Why don’t two protons bind together?

[Kostik]

TL;DR

If the nuclear force is many times stronger than the electrical force, why don’t two protons form a stable helium-2 nucleus?

If the nuclear force is many times stronger than the electrical force, why don’t two protons form a stable helium-2 nucleus?

 

[Reggid]

It does not even have to do with the electrostatic repulsion of the protons. This you can simply see from the fact that there is also no bound state of two neutrons, where the electromagnetic force does not enter at all.

The reason why you do not have a helium-2 nucleus has to do with the Pauli principle and the spin coupling of the nucleons.

Because nucleons are fermions, the state of the two-particle system must be antisymmetric under swapping the two particles. The possible isospin states are

triplet (symmetric)

$$|pp>, |nn>, \frac{1}{\sqrt{2}}\bigl(|pn>+|np>\bigr)$$
(the first is a helium-2 nucleus, the second a deuteron, and the third a deuterium nucleus (not really, as we will see later))

singlet (antisymmetric)
$$\frac{1}{\sqrt{2}}\bigl(|pn>-|np>\bigr)$$
(a deuterium nucleus)

the possible spin states are
triplet (symmetric)

$$|\uparrow\uparrow>, |\downarrow\downarrow>, \frac{1}{\sqrt{2}}\bigl(|\uparrow\downarrow>+|\downarrow\uparrow>\bigr)$$
(spin-1, all three are just related by spatial rotations)

singlet (antisymmetric)
$$\frac{1}{\sqrt{2}}\bigl(|\uparrow\downarrow>-|\downarrow\uparrow>\bigr)$$
(spin-0)

Total antisymmetry of the state implies that we have to combine either an isospion-triplet with a spin-singelt or an isospin-singlet with a spin-triplet.

Now it turns out that the spin-triplet(spin-1) is energetically preferred, while the spin-singlet (spin-0) does not lead to a bound state. Unfortunately I can not really give you an explanation of this now, I would have to look it up, but you can just see it from the fact that that there does not exist a spin-0 deuterium nucleus. All deuterium nuclei have spin 1. That means that the isospion state of the the deuterium is $\frac{1}{\sqrt{2}}\bigl(|pn>-|np>\bigr)$ and NOT $\frac{1}{\sqrt{2}}\bigl(|pn>+|np>\bigr)$.

That the spin is symmetric implies that the isospin has to be antisymmetric. But the states $|pp>$ (helium-2) and $|nn>$ (deuteron) are symmetric. So the fact that we find only deuterium nuclei with spin-1 directly tells us that we cannot have bound states of two protons or two neutrons.

 

[ChrisVer]

I don't think you can make such a claim based on just the symmetry -antisymmetry. Of course you can use it, but it will not get you to the answer. The Pauli's principle only tells you that the total wavefunction must be antisymmetric. This doesn't tell you if something is realized or not (except for if it is breaking something completely). You have also omitted the spatial momentum and thus the spatial parity which makes it a little more complicated, but the answer is in the "spin" couplings of the nucleon-nucleon interactions. The pauli's principle only let's you pick certain states to compare against each other.
What one would have to do is pick up the potential energies, get the energy eigenvalues of the configurations with $T=1$ (where nn, pp "live" in), do the same for $T=0$ (where the deuterium is) and figure out that their difference is more than 2MeV (binding energy of Deuterium)... Interested people can have a look in 6.2 here:
https://www.physics.umd.edu/courses/Phys741/xji/chapter6.pdf
about what kind of potential they could take. The big problem of "deuterium" is that a pretty weakly-bound state... If it wasn't so weakly bound, then maybe we would be able to see its "excitations" (and so even the nn/pp bound states which would be degenerate)

 

[vanhees71]

Indeed, the deuterium potential has only one bound state.

 

[snorkack]

What could be stated is what the strong force resists is above all not electromagnetic repulsion, but zero point movement.