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Exchange symmetry when isospin is concerned? 
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#1
Mar214, 07:06 PM

P: 129

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 protonneutron system;
[itex] \psi = \psi_{space} \psi_{spin} \psi_{isospin} [/itex] 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? 


#2
Mar214, 08:18 PM

P: 754

neutron and proton are fermions, so their wf has to be antisymmetric....



#3
Mar214, 08:48 PM

P: 129

I thought that only applied to identical fermions? Guess I was wrong.



#4
Mar314, 04:49 AM

Sci Advisor
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P: 4,160

Exchange symmetry when isospin is concerned?



#5
Mar414, 03:30 PM

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P: 1,948

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.



#6
May714, 11:10 AM

P: 129

Apologies for the necrobump 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? 


#8
May714, 11:42 AM

P: 305




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