# Eigenspace is an irrep of the symmetry group

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.

• what is it supposed to be an irrep of? i.e. of which group? Commented Jan 22, 2017 at 22:56
• the symmetry group, the group of all all operators that commute with the observable. Commented Jan 22, 2017 at 22:58
• Frankly, this seems to me to be a use of a weasel phrase. "This result is always true, except when it isn't" (where the latter is then termed 'accidental' degeneracy). Commented Jan 23, 2017 at 21:42

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.
• to be frank, i have been struggling with this point for quite some time, and in fact many books on group theory for and by physicists do state this, and sort of 'prove' it by excluding accidental degeneracy. but i never understood their arguments. but the other answer seems to provide a much rigorous argument. Commented Jan 23, 2017 at 12:49
• @QMechanic Yeah, I know, I forgot to mention accidental degeneracy.
– udrv
Commented Jan 23, 2017 at 16:17

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.

• so the point is that if such an eigenspace is not an irrep, one can always construct another symmetry operator which is not included . right? Commented Jan 23, 2017 at 11:03
• Yes, with the caveat that there may still be accidental degeneracy, which is probably the point QMechanic intended to cover: if the eigenspace is not an irrep, either there is purely accidental degeneracy that isn't tied to any particular symmetry, or there is some hidden symmetry. And I eventually found another nice proof, in a very readable ref: cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter6.pdf, Sec.6.2, pgs.87-88.
– udrv
Commented Jan 23, 2017 at 16:14
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