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Exchange symmetry when isospin is concerned?

by Silversonic
Tags: concerned, exchange, isospin, symmetry
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Silversonic
#1
Mar2-14, 07:06 PM
P: 130
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;

[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?
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ChrisVer
#2
Mar2-14, 08:18 PM
P: 919
neutron and proton are fermions, so their wf has to be antisymmetric....
Silversonic
#3
Mar2-14, 08:48 PM
P: 130
I thought that only applied to identical fermions? Guess I was wrong.

Bill_K
#4
Mar3-14, 04:49 AM
Sci Advisor
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Exchange symmetry when isospin is concerned?

Quote Quote by Silversonic View Post
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;

[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?
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.
dauto
#5
Mar4-14, 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.
Silversonic
#6
May7-14, 11:10 AM
P: 130
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?
Bill_K
#7
May7-14, 11:32 AM
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P: 4,160
See post #4 above. Answer hasn't changed.
Einj
#8
May7-14, 11:42 AM
P: 324
Quote Quote by Silversonic View Post
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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