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Why are fermion states anti-symmetric under exchange operator?
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[QUOTE="LCSphysicist, post: 6445303, member: 675151"] [B]Homework Statement:[/B] . [B]Relevant Equations:[/B] . Let ##L## be the state space of two identical/indistinguishable particles. Let ##L \otimes L## be the state space of the combined system formed by both particle. If the particles were distinguishable, ##LxL## would have four mutually orthogonal states: ## |\phi\rangle|\phi\rangle, |\omega\rangle|\omega\rangle, |\phi\rangle|\omega\rangle, |\omega\rangle|\phi\rangle ##. But, if the particles are bosons, we have actually three states: ## |\phi\rangle|\phi\rangle, |\omega\rangle|\omega\rangle, |\omega\rangle|\phi\rangle + |\phi\rangle|\omega\rangle## My question is for the fermions case, the only state would be ## |\omega\rangle|\phi\rangle - |\phi\rangle|\omega\rangle## But i can not understand why. (I think i understand the signal, it is because system of fermions are anti symmetric under exchange operator, right?) OBS (This is being used in the book to introduce/derive the Pauli exclusion, so it is not a good idea to answer "because Pauli exclusion (...)") [/QUOTE]
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Why are fermion states anti-symmetric under exchange operator?
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