I'm not exactly sure how to phrase my question, but I'm trying to ask 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 $R(\omega)$ 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 dont even need to have the same symmetry group. I'm more interested in whether irreducibility of one implies irreducibility of the other

I'm not exactly sure how to phrase my question, but I'm trying to ask 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 $R(\omega)$ 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 dont even need to have the same symmetry group. I'm more interested in whether irreducibility of one implies irreducibility of the other