http://arxiv.org/abs/2005.02512
If new physics contains new, heavy strongly-interacting particles belonging to irreducible representations of SU(3) different from the adjoint or the (anti)fundamental, it is a non-trivial question to calculate what is the minimum number of quarks/antiquarks/gluons needed to form a color-singlet bound state (”hadron”) with the new particle. Here, I prove that for an SU(3) irreducible representation with Dynkin label $(p,q)$, the minimal number of quarks needed to form a product that includes the (0,0) representation is $2p+q$. I generalize this result to SU($N$), with $N>3$. I also calculate the minimal total number of quarks/antiquarks/gluons that, bound to a new particle in the $(p,q)$ representation, give a color-singlet state: $n_g=\lfloor (2p+q)/3 \rfloor$ gluons, $n_{\bar q}=\lfloor (2p+q-3n_g)/2 \rfloor$ antiquarks, and $n_q=2p+q-3n_g-2n_{\bar q}$ quarks (with the exception of the {\boldmath $\overline 6$}$\sim$(0,2) and of the {\boldmath $\overline{ 10}$}$\sim$(0,3), for which 2 and 3 quarks, respectively, are needed to form the most minimal color-less bound state). Finally, I show that the possible values of the electric charge $Q_H$ of the smallest hadron $H$ containing a new particle $X$ in the ($p,q$) representation of SU(3) and with electric charge $Q_X$ are $-(2p+q)/3\le Q_H-Q_X \le 2(2p+q)/3$.
S. Profumo
Thu, 7 May 20
25/62
Comments: 9 pages, 2 figures