What is the Eigenvalue of $T^2$ ($SU(3)$ Casimir)? For example, in $SU(2)$, $\hat{S}^2|s,m_s>=\bar{h}^2 s(s+1)|s,m_s>$.
What about in $SU(3)$, $\hat{T}^2|T,m_3,m_8>=?|T,m_3,m_8>$ where $\hat{T}^2=\sum_i^8 T_iT_i $,  $T_i = \frac{\lambda_i}{2} $, $\lambda_i$ is $SU(3)$ generator.
 A: The finite dimensional unitary representations of  $\mathfrak{su}(3)$ have highest weights (eigenvalues of $\hat \lambda_3$ and $\hat \lambda_8$) that  are non-negative  integer linear combinations ${p \omega}_1+q {\omega}_2$ of the fundamental  weights
$$
{ \omega}_1 =\left(1, \frac{1}{\sqrt 3}\right),\quad {\omega}_2=\left(0, \frac{2}{\sqrt 3}\right).
$$
The  representation with highest weight ${\bf \Lambda}= p{\bf \omega}_1+q{\bf \omega}_2$ has dimension
$$
{\rm dim}(p,q)=(p+1)(q+1)(p+q+2)/2.
$$
and with $T_i= \lambda_i/2$ we have 
$$
C_2(p,q)=\sum_i T^2_i =\frac 13 (p^2+pq+q^2+3p+3q).
$$
I got this from: N.~Arisaka, On the unitary representations of SU(3) Prog. Theor. Phys. 47 1758-1781 (1972).
A: Like for $SU(2)$ the Casimirs of $SU(3)$ provide extra labels to the irreducible representations that can be useful in distinguishing irreps. that have the same dimension. Note I talk in the plural because for $SU(3)$ one can construct two Casimirs:
$$\hat{C}_{1} = \sum_{k= 1}^{8}T^{k}T^{k} \qquad \textrm{ and } \qquad \hat{C}_{2} = \sum_{klm = 1}^{8}d_{klm} T^{k}T^{l}T^{m}$$
where the $d_{klm}$ are the symmetric structure constants.
To answer your question the Casimir $\hat{C}_{1}$ acts on $SU(3)$ states labelled by weights $(n, m)$ to return the eigenvalue $\left(n^{2} + m^{2} + 3n + 3m + nm\right)/3$:
$$\hat{C}_{1} |(n, m)\rangle = \frac{1}{3}\left(n^{2} + m^{2} + 3n + 3m + nm\right)|(n, m)\rangle.$$
In particular for the states in the triplet, with labels $(1, 0)$ we get an eigenvalue $C_{1} = \frac{4}{3}$ (and it turns out that $C_{2} = \frac{10}{9}$). 
