- #1
hideelo
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Ok, so my question is "Does an irreducible representation acting on operators imply that the states also transform in an irreducible representation?" and what I mean by that is the following. If I have an operator transforming in an irreducible transformation of some group, I get a corresponding symmetry transformation on my states, is this representation acting on my states also irreducible?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
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