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I am trying to understand the Georgi chapter of tensor methods in $SU(3)$ representations, and I don't know how to resolve the tensor product of 2 matrices in a 2 heavy quark + 2 light antiquark system. Namely:

Let the tetraquark system be: $QQ\bar q\bar{q}$

So, those $Q$ heavy quarks interact and can be understood in the $3\bigotimes 3$ representations for $SU(3)$, and $\bar q\bar q$ can be understood as the $\bar 3\bigotimes\bar 3=3\bigoplus\bar 6$ representations for $SU(3)$.

The interaction diagram for the $Q$ heavy quarks is shown in the following scheme, which is an analogy to the Feynman Diagram where a gluon is understood to be exchanged between upper and lower arrows:

$i'\xrightarrow{(T_{a})^{i'}_{i}}i , {}^iQ$

$j'\xrightarrow{(T_{a})^{j'}_{j}}j , {}^jQ$

So, as usuall we consider the elements of $3$ representation as $u^i$, and therefore the elements of $\bar 3$ representation as $v_{j}$.

I must understand how does the following expression transform:

$(T_{a})^{i'}_{i}(T_{a})^{j'}_{j}\cdot(1/2)\cdot (w^iv^j+w^jv^i)=\Xi\cdot(1/2)\cdot(w^{i'}v^{j'}+w^{j'}v^{i'})$

In other words, how can one give the correct value of $\Xi$?

Note1: $T_a$ is understood to be the $a-ith$ generator of the associated Lie Algebra, which come from the definition of the Gell-Mann Matrices.

$(T_{a})^{i}_{j}\propto(\lambda)_{ij}$, in matrix notation.

Note2: I understand the Young Tableaux and they are not intended to be used here, so please don't try to explain this situation through the tables.

Also requesting information about how to understand the tensor products in these cases, since I have been checking many sources but the definitions seem not clear enough.

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    $\begingroup$ If your problem is solved, it would be helpful to future searchers for you to write an answer that you would have found helpful when you asked this question. $\endgroup$
    – rob
    Mar 29, 2016 at 23:09

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Apologies for evincing magisterial cluelessness about what your diagrams represent and what you'd want to achieve, but I'd array the standard facts on tetraquarks avoiding Young diagrams, although they are self evident in the Dynkin labelling, which I also give, next to the tensor labelling. They may be useful to what you appear to be after--but I can't tell. $$ {\bf 3}\otimes {\bf 3}\otimes {\bf \bar{3}}\otimes {\bf \bar{3}} = (2) {\bf 1} \oplus (4) {\bf 8} \oplus {\bf 10} \oplus {\bf \overline{10}} \oplus {\bf 27} ~. $$ Here, the parentheses in front of the rep label denote its multiplicity in the C-G reduction.

$$D(1,0)={\bf 3}=\xi_j, ~ D(0,1)={\bf \bar{3}}=\xi^j; \qquad D(2,0)={\bf 6}=\xi_{jk},~ D(0,2)={\bf \bar{6}}=\xi^{jk}~; $$ $$D(3,0)={\bf 10}=\xi_{jkl},~ D(0,3)={\bf \bar{10}}=\xi^{jkl} ; \qquad D(1,1)={\bf 8}=\xi_j^k, ~D(2,2)={\bf 27}=\xi_{jk}^{lm}.$$

(I do not have, nor would I waste any time on the text you seem to be trapped in. Have you tried Iachello? Cahn? Lichtenberg? Wu-Ki Tung? Wybourne? Carruthers? Gourdin? Coleman's "Aspects of symmetry"?)

Edit: In any case, from the Fierz identity of the Gell-Mann matrices I mentioned in my comment below (Okun's appendix; a straightforward consequence of their completeness when supplemented by the identity), the dot representing summation over the 8 adjoint indices a, $$ {\bf \lambda}^k _i \cdot {\bf \lambda}^l _j~ (w^i v^j+w^j v^i) =\frac{4}{3} (w^k v^l+w^l v^k)~ , $$ and analogously for the antisymmetrization of w and v, mutatis mutandis: $$ {\bf \lambda}^k _i \cdot {\bf \lambda}^l _j~ (w^i v^j-w^j v^i) =-\frac{8}{3} (w^k v^l-w^l v^k) ~ . $$

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  • $\begingroup$ Thanks for your input, but this is not exactly what I need. I want to obtain the color factors by the transformation of the casimir (in the case of singlet), for a given representation, by using only tensorial algebra and the definitions in SU(N) chapter of Howard Georgi $\endgroup$
    – Jackson
    May 10, 2016 at 17:30
  • $\begingroup$ Color? OK: one can pretend these reps are of color. But Casimir? I assume just the quadratic? What you wrote is not the Casimir, as no matrix indices are contracted. I am having trouble second guessing what it is you write down and I have no access to, nor do I wish any, of that book. Sorry.... $\endgroup$ May 10, 2016 at 18:37
  • $\begingroup$ Well, it is totally obvious that I'm referring to color, because heavy quarks are not part of flavor symmetry of SU(3). I supposed this was obvious enough, but anyways, I found the solution by approaching a heavy meson-meson scattering with non relativistic conditions. Casimir only refers to the singlet, since it's the only representation that commutes over all SU(3) generators. But I don't only need the singlet transformations $\endgroup$
    – Jackson
    May 10, 2016 at 18:41
  • $\begingroup$ ? I still don't understand what you are trying to say--of course the diagram is completely meaningless to me. What solution? Could you possibly state, clearly, in exact words, what you are seeking? The eigenvalue of the quadratic casimir is 0 for the singlet, 4/3 for the triplet, and 15/3 for the sextet. But you never wrote a casimir anywhere.... ??? $\endgroup$ May 10, 2016 at 18:53
  • $\begingroup$ The transfer matrix operator is only a proper state of the singlet. The rest of states result in a mixed representation without a color factor. $\endgroup$
    – Jackson
    May 10, 2016 at 18:57

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