Relative Angular Momentum in 2 Body Decay at Detector Level

[Naeem Anwar]

How the relative angular momentum of two particles can be detect by detector in two particle decay (center of mass frame)? I am curious about the signatures/differentiation between different relative momenta, means how one can decide that it is L=0, L=1,2,3,...?

Of course the distribution would be different, but what kind of difference exist exactly? Looking for some pictorial spirit to understand the difference.

Thanks!

 

[ChrisVer]

I think it has to do with the spherical harmonics functions $Y^m_{l}$.
There will be some particular dependence on the $\theta$ coordinate and this will help you find the $L$.
http://www.trinnov.com/wp-content/uploads/2011/11/sphericalHarm.jpg

For example an $L=0$ particle will be independent from $\theta= - \pi$ to $\pi$ (or $\cos \theta \in [-1,1]$.
An $L=1$ will have some particular dependence on theta..

I don't know maybe there are other more effective ways to do that in a detector.

 

[Naeem Anwar]

Ok got it. Thanks!

But in theoretical calculation I am not clear what is the major difference I should have to keep in mind in calculating these partial decay widths. I know in the final result there must be projection on some $Y^m_{l}$ but I am not able to get the difference from start. Do you have any idea?