# Computing the Isospin of the Deuteron

## 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:
$$I_{NN} = |NN ; I, I_3\rangle$$

## 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:

$$\frac{1}{\sqrt{2}}[|\text{NP} \rangle + |\text{PN}\rangle ]$$
and
$$\frac{1}{\sqrt{2}}[|\text{NP} \rangle - |\text{PN}\rangle ]$$

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?

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vela
Staff Emeritus
So I'm guessing isospin assumes that we're looking at two states of one particle, so we're assuming that the proton and neutron are "states" of a particle we're calling the nucleon, so the idea is that we regard the proton and neutron as identical fermions? And identical fermions (say particles a and b) have a wave function $$\psi(\vec{r_1},\vec{r_2} = A[\psi_a(\vec{r_1})\psi_b(\vec{r_2}) - \psi_b(\vec{r_1})\psi_a(\vec{r_2})]$$ Is this what you're referring to?