In quantizing a scalar field, we impose commutation relation between the field operators by hand. On the other hand, anti-commutation relation is imposed between Dirac field operators by hand. As a consequence one gets Bose-statistics (two-particle wavefunction is symmetric) in the first case and the Fermi statistics (two-particle wavefunction is antisymmetric) in the second case.

But does it really prove the spin-statistics theorem?


1 Answer 1


I write below the statement of the aforementioned theorem which assumes, as hypotheses, the validity of so-called "Wightman axioms" in the four-dimensional Minkowski spacetime.

You see that there is nothing imposed by hand. It is actually a no-go theorem. Quantizing free fields, it establishes in particular that the standard choice is the only possible.

Spin-Statistics theorem (Streater-Wightman's book Thm 4-10, adopting the signature +---):

For a general irreducible spinor field the "wrong" connection with statistics:

$$[\phi_a(x), \phi^\dagger_a(y)]_+ =0\quad \mbox{$\phi$ with integer spin}$$

$$[\phi_a(x), \phi^\dagger_a(y)]_- =0\quad \mbox{$\phi$ with half-odd integer spin}$$

and $(x-y)^2<0$, implies:

$$\phi_a(x)|vac\rangle =0\:.\quad (1)$$ In a field theory in which all fields either commute or anti commute, this also implies $\phi=\phi^\dagger=0$.

Identity (1) immediately implies that all n-point functions of the theory vanish so that the theory turns out to be trivial. $|vac\rangle$ is the unique (up to phases) normalized vector state which is Poincaré invariant.


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