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

• A
• hideelo
In summary, the irreducibility of a representation acting on operators does not necessarily imply the irreducibility of the representation acting on states. While the former may be finite-dimensional, irreducible, and unitary, the latter may be infinite-dimensional and non-unitary. This is especially true in relativistic quantum field theories, where the relevant representations are typically not irreducible. Additionally, the finite-dimensionality of the field space does not imply the finite-dimensionality of the Hilbert space.

#### 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

hideelo said:
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.

samalkhaiat said:
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?

hideelo said:
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.