Irreps of $SU(2)$ versus the irreps of $SU(3)$

Question

For the group $SU(2)$, the fundamental representation corresponds to $j=1/2$, for which the highest (eigen)value of $J_3$ is $+1/2$ and the lowest (eigen)value is $-1/2$. For $j=1$ representation, the possible values of $J_3$ are $(-1,0,1)$ and limiting values are $\pm 1$. For $j=2$, the possible values of $J_3$ are $(-2,-1,0,1,2)$ and limiting values are $\pm 2$, and so on. This follows neatly from $SU(2)$ algebra - for a fixed value of $j$, the possible values of $J_3$ are $-j$ to $+j$ in steps of unity (with highest and lowest values of $J_3$ are $+j$ and $-j$, respectively). Thus, the label $j$ of the Casimir operator $J^2$, completely determines all possible $(2j+1)$states in an irrep $j$ or in a $j$-multiplet.

Very nice!

Let us come to $SU(3)$. In this case, the fundamental representation corresponds to three states, indexed by the eigenvalues of $T_3,Y$, and these states, in terms of these labels are $(\pm\frac{1}{2},\frac{1}{3})$ and $\left(0,-\frac{2}{3}\right)$. This tells that the possible values of $T_3$ are $(-1/2,0,+1/2)$ (i.e., highest and lowest eigenvalues of $T_3$ are $+1/2$ and $-1/2$).

What determines the possible values of $T_3$ in the fundamental and in any other irrep of $SU(3)$? I have the same question about the eigenvalues of $Y$.

Well, we can consider that the basis states in the fundamental representation are $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$, and then by the action of $T_3$ and $Y$ on these simultaneous eigenstates we can find that the "coordinates" are indeed $(\pm\frac{1}{2},\frac{1}{3})$ and $\left(0,-\frac{2}{3}\right)$. But how to find/predict, like $SU(2)$, the possible values of $(T_3, Y)$ in an $(m,n)$-irrep of the $SU(3)$, using the algebra only? Any help will be greatly appreciated.

Answer 1

Yes. The strategy is to find the $SU(2)\otimes U(1)$ contents of a given $SU(3)$ irrep. Once you have this, you can just write down the weights of each $SU(2)$ piece and combined it with the $U(1)$ weight.

First, it is not so hot to think of $j$ as an eigenvalue of a Casimir. Although not incorrect, it is better to think of $j$ as the largest value $m_{max}$ for which $L_+\vert j m_{max}\rangle=0$. This pins down $m_{max}=j$. Although this is simple enough, it generalizes nicely in that labels of irreps are obtained from the eigenvalues of diagonal operators of the unique state killed by all raising operators.

In this (Dynkin) notation, $SU(3)$ irreps are labelled by two integers $\lambda,\mu$, and the labels are easier to work out that inverting the eigenvalues of Casimir operators, which are complicated functions of $\lambda$ and $\mu$.

Next, the choice of diagonal operators is not unique because there is more than one traceless hermitian diagonal $3\times 3$ matrix. In the fundamental representation, one can choose any linear combination of $$ \left(\begin{array}{ccc} 1&0&0\\ 0&-1&0\\ 0&0&0\end{array}\right)\quad \text{and} \left(\begin{array}{ccc} 0&0&0\\ 0&1&0\\ 0&0&-1\end{array}\right)\, . $$ It looks like your $Y$ is such a linear combination but most people outside of HEP use the above two.

Soo... Pick $\lambda$, $\mu$ and let $m_{13}=\lambda+\mu,m_{23}=\mu,m_{33}=0$. The SU(2) content of this $SU(3)$ irrep is obtained by first computing $m_{12}$ and $m_{22}$ so that they satisfy the "in-betweeness" condition of Gelfan'd and Zeitlin: $$ m_{13}\ge m_{12}\ge m_{23}\, ,\qquad m_{23}\ge m_{22}\ge m_{33}\, . $$ Thus, for the adjoint, with $(\lambda,\mu)=(1,1)$, we have $$ m_{13}=2\, ,\quad m_{23}=1\, ,\quad m_{33}=0 $$ so that the possible pairs $(m_{12},m_{22})$ are $$ (2,1),(2,0),(1,1),(1,0) $$ The list of $SU(2)$ irreps is now obtained by taking $\frac12(m_{12}-m_{22})$ and thus we have $J=\frac12,1,0,\frac12$. From this list you can extract the list of $T_3$ values. Following

Draayer JP, Akiyama Y. Wigner and Racah coefficients for SU 3. Journal of Mathematical Physics. 1973 Dec 1;14(12):1904-12.

the value of $Y$ is given by $$ Y=\frac{1}{3}(3 m_{12}+3m_{22}-2\lambda-4\mu) $$ so with $m_{12}$ and $m_{22}$ you can recover all values of $T_3$ and $Y$.