A Suggestive Bogus Argument
Consider the two-field operator product
where R is the matrix which rotates the spin polarization of the field by 180 degrees when one does a 180 degree rotation around some particular axis. The components of phi are not shown in this notation, has many components, and the matrix R mixes them up with one another.
In a non-relativistic theory, this product can be interpreted as annihilating two particles at positions x and −x with polarizations which are rotated by π (180°) relative to each other. Now rotate this configuration by π around the origin. Under this rotation, the two points and switch places, and the two field polarizations are additionally rotated by a . So you get
which for integer spin is equal to
and for half integer spin is equal to
(proved here). Both the operators still annihilate two particles at and . Hence we claim to have shown that, with respect to particle states: . So exchanging the order of two appropriately polarized operator insertions into the vacuum can be done by a rotation, at the cost of a sign in the half integer case.
This argument by itself does not prove anything like the spin/statistics relation. To see why, consider a nonrelativistic spin 0 field described by a free Schrödinger equation. Such a field can be anticommuting or commuting. To see where it fails, consider that a nonrelativistic spin 0 field has no polarization, so that the product above is simply:
In the nonrelativistic theory, this product annihilates two particles at x and −x, and has zero expectation value in any state. In order to have a nonzero matrix element, this operator product must be between states with two more particles on the right than on the left:
Performing the rotation, all that you learn is that rotating the 2-particle state gives the same sign as changing the operator order. This is no information at all, so this argument does not prove anything.
Read more about this topic: Spin-statistics Theorem
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