How do I find the corresponding tensor component v^ij of the six dimensional representation of SU(3) with dynkin label (2,0).
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The irreducible $SU(3)$ representations of Dynkin indices $(n,0)$ are the $n-$ symmetric tensor powers of the fundamental representation. Therefore let $e_1$, $e_2$, $e_3$ be an orthonormal basis of the fundamental representation space, then $ v_{ii} = e_i \otimes e_i $ $ v_{ij} = \frac{e_i \otimes e_j + e_j \otimes e_i }{\sqrt2}$, $i \ne j$ Identifying the fundamental representation basis with the weight vectors follows: $ e_1 = (1,0)$ $e_2 = (-1,1)$ $e_3 = (0,-1)$, The weights of the $(2,0)$ representation space can be obtained by inspection: $ v_{11} = (2,0)$ $ v_{22} = (-2,2)$ $ v_{33} = (0,-2)$ $ v_{12} = (0,1)$ $ v_{13} = (1,-1)$ $ v_{23} = (-1,0)$ There exist a lot of algorithms for the construction of group representations. One excellent reference is Slansky's seminal article: GROUP THEORY FOR UNIFIED MODEL BUILDING. |
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