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Isospin decomposition of K->ππ decay

  1. Nov 19, 2012 #1
    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:

    |\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)

    Now, my textbook says that we can decompose the decay amplitudes as follow:

    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}

    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?
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2


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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?
  4. Nov 19, 2012 #3
    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.
  5. Nov 19, 2012 #4


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    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.
  6. Nov 20, 2012 #5
    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 :biggrin:
    Thank you very much
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