# Isospin decomposition of K->ππ decay

by Einj
Tags: decay, decomposition, isospin, k>ππ
 P: 321 I'm studying the decay K->ππ and I have some doubts on the isospin decomposition. We know that the state $(\pi\pi)$ can have total isospin 0 or 2. Now, if we remember that in the isospin representation we have $|\pi^+\langle=|1,1\langle$, $|\pi^0\rangle=|1,0\rangle$ and $|\pi^-\rangle=|1,-1\rangle$, then using Clebsch-Gordan coefficients we find: \begin{eqnarray} |\pi^+\pi^-\rangle=\frac{1}{\sqrt{6}}|2,0\rangle+\frac{1}{ \sqrt{2}}|1,0\rangle+\frac{1}{\sqrt{3}}|0,0\rangle \\ |\pi^0\pi^0\rangle=\sqrt{\frac{2}{3}}|2,0\rangle - \frac{1}{\sqrt{3}}|0,0\rangle \\ |\pi^+\pi^0\rangle=\frac{1}{\sqrt{2}}(|2,1\rangle + |1,1\rangle) \end{eqnarray} Now, my textbook says that we can decompose the decay amplitudes as follow: \begin{eqnarray} A_{K^0\rightarrow \pi^+\pi^-}=A_0e^{i\delta_0}+\frac{A_2}{\sqrt{2}}e^{i\delta_2} \\ A_{K^0\rightarrow \pi^0\pi^0}=A_0e^{i\delta_0}-\sqrt{2}e^{i\delta_2} \\ A_{K^+\rightarrow \pi^+\pi^0}=\frac{3}{2}A_2e^{i\delta_2} \end{eqnarray} where A0 and A2 are the aplitude referred to the final state with I=0,2. The problem is: why the decay amplitudes don't present the same coefficient as in the Clebsch-Gordan decomposition?