Proton-proton bound state There is something unclear for me. 
We say that the deuterium is a proton-neutron bound state of orbital angular momentum L=0 and of total spin S=1. I don't understand why can't we build such a state (with total angular momentum J=1) with two protons : (1/2,1/2) and (1/2, -1/2) (I'm not sure what to do with their orbital angular momentum, can I put it zero to both of them ?) where the first term in the curled brackets is the spin s and the second is the component along Sz (going from -s to s). 
It has to do with the Pauli's principle, but I don't see where...
 A: The binding energy of the deuteron is well known: about $2.2 \,\mathrm{MeV}$.
The electrostatic energy requirement to bring two protons to strong nuclear binding distance (about $1 \,\mathrm{fm}$) is:
$$ E = k \frac{e^2}{2\,\mathrm{fm}} = \frac{2.88 \,\mathrm{eV \cdot nm}}{10^{-6} \,\mathrm{nm}}  = 2.88 \,\mathrm{MeV}\,.$$ 
So the Coulomb interaction is strong enough to make the p-p state unbound.
A: The bound state of two protons is called diproton. From wikipedia page:

Helium-2 or ${}^2\mathrm{He}$, also known as a diproton, is an extremely unstable isotope of helium that consists of two protons without any neutrons. According to theoretical calculations it would have been much more stable (although still beta decaying to deuterium) had the strong force been 2% greater.[5] Its instability is due to spin-spin interactions in the nuclear force, and the Pauli exclusion principle, which forces the two protons to have anti-aligned spins and gives the diproton a negative binding energy.
...
${}^2\mathrm{He}$ is an intermediate in the first step of the proton-proton chain reaction.

