Exchange symmetry when isospin is concerned?

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Discussion Overview

The discussion revolves around the concept of exchange symmetry in the context of isospin, particularly focusing on the wavefunction of a proton-neutron system. Participants explore whether the requirement for antisymmetry under exchange applies to these nucleons when treated as identical fermions due to their isospin properties.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Some participants assert that identical fermions must have an antisymmetric wavefunction under exchange, while identical bosons must be symmetric.
  • There is a question about whether the requirement for antisymmetry applies to protons and neutrons, which are not identical fermions, when considering isospin.
  • Some argue that isospin can blur the distinction, suggesting that protons and neutrons can be treated as identical fermions, necessitating antisymmetry in their wavefunction.
  • Others propose that treating protons and neutrons as different fermions is also valid, in which case the wavefunction does not need to be antisymmetrized.
  • One participant notes that from the perspective of the strong force, protons and neutrons are treated as identical due to their similar interactions, despite differences in mass and charge.
  • There is a clarification that the antisymmetry requirement applies specifically to identical fermions, and the discussion about neutron-proton pairs in the context of isospin remains nuanced.

Areas of Agreement / Disagreement

Participants express differing views on whether the antisymmetry of the total wavefunction applies to non-identical fermions like protons and neutrons when isospin is considered. There is no consensus on this issue, as some support the idea of treating them as identical while others do not.

Contextual Notes

The discussion highlights the complexities surrounding the treatment of nucleons in quantum mechanics, particularly regarding their identities under isospin symmetry and the implications for their wavefunctions. Limitations include the approximate nature of isospin symmetry and the differing physical properties of protons and neutrons.

Silversonic
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As far as I know identical fermions are antisymmetric under exchange. Identical bosons are symmetric under exchange. Is this fact blurred when we consider isospin? Considering the wavefunction of a proton-neutron system;

\psi = \psi_{space} \psi_{spin} \psi_{isospin}

I'm told this needs to be antisymmetric under exchange of the proton and neutron, but they are not identical fermions. Does it need to be antisymmetric because we consider isospin which does view the proton and neutron as identical fermions?
 
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neutron and proton are fermions, so their wf has to be antisymmetric...
 
I thought that only applied to identical fermions? Guess I was wrong.
 
Silversonic said:
As far as I know identical fermions are antisymmetric under exchange. Identical bosons are symmetric under exchange. Is this fact blurred when we consider isospin? Considering the wavefunction of a proton-neutron system;

\psi = \psi_{space} \psi_{spin} \psi_{isospin}

I'm told this needs to be antisymmetric under exchange of the proton and neutron, but they are not identical fermions. Does it need to be antisymmetric because we consider isospin which does view the proton and neutron as identical fermions?
Isospin is an optional convention. You can treat protons and neutrons as different fermions, in which case the wavefunction does not need to be antisymmetrized. Or, you can treat them as identical, with an extra degree of freedom whose symmetry is used to make the overall wavefunction antisymmetric.
 
From the point of view of the strong force the proton and neutron are identical. They interact with the strong force identically and can be converted into each other easily (through the weak interaction). If any of those two statements weren't true there would be little sense in treating them as identical. Note that this treatment is inexact. Protons and neutrons have different masses and charges and the conversion between them requires the production of leptons.
 
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Apologies for the necro-bump but I want to make sure I've got this correct as I'm coming back to it.

So is the antisymmetry of total wavefunction under exchange of two general fermions definitely not thing? It's definitely only for two identical fermions, e.g. two protons, or a neutron/proton when considering isospin?
 
See post #4 above. Answer hasn't changed.
 
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Silversonic said:
Apologies for the necro-bump but I want to make sure I've got this correct as I'm coming back to it.

So is the antisymmetry of total wavefunction under exchange of two general fermions definitely not thing? It's definitely only for two identical fermions, e.g. two protons, or a neutron/proton when considering isospin?

Yes, you only need to take care of symmetry/anti-symmetry when the two particles are identical. In the case of the (approximate) isospin symmetry, neutron and proton are (approximately) identical.
 

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