# Can the correspondence between irreps of $SU(N)$ and Young diagrams be interpreted as particle permutations only for the fundamental representation?

The decomposition of the $$N$$-fold tensor product of an irreducible representation of $$\mathfrak{su}(N)$$ can be conveniently found using Young diagrams and some combination rules.

If the irreducible representation in question is the fundamental one, it is represented by a single box Young diagram, and the various terms in the direct sum decomposition are $$N$$ boxes Young diagrams, corresponding to irreps of the symmetric group $$S_N$$.

This has a natural interpretation about how the system transforms under exchange of the particles: a system of $$N$$ particles that transform under the fundamental irrep of $$\mathfrak{su}(N)$$ can be decomposed into irreducible representations, and in each one of those representations the particles transform under exchange in some irrep of the symmetric group. For example two spin 1/2 decompose into the direct sum of a spin $$0$$ antisymmetric sector and a spin $$1$$ symmetric sector.

But what about non fundamental irreps? Consider two spin $$1$$ particles:

$$1\otimes 1 = 0\oplus 1\oplus 2$$

since $$1$$ is represented by a Young diagrams with two horizontal boxes, the various terms in the decomposition are $$4$$ boxes diagrams, corresponding to representations of $$S_4$$. It doesn't seem to me like we can interpret it as exchanging the two particles anymore. What is the physical interpretation then? Can we say anything about the symmetries under particles permutation in the various sectors?

The use of Young diagrams to simultaneously label the $$k$$-fold product of representations of $$U(n)$$ AND the permutation group $$S_k$$ is limited to the fundamental representation.
For $$k$$-fold product of representations of $$U(n)$$ other than then fundamental, one needs the concept of plethysm, which is a type of symmetrized product. Practical calculations are best done using Schur functions.
An example of this type of gymnastics is provided by the multiple product of $$\ell=1$$ states. In this case, one can "embed" this irrep of $$SO(3)$$ (angular momentum) inside the fundamental of $$SU(3)$$, and use the usual Schur-Weyl duality relation for the fundamental of $$SU(3)$$ and the $$SU(3)\downarrow SO(3)$$ to show that, for instance, the symmetric $$(\lambda,0)$$ irrep of $$SU(3)$$ contains symmetric states of $$L=\lambda,\lambda-2,\ldots 0$$ or $$1$$ depending if $$\lambda$$ is even or odd.
This would thus cover the $$L=0$$ and $$L=2$$ states of the example you give. The other Young diagram is for partition $$\{1,1\}$$, which is the antisymmetric (or alternating) irrep of $$S_2$$, corresponds to the $$SU(3)$$ irrep $$(0,1)$$ (in Dynkin notation) and contains only $$L=1$$ states. You can verify this is correct by also using the symmetry properties of the CG coefficients $$C_{1m_1;1m_2}^{LM}$$ which, upon interchange of $$m_1,m_2$$, pick up an overall phase $$(-1)^{L}$$,