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Penguin diagrams and CP violation

by Einj
Tags: diagrams, penguin, violation
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Einj
#1
Apr10-14, 11:02 AM
P: 305
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?
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andrien
#2
Apr10-14, 04:59 PM
P: 1,020
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.
Einj
#3
Apr10-14, 05:29 PM
P: 305
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

andrien
#4
Apr11-14, 08:31 AM
P: 1,020
Penguin diagrams and CP violation

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 ∑eq(q-q)V+/-A.Eletroweak penguins with different final states acquire different factors like eq=2/3 for q=u and eq=-1/3 for q=d.You can show now that gluonic ones only contributes to I=0 case.
Einj
#5
Apr11-14, 09:23 AM
P: 305
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?
andrien
#6
Apr11-14, 01:23 PM
P: 1,020
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.
Einj
#7
Apr11-14, 02:16 PM
P: 305
Got it! Thanks


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