Isospin decomposition of K->ππ decay


by Einj
Tags: decay, decomposition, isospin, k>ππ
Einj
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#1
Nov19-12, 12:26 PM
P: 235
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?
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Bill_K
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#2
Nov19-12, 01:47 PM
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Instead of the second line, Wikipedia gives

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

Does that help?
Einj
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#3
Nov19-12, 02:08 PM
P: 235
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.

Bill_K
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#4
Nov19-12, 03:27 PM
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Isospin decomposition of K->ππ decay


Well, how about this, the Clebsch-Gordan coefficients give

|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.
Einj
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#5
Nov20-12, 02:08 AM
P: 235
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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