An application of the

isospin formalism:

The simplest atomic

nucleus is the

deuteron ^{2}H, a proton bound to a neutron. One may
ask why there is no bound state of two protons or two neutrons, especially considering that in
larger nuclei you actually get so-called "

pairing energy" from the formation of such pairs.

The answer lies in the

Pauli exclusion principle. Nucleons are fermions and therefore the

wave function of a pair of nucleons has to be

antisymmetric. As a neutron has

isospin I

_{3}=-1/2 and a proton I

_{3}=+1/2, the pairs would have I

_{3}=-1 and +1, respectively. That means

**I** has to be 1. This is a so-called

triplet state (because with I=1 you can have I

_{3}=-1,0,+1).

Triplet states, however, are symmetric - and not allowed by the

Pauli principle.

Symmetry in this context means symmetry under particle exchange. Imagine a wave function |nn>: If we swap the two particles, nothing changes - it's symmetric! In contrast to this there is also a

singlet state |pn>-|np>. If we swap the particles here, we introduce a minus sign - the wave function is antisymmetric!

Now the thing is that you need a neutron and a proton to get a singlet state. And guess what, this is the deuteron!
It has isospin

**I**=0 and I

_{3}=0.

But in fact this is not the whole story yet. It's not enough to consider only the

isospin - after all, the total wave function also has a

spin component!

In this case however it doesn't change anything: The

nuclear force that binds the nuclei together is dependent on the spin - there is no bound state with spin 0 but only with spin 1 (ie the nuclear force is stronger if the spins are parallel). That means that the deuteron is in the spin triplet state, and the product of the (symmetric)
spin wave function and the (antisymmetric) isospin wave function is antisymmetric - as it has to be (this is just like multiplying positive and negative numbers).

A hypothetical nn or pp pair would be in a symmetric isospin state and need an antisymmetric spin state to make the total wave function antisymmetric. But there is no such state, and that's why there are no nn or pp pairs.