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There is a 6 dimensional multiplet belonging to an irreducible representation of a unitary group of rank less than 3. How does one check if the states $|i\rangle$ belong to spin 5/2 representation of $SU(2)$ or they belong to 6 dimensional representation of $SU(3)$?

For a given choice of basis states of the Hilbert space, one can, in principle, obtain a representation of any group, the dimension of the representation being the same as that of the Hilbert space. Just by knowing the basis states, I think it may not be possible to identify which group has been represented. Is my conclusion correct, or there is some point I am missing?

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If I know the explicit form of the states I would first apply the SU(2) generators to them. For example apply the Cartan generator $L_3$ and see whether you get the eigenvalues $-5/2, -3/2,...,+5/2$.

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  • $\begingroup$ Is the form the Cartan generators known unless the representation itself is known? I was confused regarding if the the Cartan generators remain invariant accross all the equivalent representations of the same group? $\endgroup$ – vnd Oct 29 '14 at 16:33
  • $\begingroup$ Well the explicit form of the algebra generators depend upon a chosen basis. They change their explicit matrix form under similarity transformation. So... what I told you to do will not work unless you're using the same basis for generators and states. Anyway, you definitely need some extra information do determine the algebra. If I give you only six states $|i\rangle$ and nothing else, you can't even say they belong to an irreducible representation. $\endgroup$ – Diracology Oct 29 '14 at 16:43
  • $\begingroup$ So we basically assume first that the states do indeed give a representation of SU(2) and check for eigenvalues with L_3. If not then we assume they give a representation of SU(3) and check eigenvalues with H_1 and H_2. $\endgroup$ – vnd Oct 29 '14 at 16:55
  • $\begingroup$ That's what I would try first. Although it might happen you don't get something useful because the states and the matrices are in different basis. $\endgroup$ – Diracology Oct 29 '14 at 16:59
  • $\begingroup$ Even in the finite group case, the same choice of basis states can give representations of identical dimensions for two non isomorphic groups. For example the binary basis (xy,xz,yz) gives 3 dimensional representations of the Tehtrahedral(T) as well as the Octahedral(O) groups. It was this point that was causing confusion. Thanks for the clarifications. $\endgroup$ – vnd Oct 29 '14 at 17:10

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