Abelian Symmetry groups and Degeneracy

Question

Why does an Abelian Symmetry group necessarily imply no degeneracy?

As an example, consider an operator $A$ such that $A^2 = I$ (essentially a representation of $\mathbb{Z}_2$) and a Hamiltonian $H$ such that $[H,A]=0$. Assume that this is the only symmetry. Now, $H$, $A$ have a simultaneous eigenbasis, and $A$ can be divided into a direct sum of irreps, each of dimension 1 (the answer here says that this is the reason for a lack of degeneracy).

However, if the different eigenstates have different eigenvalues of $A$, what prevents $H$ from having two eigenstates with the same eigenvalue? These could then be distinguished by the labels that $A$ provides.

Answer 1

Yes, it is true when assuming that both $H$ and $A$ have pure point spectrum, the case of $A$ with continuous part is a bit more complicated. (However I suspect that $A$ is both self-adjoint and unitary from your comment, so $A^2=I$ implies $\sigma(A) \subset \{\pm 1\}$, however it does not matter below.)

Consider an eigenspace $\cal{H}_E$ of $H$ with eigenvalue $\cal E$. It is invariant under $A$, so $A|_{\cal{H}_E}: \cal{H}_E \to \cal{H}_E$ is still selfadjoint and can be written into a diagonal form with respect to a basis of eigenvectors therein. If it admits $n>1$ eigenvectors in $\cal{H}_E $ (i.e. $\cal E$ is a degenerate energy level), it is easy to construct a self-adjoint operator $B : \cal{H}_E \to \cal{H}_E$ which is not diagonal with respect to the found basis of eigenvectors of $A|_{\cal H_E}$ and satisfies $B^2=I$, so that it does not commute with $A|_{\cal{H}_E}$ and every nontrivial function $f(A)$ of it. The self-adjoint operator $A'= P^\perp + PBP$, where $P$ is the orthogonal projector onto $\cal{H}_E$, is another self-adjoint operator commuting with $H$ which is not a function of $A$ and such that $A'^2=I$, thus $H$ would admit another symmetry against the hypothesis. Thus $\cal{H}_E$ has dimension $1$: no different eigenstates of $H$ with the same eigenvalue exist.

ADDENDUM. Regarding commutativity of $A'$ and $H$, if $P_e$ denotes the orthogonal projector onto the eigenspace of $H$ with eigenvalue $e$, we have $$A' H = (PBP+ P^\perp) H = \left(P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e\right) \sum_{e'} e' P_{e'}$$ $$=P_{\cal E} B P_{\cal E} \sum_{e'} e' P_{e'} + \sum_{e \neq {\cal E}} P_e\sum_{e} e' P_{e'}$$ $$ ={\cal E} P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} eP_e$$ $$= \left( \sum_{e'} e' P_{e'}\right) P_{\cal E} B P_{\cal E} + \left(\sum_{e} e' P_{e'}\right)\sum_{e \neq {\cal E}} P_e $$ $$= HA'\:.$$

Answer 2

I don't think the above argument protects you from "accidental degeneracies".

The main idea is that if each eigenvalue of $A$ has only one eigenvector associated with it, then there is no reason to expect any further symmetry in the Hamiltonian.

This does not mean that they cannot exist (I think), but they should be rare.

You however, but construction find counterexamples. Consider a spin-one half particle in vacuum. Here the two maximally commuting operators are $A=\sigma_z$ and $H=L^2\propto 1$. How ever though $A$ has eigenvalues $\pm 1$, the energy is degenerate.

Rather is think you should consider the reverse statement as the stronger one: If $A$ has irreducible representations of dimension larger than 1, then you are guaranteed to have degenerate eigenvalues of $H$ as well.