# Isospin decomposition of K->ππ decay

by Einj
Tags: decay, decomposition, isospin, k>ππ
 P: 153 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?
 Sci Advisor P: 3,447 Instead of the second line, Wikipedia gives |1,0> ⊗ |1,0> = √(2/3)|2,0> - √(1/3)|0,0> Does that help?
 P: 153 My fault. I wrote wrong in the post but I did the calculation with the correct formula. I will correct it right now. Still if you see the coefficients of the firt and second group of equation don't match.
P: 3,447

## Isospin decomposition of K->ππ decay

|0> = √(1/3)(|1,-1> + |-1,1> - |0,0>)
|2> = √(1/6)(|1,-1> + |-1,1>) + √(2/3) |0,0>

Let K0 decay into the state Φ ≡ B0|0> + B2|2>

= √(1/3)(|1,-1> + |-1,1>)(B0 + √(1/2)B2) - √(1/3)|0,0>(B0 - √2B2)

Rescaling A0 = √(1/3)B0, A2 = - √(1/3)B2 we get

Φ = (|1,-1> + |-1,1>)(A0 + √(1/2)A2) + |0,0>(A0 - √2A2), which reproduces the first two of the textbook equations.
 P: 153 Actually, if I haven't done wrong calculation, I think there are some signs that doesn't match. However I think the situation is a little more complicated as I read here: http://web.mit.edu/woodson/Public/Duarte_isospin.pdf The article talk about some fictitious particles (spurion) that must be introduced because the decay is a weak one, while the isospin is a good quantum number for strong interaction. I'll see Thank you very much

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