I'm studying the decay K->ππ and I have some doubts on the isospin decomposition. We know that the state [itex](\pi\pi)[/itex] can have total isospin 0 or 2. Now, if we remember that in the isospin representation we have [itex]|\pi^+\langle=|1,1\langle[/itex], [itex]|\pi^0\rangle=|1,0\rangle[/itex] and [itex]|\pi^-\rangle=|1,-1\rangle[/itex], then using Clebsch-Gordan coefficients we find:(adsbygoogle = window.adsbygoogle || []).push({});

\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 A_{0}and A_{2}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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Isospin decomposition of K->ππ decay

**Physics Forums | Science Articles, Homework Help, Discussion**