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 [itex]K\to (\pi\pi)_{I=0}[/itex] 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 [itex](\bar ds)[/itex] current which is an SU(3) octet and another current which is a flavor singlet. Therefore, it transform as an [itex](8_L,1_R)[/itex] representation of [itex]SU(3)_L\times SU(3)_R[/itex]. In many books I've found that this implies that it can only carry a change in isospin [itex]\Delta I=1/2[/itex]. Can anyone explain why?
Gluonic penguin operators do not differentiate between a final state uu^{-} and final state dd^{-},so they only contribute to final states of zero isospin.Those electroweak contribute for both.And that ΔI=1/2 selection rule has it's origin in nonperturbative hadronic physics,and I don't think there is any good explanation for it.It is rather an experimental data to fit the small value of I=2 to I=0.
Thank you very much for the answer! Could you explain a little bit further why they don't differential between the two final states and why this means that they only contribute to I=0 final states? Thank you very much
The gluonic penguin operators contain factor like ∑(q^{-}q)_{V+/-A},Which contribute equally to uu^{-} and dd^{-},while the electroweak part brings in factors like ∑e_{q}(q^{-}q)_{V+/-A}.Eletroweak penguins with different final states acquire different factors like e_{q}=2/3 for q=u and e_{q}=-1/3 for q=d.You can show now that gluonic ones only contributes to I=0 case.
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?
You can use ladder operators,but see simply that quark and antiquark pair can have isospin of 0 or 1,and the 1 case is already ruled out as per symmetry requirement.The state symmetric in interchange of u and d is indeed isospin zero state.