# 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?

I am curious about this:

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. Dec 19 '18 at 20:13

## 1 Answer

A quark and its antiparticle are not identical, so the theorem doesn't apply. It would apply to 2 identical antifermions though.

Also, note that the neutral pion isn't an "up + antiup or down + antidown quark" but the linear combination:

$$\frac{u \bar{u} - d \bar{d}}{\sqrt{2}}$$

• Thank you. Could you provide a source where the fact that it does apply for identical antifermions is stated? Dec 19 '18 at 3:42
• Could you also tell me why the relative orbital angular momentum of a quark+antiquark in a pion is always equal to 0? Dec 19 '18 at 3:52
• Antiquarks, positrons, antineutrinos etc... are not different kinds of particles than, respectively quarks, electrons and neutrinos. They are still fermions and obey the same laws. Antifermion just suggests that they are the conventionally designated "anti" version of the particle. If (however unlikely) we had discovered the positron before the electron we would call them the other way around Dec 19 '18 at 6:01
• The antisymmetrization requirement applies to all half-integer spin particles. It may be tough to find a source that explicitly states this for antiparticle components, but I'll look. Dec 19 '18 at 6:22
• 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. Dec 19 '18 at 9:24