Quick question about spin/parity J^P notation

[jeebs]

I'm reading something about parity violation where it was initially thought that the positive kaon decaying into 2 pions or 3 pions was actually due to two separate mesons, the tau+ and theta+.

it goes on about assigning them J^P values, where the parity is given by (-1)^J if it's spin is J. it says that the theta+ decayed to positive & neutral pions, so J^P = 0+, 1-, 2+...

i don't get this though.the pion is a quark-antiquark object, right? so it could have a spin of 0.5+0.5 = 1, or 0.5-0.5 = 0. therefore the possibilities for the total spins J of the two decay pions should be 1+1=2, or 1+0 = 1, or 0+0=0, right?
so J=0,1, or 2,
hence P = (-1)^0=1,
or (-1)^1=-1,
or (-1)^2 = 1.

So, this gives J^P = 0+, 1-, 2+ as it says it should.

but then there is the tau to 3 pion decay.
the possible spin values J are 1 + 1 + 1 = 3
1+1+0 = 2
1+0+0 = 1
0+0+0 = 0
so, the tau can have P = (-1)^3 = -1
or (-1)^2=1,
or (-1)^1=-1.
or (-1)^0=1

so surely it should have J^P = 0+, 1-, 2+, 3-, right?
Well it claims that tau has J^P values of 0-, 2-,... so I clearly am missing something. How can you have, say, a J^P = 0- if P=(-1)^0 = 1 ? shouldn't that give you J^P = 0+ ?
I don't even understand what this notation is meant to convey to me anyway.
Can anyone explain this?

 

[rkrsnan]

I don't understand all the details of this and I think your reasoning about the spin of pions is not correct. pions are spin zero particles(spins aligned in opposite direction). As I understand, for fermions, particles and antiparticles have opposite parity. so for a pion at ground state(they are not revolving around each other and hence no orbital angular momentum) parity is -1 x 1 = -1. total angular momentum is 0. so jp is 0-. (If you give an internal orbital angular momentum, total angular momentum is 1, and this orbital angular momentum gives an additional parity factor of -1. so for this excited pion jp value is 1+. But we don't need this for kaon decay analysis)
When you have two decayed pions, total spin is 0. Parity is -1 x -1 = +1. so jp = 0+
When you have three decayed pions, total spin is again 0. Parity is -1 x -1 x -1 = -1. so jp = 0-
For a kaon, spin is zero and parity is -1(because particle and antiparticle combination like a pion). so jp= 0-
so a kaon decaying into two pions violate parity.
Someone please correct me if i am wrong.

 

[jeebs]

AH yes, my reasoning about the spins of the pions was very wrong. I was thinking if they are a quark-antiquark pair, and each quark has a spin of 1/2, then they could align either both parallel or both antiparallel. to give a magnitude 1 or 0.