- #1

- 91

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For example, suppose I had a lagrangian that was ##L = \phi^\mu \phi_\mu## then I can see that that it has SO(n) symmetry in the following sense. Let ##R(\omega)## be a rotation (in the fundamental representation) then if I send ##\phi_\mu \mapsto R(\omega)_\mu^\nu \phi_\nu## the lagrangian remains invariant. Corresponding to this I get a representation acting on the states by ##R(\omega)_\mu^\nu \phi_\nu = U(\omega)^{-1} \phi_\mu U(\omega)##

Now I know that the ##R(\omega)_\mu^\nu## is in the fundamental so that is necessarily an irreducible representation. However can I somehow conclude that the ##U(\omega)## representation is irreducible as well?

P.S. I know that in general states and operators don't even need to have the same symmetry group. I'm more interested in whether irreducibility of one implies irreducibility of the other