# Penguin diagrams and CP violation

by Einj
Tags: diagrams, penguin, violation
 P: 328 Hi everyone. I have been studying CP violation in kaon systems. I would like to know is someone can explain why, to leading order, strong penguin diagrams (i.e. involving a gluon) only contribute to the $K\to (\pi\pi)_{I=0}$ amplitude, while the isospin 2 amplitude is given by the electro-weak penguins (i.e. involving a photon or a Z boson). Thank you very much Edit: I can add more information to the question. Take the gluon penguin diagram. It contains a $(\bar ds)$ current which is an SU(3) octet and another current which is a flavor singlet. Therefore, it transform as an $(8_L,1_R)$ representation of $SU(3)_L\times SU(3)_R$. In many books I've found that this implies that it can only carry a change in isospin $\Delta I=1/2$. Can anyone explain why?
 P: 328 I'm sorry to bother you but that's exactly what I can't show. Edit: I found the following properties on Donoghue's book. Could you tell me if I'm doing correctly? The action of the isospin rising and lowering operators on the quark fields are: $$I_+d=u,\;\; I_+\bar u=-\bar d,\;\;I_-u=d,\;\;I_-\bar d=-\bar u,$$ all the others are zero. Therefore, we have, in the case of QCD penguins: $$I_+(\bar u u+\bar d d)=(I_+\bar u)u+\bar u(I_+u)+(I_+\bar d)d+\bar d(I_+d)=\bar d u-\bar d u=0$$ and $$I_-(\bar u u+\bar d d)=(I_-\bar u)u+\bar u(I_-u)+(I_-\bar d)d+\bar d(I_-d)=-\bar u d+\bar u d=0.$$ Therefore, since both the lowering and rising operators give zero it must be an I=0 operator. Is that correct?