
Which is the general representation $(n,m)$ of $SU(3)$ and how can I prove that $$C_{2}^{su(3)} = \frac{1}{3}\left(\frac{1}{3}(n^2+m^2+n\cdot m)+n\cdot m\right)\delta_{ij}~? $$

and the other question is why we need to define the cubic Casimir operator for $SU(3)$.
I know that the usual way is to use the basis $$\lbrace{T_{3},T_{8},F^{2}_{1},F_{2}^{1},F_{1}^{3},F_{3}^{1},F_{2}^{3},F_{3}^{2}\rbrace} $$
and then remembering that $$C_{2} = g^{ij}e_{i}e_{j}\Rightarrow C_{2}^{su(3)} = \frac{1}{3}\left(T_{3}^{2}+T_{2}^{2}+F_{1}^{2}F_{2}^{1}+F_{2}^{1}F_{1}^{2}+F_{1}^{3}F_{3}^{1}+F_{3}^{1}F^{1}_{3}+F_{2}^{3}F_{3}^{2}+F_{3}^{2}F_{2}^{3}\right) $$
but I can't see how to get the first expression. In $SU(2)$ we used its complex extension $sl(2,\mathbb{C})$ defining the new basis $$\lbrace{T_{+},T_{},T_{3}\rbrace} $$
where $T_{i}=ie_{i}$ and $T_{\pm} = \frac{1}{\sqrt{2}}(e_{i}\pm ie_{i})$ and working a bit we can find that $$C_{2}^{su(2)} = \frac{1}{2}(T_{+}T_{}+T_{}T_{+}+T_{3}^{2}) $$ and using the eigenvalues of the basis elements we finally prove that $$C_{2}^{su(2)}j>= \frac{1}{2}m(m+1)j> $$ I think that it's something analogous maybe? the pdf's that I've been reading on internet are so unclear 🙁
Any help is very grateful!
Best Answer
Take your expression for $C^2$ in terms of operators and reexpress your operators in terms of raising and lowering operators, then act with your new expression for $C^2$ on the highest weight state. The raising operators will kill the state, or if they appear on the left you can commute the raising operators to the right so they kill the highest weight state. You will be left with diagonal operators, from which you may easily extract the eigenvalue (which is nontrivial). Basically I'm saying in words to repeat the procedure for SU(2), i.e. indeed go to the complex entension.
In general irreps of $su(3)$ are labelled by two nonnegative integers ($m$ and $n$ in your case). For the irrep $(m,n)$, the highest weight is $m w_1+n w_2$, where $w_1$ and $w_2$ are the fundamental weights. The benefit of this notation is that it uniquely identifies the irrep, as opposed to using the dimension as a label (which is not unique).
As to why we need to define a cubic Casimir for SU(3): we don't "need" to define it; it is a statement of fact that there is such an invariant operator. It's use is not so common because we rarely have observables that are cubic in the generators, but its eigenvalue is used as additional label when the eigenvalue of $C^2$ is degenerate or not enough for labelling purposes.