Selection rule of Kaon decay to pions in Peskin QFT page 611, he said the isospin change for 
$${K^0} \to {\pi ^ + }{\pi ^ - }$$ is 1/2,
while isospin change for 
$${K^ + } \to {\pi ^ + }{\pi ^0}$$ is 3/2. Why? 
If ${I_3}({K^ + }) = 1/2$,${I_3}({K^0}) =  - 1/2$, ${I_3}({\pi^0}) = 0$,${I_3}({\pi^ +}) = +1$,${I_3}({\pi^ -}) = -1$. Then the isospin change for above process obviously not match the above result. Where I goes wrong?
 A: The 3 pions can be considered as 3 states of the same particle, the isospin being used to label the 3 states. Since pions are bosons, the total wave function must be symmetric (Pauli principle). The total wave function is the (tensorial) product of space-wave function, spin wave-function and isospin wave-function. Spin wave-function is symmetric since pions have spin 0. For spatial wave-function, because of angular momentum conservation and knowing that the spin of the kaons is 0, we have 0 = L+S = L+0 => L=0. Hence, it gives a symmetric contribution  $(-1)^0$. So necessarily the isospin wave-function is symmetric. Thus, the total isospin must be an even number (look at the Clebsch Gordon coefficients for $1 \times 1$.) The addition of 2 isospins 1, gives 0,1 or 2. The 2 possible solutions are then 0 or 2. 
For the $\pi^+ \pi^0$ system it has $I_3 = 1$, and thus it excludes $I=0$, validating the only other possibility $I=2$. We conclude that for the transition $K^+ \to \pi^+ \pi^0$ the isospin change is 3/2.  
For the $\pi^+ \pi^-$ system, $I_3=0$ and thus both $I=0$ and $I=2$ are possible. I don't see an argument to definitely exclude the latter possibility.
A: Since the two pions are in an L=0 state, they cannot have isospin 1.
Therefore $\pi^+\pi^0$ must be an on isospin 2 state.
