# Isospin In Kaon Decay

The decay $K \to \pi \pi$ at zero-strong interaction level is calculated by considering the matrix element of the operator $Q_2 = (\bar{s}u)_{V-A} (\bar{u}d)_{V-A}$ for two kinds of processes: Final Isospin = 0, 2 as $<Q_2>_0$, $<Q_2>_2$. How do we incorporate these isospin states of the two final pions in the calculations of those matrix elements ?

In the context of Chiral Perturbation Theory (Low energy processes), we have strangeness changing Lagrangian that can also be used to calculate the above mentioned process ($\Delta S = 1$), this interaction Lagrangian is given by: $$L_{\Delta S = 1} = G_8 {L_\mu L^\mu}_{23} + h.c$$, where $L_\mu = i f^2 U\partial_\mu U^\dagger$ and $U = Exp[ i\sqrt{2} \Phi/f ]$

Where $\begin{equation} \Phi = \begin{bmatrix} \pi^0/\sqrt{2} & \pi^+ & K^+\\ \pi^- & -\pi^0/\sqrt{2} & K^0\\ K^- & \bar{K}^0 & 0 \\ \end{bmatrix} \end{equation} {}\\$

I have neglected the "$\eta_8$" meson.

In this picture, when we expand the Lagrangian up to 3-meson fields, that is terms that contain $K \pi \pi$ we obtain the terms necessary for tree level calculation of the decay mentioned above. Let's say one such term is of the form ${ \cal L } _1 = G_8 \partial_\mu K^0 \partial^\mu \pi^+ \pi^-$, I can simply calculate $$A_{K^0(k) \to \pi^+(p_1) \pi^-(p_2)} = < \pi^+ (p_1) \pi^- (p_2) | \partial_\mu K^0 \partial^\mu \pi^+ \pi^- | K^0 (k) >$$

But this doesn't consider the isospin of the final state ! How do we introduce the isospin ? Do we consider the fact that pions have isospin 1 and then we consider the addition of two isospin "1"s to get total isospin 0 and 2 states ? In that case how do I do that ? What will be the form of the state on which I apply the operator ${\cal L} _1$ let's say ? $$|\pi^+ \pi^->  = \alpha |pi^-> |pi^+> + \beta |pi^+> |pi^->$$ of this form ? Then how do I calculate these $\alpha$ and $\beta$ ? How do we add two isospins of value "1" ?

Any help will be highly appreciated.