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Using second quantization for scalar field, spinor field and vector fields, we can get commutation and anticommutation relations for the birth and destruction operators of the fields, which leads us to the Bose or to Fermi statistics. Is it possible to expand these results on a field of arbitrary spin (integer or half-integer), using the basic idea that each field can be built by combination of spinor $\frac{\hbar }{2}$ fields?

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More on spin-statistics theorem: physics.stackexchange.com/q/23338/2451 and links therein. –  Qmechanic Aug 19 '13 at 11:09
    
The normal proofs require certain assumptions. Although the list of assumptions isn't unique, one possible list is given here: en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem#Proof . Are you talking about devising a different method of proof based, say, on this list of assumptions, or are you talking about replacing one of the assumptions with some other assumption? –  Ben Crowell Aug 19 '13 at 13:53

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Suppose we assume that an object's statistics depend only on its spin and not on whether the object is composite or fundamental. This assumption seems natural, since if it failed, it would be too good to be true -- it would give us a way of finding out about the internal structure of any particle, at all scales, without having to build particle accelerators.

Given that assumption, the full theorem follows directly from the spin-1/2 case. Any spin can be realized by coupling spin 1/2's. Given that spin 1/2 has an eigenvalue of $-1$ under particle exchange, coupling $n$ of them produces a composite system that has an eigenvalue of $(-1)^n$.

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"...Any spin can be realized by coupling spin 1/2's...", - how exactly does this go? By straight sum or multiplying of spinor representations of the Lorentz group? And how that save (or change) anticommutation relations? –  PhysiXxx Aug 19 '13 at 20:51
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how exactly does this go? There's nothing fancy going on. For example, a deuteron is made out of two spin-1/2 particles, coupled to spin 1. And how that save (or change) anticommutation relations? It doesn't change the commutation relations. It's just that the particle-exchange operator operates on every particle in the system. –  Ben Crowell Aug 19 '13 at 21:26
    
"...There's nothing fancy going on...", - so, one of deuteron's state, of course, can't be interpreted as representation $\left( \frac{1}{2}, 0\right) \times \left( \frac{1}{2}, 0\right)$? "...It doesn't change the commutation relations...", - thank you, I understand this now. –  PhysiXxx Aug 20 '13 at 9:44

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