Subject: Follow up to ID: 1270176523.Ph for Steve Nelson please!

Dear Sir,
I now understand how the angular momenta can be coupled as vectors and thus why 
the neutron to proton decay is of mixed type. However, the first question that I 
asked was not to do with the antimony example, but rather to do with the 
statement 13 lines up the bottom on page 291, which quotes, 
'The coupling of S = 0 with l = 1 for the fermi decays gives total angular 
momentum of one unit carried by the beta decay, so that delta I = 0 or 1 (but 
not 0 to 0).'
Even taking into consideration the vector coupling of momenta, I don't see how 
if a decay must be forbidden i.e. orbital ang.momentum =1, can I = 0, without 
the spins changing. Also, the coupling of I should only give 1, right? as l-s 
and l+s both equal 1. 
this is my thought process: if L must be 1, then parity will change, so if we 
want I = 0, we will need that s=1, so that the L and S can cancel vectorially to 
give I=0 and at the same time the parity also changes, such as illustrated by 
the antimony example. However, If L must be 1, and we still want I=0 with s=1, 
then this is not possible since the total momentum change is just 1. 

Thanks!

Re: Follow up to ID: 1270176523.Ph for Steve Nelson please!