The nucleus of an atom of deuterium, containing one proton and one neutron. They are formed chiefly by ionizing dueterium and are used as projectiles to produce nuclear reactions after accumulating high energies in particle accelerators.

See also: triton

Greek word meaning "the second". This is often used by fraternities and sororities when naming their chapters. After going through the entire alphabet, chapters start over with "alpha deuteron", meaning "the second alpha" and go from there.

Chapters can also use the same syntax for naming their pledge classes.

An application of the isospin formalism:

The simplest atomic nucleus is the deuteron 2H, 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 I3=-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=-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 I3=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.

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