In Weinberg's textbook on QFT(google book preview), he discussed the phase acquired after interchanging particle labels in the last paragraph of page 171 and the footnote of page 172. It seems he's suggesting interchanging particles of same species but different spin states will only bring a phase determined by convention, that is, the phase does not have to be ±1. I'm having a hard time understanding this, because I was taught that interchange of identical particles must give a phase of ±1 and took it for granted. Besides, for fermions antisymmetry seems to be an inevitable consequence of Pauli exclusion principle：(adsbygoogle = window.adsbygoogle || []).push({});

Exclusion principle tells us the square of a fermion creation or annihilation operator must be zero, so [itex]a^2_m=0,\ a^2_n=0,\ (a_m+a_n)^2=0[/itex](assuming no superselection rule on m, n so that the 3rd operator is well defined), and we can easily see this implies [itex]a_ma_n+a_na_m=0[/itex]

Is Weinberg actually treating particles of same species but different spin states as distinguishable particles? If so can I take it further and conclude particles of same species, same spin states but different momenta are also distinguishable, so that the interchange phase is also conventional?

Cross-posted:http://physics.stackexchange.com/qu...-of-identical-particles-in-weinbergs-qft-book

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# About interchange phase of identical particles in Weinberg's QFT book

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