Why can we not produce a "giant" nucleus?

[wacki]

Apart from the giant nucleus. Why is it not possible to have a 2 neutron bound state. (I'm not sure if that should be asked in a separate thread, but it is so closely related). For the 2 neutron bound state there is no electric repulsion and if they have opposite spin they could both sit in the ground state. Shouldn't that be extremely stable?

 

[Nik_2213]

Seems not. Nature can be inconvenient...
FWIW, see...
https://en.wikipedia.org/wiki/Tetraquark
https://en.wikipedia.org/wiki/Pentaquark

Could be said that, without a deutron's proton to supply the relevant 'glue', any di-neutron is a 'hexa-quark' which promptly falls apart to two single neutrons, which then decay in the usual way...

 

[vanhees71]

Already the nucleon-nucleon potential is a quite complicated beast. Particularly it's spin-dependent. Model calculations, e.g., based on effective low-energy effective descriptions of the nuclear two-body force (the most modern version is based on a systematic expansion using chiral effective theory; historically there are effective one-boson exchange models like the Walecka model too), show that the two-nucleon system has only one bound state, namely the deuteron with $S=1$ and total angular momentum $J=1$. That implies that the orbital angular momentum must be $L=0$ or $L=2$, and indeed the bound state is a mixture of both, dominated by $L=0$ but with a small admixture of the $L=2$ state to account for the non-vanishing electrical quadrupole moment.

Now consider two protons or two neutrons. Since they are identical fermions the only bound state possible for the pn system (the deuteron) is not allowed for these systems, because for them the two-body wave function must be antisymmetric. The spatial part and spin of the bound state is symmetric, and this cannot be for pp or nn. Thus neither the pp nor the nn system have any bound state.