# How to build tetra-quark mesons symmetry group?

I was thinking about the issue while reviewing my group theory notes. One can construct mesons with a nonet as an octet and a singlet, $SU(3)\otimes SU(3) = 8\oplus \bar{1}$. In a same way but for $qq$ interactions it is $6\oplus \bar{3}$. How about i.e, $(ud\bar{u}\bar{d})$ bound state or i.e, $(ud\bar{s}\bar{d})$ ?

So, suppose the meson is bounded as diquark-diantiquark pair. Then I guess sextet and triplet may remain to refer $qq$ diquark and then I try to figure out the di-antiquark part, I think that part should be a singlet but actually not quite sure.

Anyway, these are just a naive assumptions of mine. I'd be appreciated if someone help me to figure out the -thfold Way to exotic multi-quarks.

• Don't you just take the $\mathbf{3}\otimes\mathbf{3}\otimes\bar{\mathbf{3}}\otimes\bar{\mathbf{3}}$ and decompose it to get what the $ud\bar u\bar d$ might be? – ACuriousMind Jan 22 '16 at 13:49
• @ACuriousMind I think so but I can not figure out the multiplets of $(6\oplus \bar{3})\otimes \bar{3}\otimes \bar{3}$ – aQuestion Jan 22 '16 at 14:19
• It might be tedious, but that's just drawing the Young diagrams correctly. – ACuriousMind Jan 22 '16 at 14:22
• @ACuriousMind yeah, that's the point I got lost. – aQuestion Jan 22 '16 at 14:25

Your $SU(3)\otimes SU(3)={\bf 1}\oplus {\bf 8}$ above is a chimaeric typo from hell.

OK, I'll just give you the self-evident answers, but they would be meaningless junk numbers if you failed to reproduce them directly on the basis of your SU(3) text or the WP article which explains the rules and the Dynkin representation notation, D(p,q), which connects to the Young tableaux directly, but which I will nevertheless not use. I'll just use the dimensionality of the rep as its sole label, $D(1,0)={\bf 3}, ~D(2,0)={\bf 6},~ D(3,0)={\bf 10},~ D(1,1)={\bf 8}, ~D(2,2)={\bf 27}$, etc, and the numbers in parenthesis before each to denote the multiplicity of such multiplets in the C-G reduction.

So these are the answers you should be able to work out in less than an hour: $${\bf 3}\otimes {\bf 3}\otimes {\bf \bar{3}}\otimes {\bf \bar{3}} = ( {\bf 6}\oplus {\bf \bar{3}})\otimes ({\bf 3}\oplus {\bf \bar{6}})= ({\bf 1}\oplus {\bf 8}) \otimes ({\bf 1}\oplus {\bf 8}) ~.$$ So you have two options/pathways to get to your answers, and it is highly recommended you do both, for consistency.

Firstly, work out $({\bf 1}\oplus {\bf 8}) \otimes ({\bf 1}\oplus {\bf 8})= {\bf 1}\oplus {\bf 8} \oplus {\bf 8} \oplus ({\bf 8} \otimes {\bf 8})$ and ${\bf 8} \otimes {\bf 8}={\bf 1}\oplus {\bf 8} \oplus {\bf 8} \oplus {\bf 10} \oplus {\bf \overline{10} } \oplus {\bf 27},$ and check the arithmetic of the dimensionalities, so you have not left anything out.

Alternatively, work out, perhaps more simply, ${\bf 6} \otimes {\bf 3}= {\bf 8}\oplus {\bf 10}$, its obvious conjugate, ${\bf \bar{3}}\otimes {\bf 3}= {\bf 1} \oplus {\bf 8}$, and ${\bf 6}\otimes {\bf \bar{6}}= {\bf 1}\oplus {\bf 27} \oplus {\bf 8}$.

So, distributing terms in the above, the final answer is $${\bf 3}\otimes {\bf 3}\otimes {\bf \bar{3}}\otimes {\bf \bar{3}} = (2) {\bf 1} \oplus (4) {\bf 8} \oplus {\bf 10} \oplus {\bf \overline{10}} \oplus {\bf 27} ~.$$ Check conservation of states, as always--elt. school arithmetic.

Note the allowed isospins are then I=0, 1/2, 1, 3/2, 2, and the hyper charges, Y= 0, $\pm 1, ~\pm 2$, The last two hypercharches, $\pm 2$, are exotic for plain $\bar{q}q$ mesons, and so are I= 3/2, 2 !!!

Further note over 3/4 of all states are in self-conjugate representations.