angular momentum. Are you familiar with that?

There is a sort of symmetry between protons and neutrons as far as the strong force goes, but it’s not independent of them. The implication here is that a “nucleus” of two neutrons would be stable iff a nucleus of two protons were. In both cases, it’s a strong force effect that prevents stability, not an electromagnetic one.

I’ve heard and read some about isospin though I can’t say I really understand it.

Certainly at some point the electromagnetic force must come into play since there are many more stable isotopes with n neutrons and p protons than there are with p neutrons and n protons when n > p. Is this because when you pack more nucleons into the nucleus they must be further apart (on average) and the strong force is short range?

Two relevant facts:
(1) The strong interaction has spin dependent terms, meaning that the relative spins of the two particles affects the binding energy.
(2) Two identical particles (neutrons or protons) can only form a bound state if it is overall antisymmetric (Pauli exclusion). For these spin-(1/2) particles, this requires that the bound state have total spin S=0.

The spin-dependent terms of (1) are such that S=0 states suffer a penalty (or, if you’d rather, S=1 states get a bonus), and this breaks the binding energy bank, making n-n unstable. A corollary is that any deuteron you find is in an S=1 (“spin triplet”) state. This is okay for n-p because the overall state need not be antisymmetric since the proton and neutron aren’t identical particles.

Note that spin is not isospin. Isospin is an approximate symmetry that tells us that any conclusions we make for neutrons holds for protons as well (when Coulomb effects are corrected for). So, from the above information, you can conclude that the proton-proton state is unbound even without the effect of Coulomb repulsion.

Here’s a wikipedia article on Tetraneutrons, four neutrons bound together with no protons. There was a thread that touched on these, and I got an explanation why they aren’t stable. I’ll see if I can find it.

Here’s the thread, my question is post #49, and the answer by Pasta immediately follows.

Whadya know? When I was typing the above post, I was saying to myself, “Hmm, I have a feeling I might have answered this question before.” I guess that’s why!

Yes, I believe this is correct.