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  Does the Flavor symmetry forbid uucc,ss?

+ 3 like - 0 dislike
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This question comes from the reading of this paper https://arxiv.org/abs/1201.6510
They suppose a flavor symmetry group GF=U(3)q×U(3)d×U(2)d which acts on the three LH quarks qL, three RH quarks uR and the two lightest RH quarks dR

Then, I suppose that qL=(q1L,q2L,q3L) transforms in the fundamental of U(3), where the qiL are the SU(2)W quarks doublets of the SM, for example q1L=(u,d)L and q2L=(c,s)L.

At the page 5, they state that

processes with other families in the final states but having u, d in the initial state,
such as uuss,cc, do not arise from the four-quark operators of A.1.1 due to the flavor symmetry GF

where the list of four-quark operators is given in the appendix.

My question is: why the flavor symmetry should forbid processes uu to the final states ss,cc?

For example, one operator they list is O(1)qq=(ˉqLγμqL)(ˉqLγμqL) which contains, for example, the interactions (ˉq1Lγμq1L)(ˉq2Lγμq2L)=((ˉuLγμuL)+(ˉdLγμdL))((ˉcLγμcL)+(ˉsLγμsL))=(ˉuLγμuL)(ˉcLγμcL)+(ˉuLγμuL)(ˉsLγμsL)+(ˉdLγμdL)(ˉcLγμcL)+(ˉdLγμdL)(ˉsLγμsL)

So, I expect to take the processes uLuLsLsL,cLcL into account.

asked Jun 14, 2016 in Theoretical Physics by apt45 (50 points) [ no revision ]
reshown Jul 27, 2016 by dimension10

I guess uuss,cc is not the same as uˉusˉs,cˉc. The former changes flavor by 2 units, while the latter does not change flavor.

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