Pions, parity, spin Pions have odd parity ($P=-1$) which means their wavefunction is anti-symmetric $\psi(x)=-\psi(-x)$. According to Spin-Statistics theorem fermions (spin 1/2 particles) have anti-symmetric wavefunctions. Just looking at the wavefunction it seems that pions are fermions. However, we know that pions have spin 0 or spin 1 and thus are bosons.
There must be something wrong in the above chain of conclusion. What is it?
 A: Fermion wavefunctions are antisymmetric under the interchange of two particles. Spatial inversion flips the spatial coordinate, but does not interchange particles.
In other words, let's say we have a two particle wave function, $\psi(x_1, x_2)$ (where $x_1$ is the position of particle 1, and $x_2$ is the position of particle 2).
Being odd under parity says:
\begin{equation}
\psi(x_1,x_2) = - \psi(-x_1,-x_2).
\end{equation}
Being odd under interchange of particles says
\begin{equation}
\psi(x_1, x_2) = - \psi(x_2, x_1).
\end{equation}
Thus parity and statistics are independent properties. In particular, it is perfectly consistent to have a parity odd boson.
(Things get a little more interesting if you have spin, because parity also affects the polarization, but that seems like a more complicated question than what you asked).
A: The pion is a pseudoscalar particle, which behaves like a scalar, except that it changes sign under a parity inversion while a true scalar does not.
For details, see this post by @Luboš Motl and links there:
What is a Pseudoscalar particle?
A: well, fermions' "spatial wave function" can also be antisymmetric. I think it's the whole wave function(spin+spatial) that matters.
