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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?

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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).

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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?

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well, fermions' "spatial wave function" can also be antisymmetric. I think it's the whole wave function(spin+spatial) that matters.

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