# Can one prove the full spin-statistics theorem from the spin 0, 1/2 and 1 cases?

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 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 at 13:53
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 $(-1)^n$.
"...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 at 9:44