Does an irreducible representation acting on operators imply....

[hideelo]

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

 

[samalkhaiat]

can I somehow conclude that the $U(\omega)$ representation is irreducible as well?

No. Irreducibility of $R$ does not imply that of $U$. In fact, in relativistic QFT: $$\begin{align*}U : & \ T(4) \rtimes SL(2, \mathbb{C}) \to U( \mathcal{H}) \\ & \ \ \ \ \ \ ( a , A ) \mapsto U( a , A) , \end{align*}$$ the representation $U(a,A)$ is faithful, unitary and infinite-dimensional but not irreducible. While the representation $$D : SL(2, \mathbb{C}) \to GL( V^{(j_{1} , j_{2})})$$ is non-unitary, finite-dimensional and irreducible.

 

[hideelo]

No. Irreducibility of $R$ does not imply that of $U$. In fact, in relativistic QFT: $$\begin{align*}U : & \ T(4) \rtimes SL(2, \mathbb{C}) \to U( \mathcal{H}) \\ & \ \ \ \ \ \ ( a , A ) \mapsto U( a , A) , \end{align*}$$ the representation $U(a,A)$ is faithful, unitary and infinite-dimensional but not irreducible. While the representation $$D : SL(2, \mathbb{C}) \to GL( V^{(j_{1} , j_{2})})$$ 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 $U(\omega)$ representation would not imply irreducibility of the $R(\omega)$ representation. Am I understanding you correctly?

 

[samalkhaiat]

Am I understanding you correctly?

No.
Is $U( \omega )$ 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 $SO(n)$ 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) $$U^{\dagger}(g) \varphi_{a} U(g) = D_{a}{}^{b}(g) \varphi_{b} , \ \ a = 1,2, \cdots , n$$
$D : G \to GL (V^{n})$ is a finite-dimensional irreducible representation of the “symmetry” group $G$ (since we always take our fields to be irreducible) and $U : G \to U( \mathcal{H})$ is the corresponding (mostly infinite-dimensional) unitary representation of $G$ in the Hilbert space $\mathcal{H}$.
So, the irreducibility of $D$ does not imply that $U$ is also irreducible(in QFT, the relevant $U$'s are not irreducible). Also, the field space $V^{n}$ is a finite-dimensional vector space, this does not mean that the Hilbert space $\mathcal{H}$ is also finite-dimensional.