# Is the spin-statistics theorem true for antifermions?

The spin-statistics theorem says that having a system of identical fermions, the total wavefunction is antisymmetric with respect to exchange of any two fermions.

My question is, does this hold for antifermions too?

The neutral pion is formed by up + antiup or down + antidown quarks. Do these constituents follow the spin-statistics theorem inside the neutral pions? i.e., does the wavefunction of the neutral pion have to be antisymmetric under exchange of the quark and the antiquark?

• The spin-statistics theorem says that having a system of identical fermions, the total wavefunction is antisymmetric with respect to exchange of any two fermions. Your statement of the theorem isn't correct. A fermion is (by definition) a particle obeying Fermi-Dirac statistics. Fermions have antisymmetric wave functions. The theorem says that half-integer spin particles are fermions, integer spin ones are bosons. – Elio Fabri Dec 19 '18 at 20:13

$$\frac{u \bar{u} - d \bar{d}}{\sqrt{2}}$$
• I'm not sure what you mean by "relative", but a pion's orbital angular momentum doesn't have to be 0. By definition, a pion is a $J^{P\,C}=\mathrm{(even)}^{-+}$ light unflavored meson with I=1. This naively decomposes to S=0, L=even, and anything matching it is defined as a pion. Check out the pi(2)(1670). Its actual quark content is a matter of debate. – alexchandel Dec 19 '18 at 9:24