Taking the tensor product of two spin 1 representations of $su(2)$ yields $$1 \otimes 1 = 0 \oplus 1 \oplus 2.$$
What changes if instead we take the symmetric tensor product $1 \odot 1$ of these two representations? I was thinking of using the tensor methods but I am unsure how they work for $su(2)$... (Also was hoping there was a simpler way to understand this).
The procedure is simple when coupling multiple copies of the fundamental representation of a unitary group as once can read of the (permutation) symmetry from the shape of the corresponding Young diagram (this is basically Schur-Weyl duality). The fundamental of $SU(2)$ is $s=1/2$ so this does not quite work here.
However there is a trick: the fundamental of SU(3) contains $\ell=1$ exactly once so the repeated coupling of $\ell=1$ irreps will break up in terms of permutation symmetry like the repeated coupling of fundamental of SU(3). Thus the symmetric part of two copies of the fundamental of SU(3), written in Dynkin notation as $(1,0)\otimes (1,0)=(2,0)\oplus (0,1)$ is $(2,0)$.
A basis for the irrep $(2,0)$ is actually the six 3D harmonic oscillator states with $n=2$ quanta, and from elementary quantum mechanics we know this contains angular momentum states with $\ell=2$ and $\ell=0$. (For completeness, the $(0,1)$ part is antisymmetric, is of dimension $3$ and contains $\ell=1$ states.) Thus using this trick one finds that the symmetric part of $1\otimes 1$ is $2\oplus 0$; the antisymmetric part contains $\ell=1$ since the antiquark $(0,1)$ branches to the $SO(3)$ irrep $\ell=1$. (It must have the same angular momentum content as its conjugate irrep $(1,0)$.)
This extends to multiple $n$-fold couplings of $\ell=1$: the symmetric part of the corresponding $SU(3)$ coupling will be the irrep $(n,0)$, which contains $\ell=n,n-2,n-4,\ldots, 1$ or $0$ depending if $n$ is even or odd.
Now there is also a purely dimensional argument that also works although there’s a long of counting and it can be applied with difficulty to more than two couplings. The product $1\otimes 1$ is of dimension $9$. The highest weight state $\vert \ell=2, m=2\rangle=\vert 1,1\rangle\vert 1,1\rangle$ is certainly symmetric so all $\ell=2$ states - there are $5$ of those - will be symmetric. This leaves us with $4$ states to account for. It’s not hard to see that $$ \vert 1,1\rangle \vert 1,-1\rangle - \vert 1,0\rangle \vert 1,0\rangle +\vert 1,-1\rangle \vert 1,1\rangle $$ is actually proportional to the $\ell=0$ state and clearly symmetric. The remaining states must be antisymmetric, and by elimination this only leaves the $\ell=1$ states.
Finally, there is another method which works for the coupling of two identical irreps: look for the symmetry of the Clebsch- Gordan coefficient under interchange of labels, i.e. compare $\langle \ell m_1;\ell m_2\vert JM\rangle$ and $\langle \ell m_2; \ell m_1\vert JM\rangle$. The phase obtained in interchanging $m_1$ and $m_2$ is $(-1)^J$ whenever $\ell$ is integer, so that states will be symmetric when $J$ is even and antisymmetric when $J$ is odd. Of course this agrees with previous method for case when $\ell=1$ and $J=0,1$ or $2$.
0 \oplus 2? $\endgroup$