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## Homework Statement

The deuteron is mostly a bound state of a proton and neutron with orbital angular momentum L=0 and spin S=1. To a good approximation we can neglect the proton-neutron mass difference and electromagnetic interactions, and treat the proton and neutron as two isospin components of the same I = 1/2 fermion, the nucleon.

Show that the isospin of the deuteron is I=0.

## Homework Equations

Isospin Proton: 1/2

Isospin Neutron: -1/2

And the isospin of two nucleons:

[tex] I_{NN} = |NN ; I, I_3\rangle [/tex]

## The Attempt at a Solution

I know the isospin of the neutron is -1/2 and the isospin of the proton is 1/2. I've narrowed down the possible isospin states to the following:

[tex] \frac{1}{\sqrt{2}}[|\text{NP} \rangle + |\text{PN}\rangle ] [/tex]

and

[tex] \frac{1}{\sqrt{2}}[|\text{NP} \rangle - |\text{PN}\rangle ] [/tex]

I know which one it needs to be (i.e. |0,0>), but I am feeling unsure of my reasoning as to why it should be explicitly that state and not |1,0>. My guess is that because of S=1 and L=0, this means the nucleus is in the S=0 state which means the nucleons are in the singlet configuration and this transfers over to the isospin (i.e. the |0,0> isospin state corresponds to the singlet configuration). Does anyone know if this roughly correct reasoning?