Simple/elementary explanation for $\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$?

Question

I am preparing a talk on the Eightfold Way, and am attempting to explain the spectra of the light mesons/baryons via representation theory. It will be delivered to students who have never seen representation theory before. I understand the arrangement of particles can be explained by reps of $\mathrm{SU}(3)$, e.g. the light mesons ($\mathrm{u}$, $\mathrm{d}$, $\mathrm{s}$) correspond to the representation $\mathbf{3} \otimes \mathbf{\bar{3}}$ of $\mathrm{SU}(3)$, which I understand decomposes as $\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$.

Is there a simple and elementary way to derive this decomposition, i.e. without needing to know a lot of Lie groups/rep theory?

I've seen Young diagrams used as a tool, but still have not been able to understand how they work. If someone could give a self-contained explanation, or direct me to where I could find one, that would be great. I know group theory and surface-level stuff, and am willing to blindly accept some facts [e.g. that there exist representations $\mathbf{1}$, $\mathbf{3}$, $\mathbf{6}$, $\mathbf{8}$ of $\mathrm{SU}(3)$], but I'd like to give some motivation for the formula $\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$ without just handwaving it.

Same for the baryons: $\mathbf{3} \otimes \mathbf{3} \otimes \mathbf{3} = \mathbf{10} \oplus \mathbf{8} \oplus \mathbf{8} \oplus \mathbf{1}$.

Answer 1

$SU(3)$ has 2 fundamental representations, which I denote $t_a$ ($\bar{3}$) and $t^a$ ($3$). The transformation rules under an element $S$ of $SU(3)$ are $t^a\mapsto {S^a}_bt^b$ and $t_a\mapsto {S^H_a}^b t_b$, where $S^H$ denotes the conjugate transpose matrix. Being unitary, $S$ preserves ${\delta^a}_b$ and being special, it preserves the antisymmetric tensors $\epsilon^{abc}$ and $\epsilon_{abc}$. $3\otimes3$ then is the mixed tensor ${T^a}_b$, which breaks down into an irreducible 8 dimensional trace free component ( ${T^a}_b{\delta^b}_a=0$) and a one dimensional trace ${\delta^a}_b$.

$3\otimes3\otimes3$ becomes the tensor $T^{abc}$. This tensor breaks down into a fully symmetric tensor $T^{(abc)}$, which can be shown to be 10 dimensional (the number of ways 3 can be expressed as a sum of 3 non-negative integers), the one dimensional totally antisymmetric tensor, and two 8 dimensional tensors. The 8 dimensional tensors can be written as $T^{abc} = {T^a}_d\epsilon^{dbc}$ and $T^{abc} = {T^c}_d\epsilon^{dab}$, where, again, ${T^a}_b$ is trace free.

In general, irreducible representations can be written as ${T^{(abc...f)}}_{(ghi...p)}$, where "()" denotes symmetrization and $T$ is trace free.

Answer 2

The vector space of the representation $3$ is just $\mathbb C^3$, which you can write as column vectors $v$. The group then acts on it as $v'=Uv$. The vector space of the representation $\bar 3$ is the adjoint (Hermitian conjugate), which gives row vectors $w$, and the group acts on them as $w' = w U^\dagger$. Now the tensor product of a row vector and a column vector can be mapped onto the outer product $vw$, which is just a $3\times 3$ matrix. That is, you can represent $3\otimes\bar 3$ as $3\times 3$ matrices $A$, which the group operates on as $A'=UAU^\dagger$. That vector space is 9-dimensioal (because the matrix has 9 entries). On that matrix space you can define the Hilbert-Schmidt scalar product $\langle A,B\rangle = \operatorname{Tr}(A^\dagger B)$ (note that I'm using here the physicist's convention that the first argument of the scalar product is the antilinear one). It is easily checked that this is invariant under the group operation.

Now you immediately see that the identity matrix is invariant: $$UIU^\dagger = UU^\dagger = I$$ Thus the multiples of the identity matrix form a one-dimensional representation of $SU(3)$.

The vector space orthogonal to this is, obviously, $8$-dimensional and consists of the traceless matrices, as the orthogonality condition reads $0 = \operatorname{Tr}(I^\dagger A) = \operatorname{Tr} A$.

I don't see an immediately obvious way to see that this is irreducible, but it is at least plausible, given that there exists an irreducible representation of dimension 8.

I don't know a similarly simple way to see the decomposition of $3\otimes 3\otimes 3$, though.