Eigenspace is an irrep of the symmetry group

Question

consider an observable, say the Hamiltonian,and the symmetry group of it, with accidental degeneracy excluded. now decompose the state space into the direct sum of eigenspaces, namely each spanned by the set of eigenvectors with the same eigenvalue.

It is often stated (e.g. A. Zee, "group in a nut", p. 163-164), that such an eigenspace is an irrep. I understand it is necessarily a representation of the group, but why it is an irrep? Also is the converse true that an irrep of the group is necessarily such an eigenspace?

To clarify the question, consider the hamiltonian of a system which has only one symmetry, $SO(3)$. then it is stated that each of the subspaces of the state space spanned by eigenvectors with the same energy eigenvalue and total angular momentum forms an irrep.

Answer 1

One of the problems with the informal casual style of Ref. 1 is that it is difficult to extract precise statements. Ref. 1 makes several imprecise or even wrong statements because context and assumptions are lacking from the pertinent paragraph.

1. Let us for simplicity assume that the Hilbert space $V$ of the system is finite dimensional, and leave it to the reader to generalize to infinite-dimensional Hilbert spaces.

2. Let $H\in{\rm End}(V)$ be a linear diagonalizable operator on $V$.

3. Let $$V_{\lambda}~:=~{\rm ker} (H-\lambda {\bf 1}_V)~\subseteq~V$$ be an eigenspace for $H$; where $\lambda\in \mathbb{C}$.

4. Let $$G~:=~GL(V)\cap \underbrace{{\rm span}(H)^{\prime}}_{\text{commutant}} ~\subseteq~ {\rm End}(V)$$ be the set of invertible operators that commute with $H$.

5. Then it is easy to check that $G$ is a group and that $V_{\lambda}$ is a representation of $G$ (or of any of its subgroups). OP is right that $V_{\lambda}$ does not need to be an irreducible representation.

References:

1. A. Zee, Group Theory in a Nutshell for Physicists, 2016, p. 163-164.

Answer 2

An eigenspace of the Hamiltonian is not only a rep of the symmetry group G, but also a shared eigenspace for a complete set of observables/symmetry generators $\Omega$.

Suppose some eigenspace of the Hamiltonian is not one irrep of $G$, but a direct sum of two distinct irreps $A$ and $B$. If $\Pi_A$ and $\Pi_B$ are corresponding subspace projectors, it is always possible to define an observable $O = \lambda_A \Pi_A + \lambda_B \Pi_B$ whose eigenvalues distinguish between $A$ and $B$, and which necessarily commutes with the Hamiltonian, with the complete set $\Omega$, and with all transformations in $G$. But then $\Omega$ alone is no longer a complete set of observables/symmetry generators. The complete set must now include $O$ and the original eigenspace splits into distinct eigenspaces A and B, each an irrep of the augmented symmetry group, etc.