- #1
Einj
- 470
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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:
\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?
\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?
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