$SU(3)$ irreducible representations with tensor method

Question

I am dealing with the tensor product representation of $SU(3)$ and I have some problems in understanding some decomposition.

1) Let's find the irreducible representation of $3\otimes\bar{3}$

we have that this representation trasforms like

$${T^\prime}^i_j=U^i_k {U^{\dagger}}^l_j T^k_l $$

hence I observe that $$Tr(T)=\delta^j_iT^i_j$$ is an invariant and so

$$T^i_j=\left(T^i_j-\frac{1}{3}\delta^j_iT^i_j\right)+\frac{1}{3}\delta^j_iT^i_j$$

allows me to write $$3\otimes\bar{3}=8\oplus1$$ Here comes my questions: I have heard that this $8$ representation is an "$8_{MA}$" where MA is for "mixed-antisymmetric". The meaning of "mixed-antisymmetric" shold be: "the tensor $\left(T^i_j-\frac{1}{3}\delta^j_iT^i_j\right)$ should be antisymmetric for an exchange of 2 particular indexes but not for a general exchange of 3 indexes". What does this mean? I see only 2 index in that tensor.

2) Consider this representation: $$3\otimes3\otimes3=3\otimes(6\oplus\bar{3})=3\otimes6_S\oplus3\otimes\bar{3}=3\otimes 6_S\oplus8_{MA}\oplus1$$

and now on my notes I have $$3\otimes6_S=10_S\oplus8_{MS}$$

Where "MS" is for "mixed symmetric": symmetric for an exchange of 2 particular indexes but not for a general exchange of 3 indexes.

I could not demonstrate this last decomposition using tensor method. I started noticeing that: $$3\otimes6_S=q^iS^{k,l}$$ where $S^{k,l}$ is a symmetric tensor But then I am not able to proceed in demonstrating the above decomposition (note: I would like to demonstrate this decomposition using only tensor properties, not Young tableaux). I tried to look on Georgi, Hamermesh, Zee and somewhere online but I have not found any good reference which explains well this representatin decomposition...

EDIT: the demonstration should not include the use of Young diagrams...my professor started the demonstration by writing $\epsilon_{\rho,i,k}q^i S^{k,l}=T'^l_\rho=8_{MS}$ and then stopped the demonstration.

Answer 1

Since this question looks like homework we will be somewhat brief. OP's notes are apparently describing the symmetry of the corresponding Young diagram for each $SU(3)$ irrep. Each box corresponds to an index. Roughly speaking, indices in same row (column) are symmetric (antisymmetric), respectively.

Examples:

1. A single box $[~~]$ corresponds to the fundamental irrep ${\bf 3}$.

2. Two boxes on top of each other $\begin{array}{c} [~~]\cr [~~] \end{array}$ is the anti-fundamental irrep $\bar{\bf 3}$ if we dualize with the help of the Levi-Civita symbol $\epsilon^{ijk}$. Here we adapt the sign convention $\epsilon^{123}=1=\epsilon_{123}$.

3. The tensor product ${\bf 3}\otimes{\bf 3}\cong\bar{\bf 3}\oplus{\bf 6}_S$ corresponds to $$ [~~]\quad\otimes\quad[a]\quad\cong\quad\begin{array}{c} [~~]\cr [a] \end{array}\quad\oplus\quad\begin{array}{rl} [~~]&[a] \end{array}$$
   or $T^{ij}=\epsilon^{ijk}A_k+S^{ij}$, where $A_k:=\frac{1}{2}T^{ij}\epsilon_{ijk}$.

4. The tensor product $\bar{\bf 3}\otimes{\bf 3}\cong{\bf 1}\oplus{\bf 8}_M$ corresponds to $$\begin{array}{c} [~~]\cr [~~] \end{array}\quad\otimes\quad[a]\quad\cong\quad\begin{array}{c} [~~]\cr [~~]\cr [a] \end{array}\quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~] \end{array}$$ or $T^i{}_j=S\delta^i_j+M^i{}_j$, where $S:=\frac{1}{3}T^i{}_i$, and ${\rm Tr}M=0$.

5. The tensor product ${\bf 6}_S\otimes{\bf 3}\cong{\bf 8}_M\oplus{\bf 10}_S$ corresponds to $$\begin{array}{rl} [~~]& [~~] \end{array}\quad\otimes\quad[a]\quad\cong\quad\begin{array}{rl} [~~]&[~~]\cr [a] \end{array}\quad\oplus\quad\begin{array}{rcl} [~~]& [~~] & [a] \end{array}$$ or $T^{ij,k}=\left\{M^{i}{}_{\ell}\epsilon^{\ell jk}+(i\leftrightarrow j)\right\} +S^{ijk}$, where $M^i{}_{\ell}:=\frac{1}{3}T^{ij,k}\epsilon_{jk\ell}$, and ${\rm Tr}M=0$.

References:

1. H. Georgi, Lie Algebras in Particle Physics, 1999, Section 13.2.

2. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 6.5.

Answer 2

Since the accepted answer treats the problem through the point of view of Young diagrams and I was having some problems with tensor methods myself, I believe it is worth to add the tensor approach in here.

Mixed Antisymmetric

Concerning the meaning of $MA$, recall that $SU(3)$ representations can be characterized in terms of "upstairs symmetrized indices" and "downstairs symmetrized indices" (see, for example, the beginning of Chapter V.2 of Zee's Group Theory in a Nutshell for Physicists). In other worlds, while $(𝑇^𝑖_𝑗−\frac{1}{3}𝛿^i_j 𝑇^k_k)$ does have only two indices, we could also raise the lower index with a Levi-Civita symbol, hence obtaining $(𝑇^𝑖_𝑗\epsilon^{jkl} −\frac{1}{3}\epsilon^{ikl}𝑇^k_k)$, which is now explicitly antisymmetric in two indices. The lesson here is that in $SU(3)$, thanks to the peculiar existence of a Levi-Civita symbol with three indices, we can write every tensor in terms of tensors with symmetrized indices, being these "upstairs" or "downstairs".

Computing $3 \otimes 6_S$

Let us then proceed to use tensor methods to write the tensor product of the representations. We have here a vector $q^i$ transforming in the fundamental representation, $3$, and a symmetric tensor $S^{jk}$ transforming in the $6_S$ representation. As you noticed, an object transforming in the $3 \otimes 6_S$ representation transforms as $q^i S^{jk}$. Since the representations of $SU(3)$ are characterized by symmetric tensors with indices upstairs or downstairs, our goal will be to separate the antissymetric parts in each index combination.

We can't antisymmetrize the indices of $S_{jk}$, since it is a symmetric tensor. We are left to antisymmetrize the index of $q^i$ with each of the indices of $S_{jk}$. Hence, we employ the Levi-Civita symbol to get

$$q^i S^{jk} = \frac{1}{3}\epsilon^{ijl}\epsilon_{lmn}q^mS^{nk} + \frac{1}{3}\epsilon^{ikl}\epsilon_{lmn}q^mS^{jl} + \left(q^i S^{jk} - \frac{1}{3}\epsilon^{ijl}\epsilon_{lmn}q^mS^{nk} - \frac{1}{3}\epsilon^{ikl}\epsilon_{lmn}q^mS^{jl}\right)$$

The coefficients are chosen so that the last term is fully symmetric in the three indices. Another possibility would be to pick the ansatz $S^{ijk} = q^i S^{jk} - a\epsilon^{ijl}\epsilon_{lmn}q^mS^{nk} - b\epsilon^{ikl}\epsilon_{lmn}q^mS^{jl}$ and impose the condition that $S^{ijk}$ is symmetric in all indices to get the correct coefficients.

Either, now the first term is symmetric under $j \leftrightarrow k$ and has no definite symmetry when $i$ is considered. The last term corresponds to a fully symmetric tensor, which has $10$ independent components, meaning it is a $10_S$. Hence, the first term has $8$ independent components and corresponds, as a consequence, to an $8_{MS}$, the $MS$ coming from mixed symmetry.

References

1. A. Zee, Group Theory in a Nutshell for Physicists (Princeton University Press: Princeton, 2016). Chapter V.2.