# Decay of a particle

Would someone please explain the following found on P. 125 of these notes?

On the other hand, two $π^0$’s cannot be in an $l = 1$ state. The reason for this is that pions are bosons and so the wavefunction for two identical pions must be symmetric under interchange, whereas the wavefunction for an $l = 1$ state is antisymmetric if we interchange the two pions. This means that the decay mode $$\rho^0\to \pi^0+\pi^0$$ is forbidden.

I don't understand why the wavefunction of $l=1$ must be antisymmetric. Perhaps I have forgotten something?

Thanks.

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The (orbital) wave function for $l=1$ doesn't just "have to be" antisymmetric. It demonstrably "is" antisymmetric. The relevant part of this wave function is completely determined, it's a particular function, so we may see whether it's symmetric or antisymmetric and indeed, it's the latter.
Two particles – in this case two pions – orbiting each other are described by a wave function of the relative position $$\vec r = \vec r_1 - \vec r_2$$ Writing $\vec r$ in spherical coordinates, a well-defined $l,m$ means that the wave function factorizes to $$\psi (\vec r) = \psi_r(r)\cdot Y_{lm}(\theta,\phi)$$ The angular dependence simply has to be given by $Y_{lm}$ because $Y_{lm}$ is, up to the overall normalization, the only wave function with the right angular momentum and its $z$-component being $l,m$.
But $Y_{lm}$ is easily seen to be an odd function of $\hat r$ for odd $l$ and even function of $\hat r$ for even $l$. In particular, $Y_{00}$ is a constant while $Y_{10}$ and $Y_{1,\pm 1}$ are proportional to $z$ and $x\pm iy$ on the unit sphere, respectively.
These functions proportional to $x,y,z$ are clearly odd functions of $\hat r$, so they change the sign under $\vec r\to -\vec r$ which is the sign flip equivalent to $\vec r_1\leftrightarrow \vec r_2$. This odd parity is also called antisymmetry.