Timeline for Angular momentum in annihilation $n\overline{n} \rightarrow \pi^0 \pi^0$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2016 at 14:04 | vote | accept | nabla | ||
| Jul 1, 2016 at 8:33 | comment | added | Paganini | 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? | |
| Jun 30, 2016 at 8:27 | comment | added | nabla | One last "why", I promise ;) Why interchanging the position of the 2 bosons is equivalent of changing $Y_l^m \rightarrow Y_l^m(\pi-\theta, \pi + \phi)$? | |
| Jun 18, 2016 at 15:53 | comment | added | Paganini | 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 $(-1)^l$. | |
| Jun 18, 2016 at 9:17 | comment | added | nabla | 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? | |
| Jun 18, 2016 at 7:53 | history | answered | Paganini | CC BY-SA 3.0 |