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  Does the Flavor symmetry forbid $uu\rightarrow cc,ss$?

+ 3 like - 0 dislike

This question comes from the reading of this paper https://arxiv.org/abs/1201.6510
They suppose a flavor symmetry group $G_F = U(3)_q\times U(3)_{d} \times U(2)_{d}$ which acts on the three LH quarks $q_L$, three RH quarks $u_R$ and the two lightest RH quarks $d_R$. 

Then, I suppose that $q_L = (q_L^1,q_L^2,q_L^3)$ transforms in the fundamental of $U(3)$, where the $q_L^i$ are the $SU(2)_W$ quarks doublets of the SM, for example $q_L^1= (u,d)_L$ and $q_L^2= (c,s)_L$.

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 $uu\rightarrow ss, cc$, do not arise from the four-quark operators of A.1.1 due to the flavor symmetry $G_F$

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 $\mathcal{O}_{qq}^{(1)} = (\bar{q}_L\gamma^\mu q_L)(\bar{q}_L\gamma^\mu q_L)$ which contains, for example, the interactions $$
\begin{align}(\bar{q}^1_L\gamma^\mu q^1_L)(\bar{q}^2_L\gamma^\mu q^2_L) &= \left((\bar{u}_L\gamma^\mu u_L) + (\bar{d}_L\gamma^\mu d_L)\right)\left((\bar{c}_L\gamma^\mu c_L) + (\bar{s}_L\gamma^\mu s_L)\right) \\&=(\bar{u}_L\gamma^\mu u_L)(\bar{c}_L\gamma^\mu c_L) +(\bar{u}_L\gamma^\mu u_L) (\bar{s}_L\gamma^\mu s_L) + (\bar{d}_L\gamma^\mu d_L)(\bar{c}_L\gamma^\mu c_L) + (\bar{d}_L\gamma^\mu d_L)(\bar{s}_L\gamma^\mu s_L)

So, I expect to take the processes $u_L u_L \rightarrow s_L s_L, c_Lc_L$ into account.

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

I guess \(u u \to ss, cc\) is not the same as \(u \bar u \to s \bar s, c \bar c\). The former changes flavor by 2 units, while the latter does not change flavor.

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