Does an irreducible representation acting on operators imply...

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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 dont even need to have the same symmetry group. I'm more interested in whether irreducibility of one implies irreducibility of the other
 

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samalkhaiat
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can I somehow conclude that the ##U(\omega)## representation is irreducible as well?
No. Irreducibility of [itex]R[/itex] does not imply that of [itex]U[/itex]. In fact, in relativistic QFT: [tex]\begin{align*}U : & \ T(4) \rtimes SL(2, \mathbb{C}) \to U( \mathcal{H}) \\ & \ \ \ \ \ \ ( a , A ) \mapsto U( a , A) , \end{align*}[/tex] the representation [itex]U(a,A)[/itex] is faithful, unitary and infinite-dimensional but not irreducible. While the representation [tex]D : SL(2, \mathbb{C}) \to GL( V^{(j_{1} , j_{2})})[/tex] is non-unitary, finite-dimensional and irreducible.
 
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No. Irreducibility of [itex]R[/itex] does not imply that of [itex]U[/itex]. In fact, in relativistic QFT: [tex]\begin{align*}U : & \ T(4) \rtimes SL(2, \mathbb{C}) \to U( \mathcal{H}) \\ & \ \ \ \ \ \ ( a , A ) \mapsto U( a , A) , \end{align*}[/tex] the representation [itex]U(a,A)[/itex] is faithful, unitary and infinite-dimensional but not irreducible. While the representation [tex]D : SL(2, \mathbb{C}) \to GL( V^{(j_{1} , j_{2})})[/tex] is non-unitary, finite-dimensional and irreducible.
I think you're saying that the implication does not go the other way. i.e. in my example irreducibility of the [itex]U(\omega)[/itex] representation would not imply irreducibility of the [itex]R(\omega)[/itex] representation. Am I understanding you correctly?
 
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samalkhaiat
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Am I understanding you correctly?
No.
Is [itex]U( \omega )[/itex] in your example finite-dimensional or (the important) infinite-dimensional unitary representation? The relevant “Hilbert” spaces in QFT’s are infinite-dimensional.
For some reason you considered [itex]SO(n)[/itex] which is a compact group. A compact group has, among others, also finite-dimensional, irreducible, unitary representations; however, it does not have infinite-dimensional, irreducible unitary representation.
In the transformation law (of a relativistic field theory) [tex]U^{\dagger}(g) \varphi_{a} U(g) = D_{a}{}^{b}(g) \varphi_{b} , \ \ a = 1,2, \cdots , n[/tex]
[itex]D : G \to GL (V^{n})[/itex] is a finite-dimensional irreducible representation of the “symmetry” group [itex]G[/itex] (since we always take our fields to be irreducible) and [itex]U : G \to U( \mathcal{H})[/itex] is the corresponding (mostly infinite-dimensional) unitary representation of [itex]G[/itex] in the Hilbert space [itex]\mathcal{H}[/itex].
So, the irreducibility of [itex]D[/itex] does not imply that [itex]U[/itex] is also irreducible(in QFT, the relevant [itex]U[/itex]'s are not irreducible). Also, the field space [itex]V^{n}[/itex] is a finite-dimensional vector space, this does not mean that the Hilbert space [itex]\mathcal{H}[/itex] is also finite-dimensional.
 

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