Deuteron (2H) is composed of a neutron (spin-1/2) and a proton (spin-1/2), with a total spin of 1, which is a boson. Therefore, it is possible for two deuterons to occupy the same quantum state.

However, the protons and neutrons inside are fermions, so they proton/neutron shouldn't be able the share states due to the exclusion principle. Why it does not happen?


2 Answers 2


The exclusion principle is for identical fermions; but as proton and neutron are different they can occupy the same quantum state in the deuteron: that's the key point.

That's why you can have a bound state proton-neutron (the deuteron) and no neutron-neutron.

Look at my answer to this question for more details. The main problem is that to have a bound state, you need a positive contribution in the exchange integral, requiring the nucleus to have parallel spins. This is possible in the case of the deuteron.


I think you also mean: "if two deuterons are in the same state, which is allowed because they are boson, then the proton/neutron composing these deuterons should also be in the same state".

The problem is that the exclusion principle does not work like that. If you consider the state vector of the deuteron you cannot after "go down one level" if you consider the deuteron as a boson and consider the individual nucleus. That's because the stage vector of the deuteron is not the one of a neutron + the one of a proton; it is a linear combination of both.

If you try to construct a global state composed of two neutrons and two protons then you will see that you cannot have a bound state where the spins are all parallel. You will have to construct a state where the spins of the protons are anti parallel and the same for the neutrons. At the end you will have a nice state vector the Helium, but that's a more complicated story.


This one state for deuteron is not a single state for both protons and neutrons.


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