I am looking at the tensor product of $n$ spin halfs (fundamental of $SU(2)$): \begin{equation} \left(\frac{1}{2}\right)^n = \frac{1}{2} \times \frac{1}{2} \times ... \times \frac{1}{2} = \frac{n}{2} + ... \end{equation} For example, for $n = 3$ \begin{equation} \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \times \left(1 + 0\right) = \left(\frac{3}{2}\right)_1 + \left(\frac{1}{2}\right)_1 + \left( \frac{1}{2} \right)_0 \end{equation} where lower indexes on the right hand side specify from which tensor product representation came from. My question is how do I read off if a certain term on the right is a symmetric, antisymmetric or a mixed symmetry representation of $SU(2)$?

I came up with two ways of approaching the problem.

  1. First is Young Tableaux. As far as I understand rows of Young Tableaux correspond to symmetric indexes while columns to antisymmetric ones. So, for example, a diagram with just one row (like one that corresponds to the highest spin $\frac{3}{2}$ from the example - three boxes in a row) describes a totally symmetric representation. A diagram that has one column (maximum 2 boxes since we are dealing with $SU(2)$) describes totally antisymmetric representation. Everything that has both rows and columns is a mixed symmetry representation. Following this logic it looks like only the highest representation in the product can be totally symmetric, singlet is antisymmetric and everything in between has mixed symmetry. So, for the example \begin{equation} \left(\frac{3}{2}\right)_1: \text{symmetric rep}\\ \left(\frac{1}{2}\right)_{0,1}: \text{mixed symmetry reps}\\ \end{equation} Is this is a correct logic?

  2. Another way to think about this is the following. Let's build representation the way we do in QM starting from the highest one, when all the spins are up. Then all components of the highest multiplet can be generated by applying lowering operators to the highest state. So, all the relative signs in the expressions are "+" and thus the highest representation is always symmetric as 1. also suggests. Then all the other representations must be constructed by writing down expressions orthogonal to components of the highest multiplet, so they will necessary get some "-" signs and as a result must be either antisymmetric or have mixed symmetry, but can not be totally symmetric.

Is this correct logic? Is it true that only the highest representation in the product can be symmetric and all the others have mixed symmetry or antisymmetric?


1 Answer 1


You need to use Schur-Weyl duality, i.e. look at the $n$-box Young diagram, obtain the value of $J$, and they read off the permutation symmetry from the diagram. To be specific, the partition $(\lambda_1,\lambda_2)$ corresponds to $J=\frac{1}{2}(\lambda_1-\lambda_2)$ (for $SU(2)$ the diagrams have at most two rows), and states of these irreps (there will be more than $1$) will also carry a representation of $S_n$ of this type (usually mixed symmetry).

So... you can only a symmetric irrep when the diagram contains a single row (i.e. $\lambda_2=0$) and all the other ones will be of mixed symmetry except for $(1,1)$, which contains a single column and which is fully antisymmetric. Note that a rectangular diagram like $(n/2,n/2)$ will not be fully antisymmetric as the symmetrizer acts non-trivially on any valid tableau of this shape.

Thus, yes it is tru that only the ``stretched" representation corresponding to the tableau $(n,0)$ is symmetric and all the others have mixed symmetry.

  • $\begingroup$ Could you please clarify how to read the permutation symmetry from the diagram? I see that formula $J = \frac{1}{2} (\lambda_1 - \lambda_2)$ works but I do not see how it helps to make a conclusion about the symmetry of the representation. $\endgroup$
    – noob
    Commented Aug 7, 2020 at 19:00
  • $\begingroup$ For the partition $(\lambda_1,\lambda_2)$, the $S_n$ representation is just that corresponding to this partition. Basis states are constructed - for example - using Young symmetrizers and will contain products of both symmetric and antisymmetric operators as there is more than one row or one column. The mixed symmetry under permutation is uniquely determined from the irrep of $S_n$ and thus the Young diagram. $\endgroup$ Commented Aug 8, 2020 at 17:36

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