Decay and parity

1. Jan 7, 2009

erwinscat

Hello all ! I was trying to solve the following problem : you have a particle with spin J=1 and unknown parity that decays into 2 indentical particals of spin = 1/2

We want to know how much the angular momentum of the final state is and also the final total spin.

Now, my reasonment was the following.

Stot_final = either 0 or 1 since it is the vector sum of the spin of both identical final particles.

Knowing that |L-S|<=J<= L+S we have to cases but I assumed that J=1 is conserved in the decay but it wasn't told so ...

we would have

for S=0 => L= 1
for S=1 => L= (0,1,2)

knowing that the final particles are indentical they must have a antisymmetric wave function due to Fermi Dirac...and so L must be odd . => L=1

but we still have to cases (S,L)= (0,1) or (1,1) if J=1 was to be conserved I'd say th only possible is (0,1) but the solutions say the only possibile solution knowing that J=1 is (1,1) ..

Conclusion, I don't know where I'm mistaking !

Any help would be appreciated ! Maybe I shouldn't assume that J was conserved...

Erwin

2. Jan 7, 2009

clem

J is conserved and equals 1.
But,
"knowing that the final particles are indentical they must have a antisymmetric wave function due to Fermi Dirac...and so L must be odd . => L=1"
is wrong.
The spinXorbital wave function must be antisymmetric.
This means that S=0 goes with L=0 or 2, neither of which can give J=1.
Therefor S=1, and L=1, with 1+1=1.

3. Jan 8, 2009

erwinscat

Hi and thanks a lot for the answer !

There is still something I don't get .. how do you get L=0 or L=2 with S=0 and more how do you get J=1 = L+S = 1+1 ?? if L=1 and S= 1 shouldn't you get in this case 0 or 2 . (L-S or L+S)?

Isn't this correct :

|L-S|<= 1<= L+S

if S= 0 then you have L=1
if S=1 then you have L= 0,1,2 ?

I guess not since you just said so ...but why isn't that so , where is my mistake ?