Multispin Casimir operators

Question

If we add together $N$-number of half spins, is it possible to tell the structure of the multiplets of the total spin states immediately? Is there a formula which quickly tells the structure of the eigenstates of the Casimir operator? For example, if we add together two half spins, there is a singlet state and a triplet state with total spins 0 and 1, respectively.

Answer 1

Yes it is possible. This is a result that uses Schur-Weyl duality. The n-fold tensor product of spin-1/2 states decomposes as the sum of irreps of the symmetric group $S_n$ of n particles, and the multiplicity of the spin-j irrep is the multiplicity of one of the $S_n$ irrep.

To give an example, suppose $n=5$. Write down all the partitions of $5$ having at most two parts (this is because we are dealing with the group $SU(2)$; for $SU(N)$ you would keep partitions with at most $N$ parts). These partitions are $(5,0),(4,1)$ and $(3,2)$. The possible values of $j$ corresponding to the partition $(\lambda,\mu)$ is $j=\frac12 (\lambda-\mu)$ so here we have $j=5/2,3/2$ and $1/2$.

There remains to determine how many times each irrep occurs. For this, you use the dimension of the corresponding $S_5$ irrep. The dimensions are: \begin{array}{cc} (5,0)&1\\ (4,1)&4\\ (3,2)&5\, . \end{array} These dimension rules are best computed using hook length formula. Thus the irrep with $j=5/2$ occurs once, the irrep with $j=3/2$ occurs $4$ times and the irrep with $j=1/2$ occurs $5$ times. You can verify the dimensionality: you have $2^5=32$ product states and you have $6+4\times 4+5\times 2=32$ states when decomposed in irreps.

This also works for $n$-fold tensor products of the fundamental of $SU(N)$ (well...$GL(N)$ really). Thus, for the $5$-fold product of the $(1,0)$ of $SU(3)$ (sometimes called the $\boldsymbol{3}$), we have now the partitions $(5,0,0),(4,1,0),(3,2,0),(3,1,1),(2,2,1)$ and the Dynkin labels for the partitions $(\lambda_1,\lambda_2,\lambda_3)$ are $(\lambda_1-\lambda_2,\lambda_2-\lambda_3)$. Thus the irreps $\{(5,0),(3,1),(1,2),(2,0),(0,1)\}$ occur $\{1,4,5,6,5\}$ times respectively.

If you are taking repeated tensor products of a representation of $SU(r)$ that is not the fundamental, then you have to use plethysms, which amounts to embedding your representation (of - say - dimension $q$) inside $SU(q)$, decompose the tensor product of $SU(q)$ using Schur-Weyl, and undo the embedding to come back to your original $SU(r)$ irreps. The only I know of doing this uses Schur functions.