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Computing the Isospin of the Deuteron

  • Thread starter Estartha
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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?
 

Answers and Replies

  • #2
vela
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I don't think that's correct. It's been awhile, so I may be totally wrong here. You're considering the proton and neutron to be identical particles, so what symmetry requirement has to be met by the deuteron's state?
 
  • #3
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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 [tex] \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})] [/tex] Is this what you're referring to?
 
  • #4
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Okay, I think I see it now. If we're using the isospin representation, then we have to regard the proton and neutron as indistinguishable fermions which means the deuteron's wave function is antisymmetric. Thank you for your help!
 

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