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Consider the annihilation of a neutron by an anti-neutron $$ n\overline{n} \rightarrow \pi^0 \pi^0 $$ so that the initial relative angular momentum is zero. Because the spin of neutrons is $1/2$, $J_i$ can take the values $0, 1$.

Now, in pions the spin is zero, so any angular momentum in the rhs should be in the form of $L_f$, which can only take (by conservation of total momentum) the values $0$ or $1$. By some sort of symmetry on bosons wavefunctions, it can be seen that $L_f = 0$, but I don't see why this is the case. Can you shed some light into it?

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You got it almost right. The symmetry of the wavefunction describing a system made of 2 identical boson must be even by the interchange of the 2 bosons because of the Pauli principle. Interchanging the 2 bosons position introduces a factor $(-1)^l$ with $l$ the angular momentum quantum number. Thus among the 2 values allowed by the conservation of angular momentum ($l=0,1$), only $l=0$ is allowed.

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  • $\begingroup$ This is the point I don't see. Why does the interchange of bosons induce the $(-1)^l$ factor in the spacial part of the wavefunction? $\endgroup$
    – nabla
    Commented Jun 18, 2016 at 9:17
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    $\begingroup$ because the eigenfunctions of the orbital angular momentum are the spherical harmonics $Y_l^m(\theta,\phi)$. Interchanging the position of the 2 bosons is equivalent as doing $Y_l^m(\pi-\theta,\pi+\phi) = (-1)^l Y_l^m(\theta,\phi)$, and hence the extra factor $(-1)^l$. $\endgroup$
    – Paganini
    Commented Jun 18, 2016 at 15:53
  • $\begingroup$ One last "why", I promise ;) Why interchanging the position of the 2 bosons is equivalent of changing $Y_l^m \rightarrow Y_l^m(\pi-\theta, \pi + \phi)$? $\endgroup$
    – nabla
    Commented Jun 30, 2016 at 8:27
  • $\begingroup$ Consider the barycenter of the 2 bosons: Let be $\theta,\phi$, the coordinates of one of the 2 bosons with respect to the barycenter. If you swap the 2 bosons, the one which was at $\theta,\phi$ moves to $\pi-\theta, \pi+\phi$ and hence the effect on $Y_l^m$. Does that answer your question? $\endgroup$
    – Paganini
    Commented Jul 1, 2016 at 8:33

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