An application of the isospin
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
=-1/2 and a proton I3
=+1/2, the pairs would have I3
=-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 I3
Triplet states, however, are symmetric - and not allowed by the Pauli
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 I3
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
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.