4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$A' H = (PBP+ P^\perp) H = \left(P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} P_e\right) \sum_{e'} e' P_{e'}$$ $$=P_{\cal E} B P_{\cal E} \sum_{e'} e' P_{e'} + \sum_{e \neq {\cal E}} P_e\sum_{e} e' P_{e'}$$ $$ ={\cal E} P_{\cal E} B P_{\cal E} + \sum_{e \neq {\cal E}} eP_e$$ $$= \left( \sum_{e'} e' P_{e'}\right) P_{\cal E} B P_{\cal E} + \left(\sum_{e} e' P_{e'}\right)\sum_{e \neq {\cal E}} P_e $$ $$= HA'\:.$$

$\endgroup$
  • $\begingroup$ I am unable to see why $A'$ is self adjoint, or should the definition include $PBP$ instead of just $BP$ (or is $PB=BP=B$)? $\endgroup$ – Primo-uomo Jul 3 '17 at 15:02
  • $\begingroup$ Right, I forgot a factor $P$... corrected. Actually, it was not necessary since $Ran(B) \subset \cal H_E$, it holds $PB=B$... $\endgroup$ – Valter Moretti Jul 3 '17 at 15:24
  • $\begingroup$ As a clarification, this argument could be extended to a general Abelian group simply because the key point here, is that one can find a non-commuting operator if n>1, yes? In addition to this, could you please explain how the commutation relations between $A$ and $H$, and $P$ come to be? Thank you so much! $\endgroup$ – Primo-uomo Jul 3 '17 at 15:29
  • $\begingroup$ Yes!... I added a comment regarding commutativity $\endgroup$ – Valter Moretti Jul 3 '17 at 15:43
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I was about to suggest the counter-example, but I thought it might be facetious and specific to this case, since the Casimir is just the Identity for the spin-1/2 representation. The reverse statement is true, no doubt about it. It's just that many answers seemed to suggest that an Abelian symmetry group precludes the possibility of degeneracy, and I wanted to know if this was true. Could you please elaborate on "accidental degenracies"? $\endgroup$ – Primo-uomo Jul 3 '17 at 9:52
  • $\begingroup$ Wrong: for instance also $\sigma_x$ commutes with $H$: there is another symmetry against the hypothesis that $A$ is the only symmetry. $\endgroup$ – Valter Moretti Jul 3 '17 at 14:26
  • $\begingroup$ @ValterMoretti Also $\sigma_y$ commutes with $H$, but neither commutes with $A=\sigma_z$...so i'm not sure how they relate to the statements about irreps of $A$? Is the elaboration in your answer? $\endgroup$ – Mikael Fremling Jul 3 '17 at 15:48
  • $\begingroup$ I mean $\sigma_y$ is another symmetry since it commutes with $H$ but it is not function of $\sigma_x$ (otherwise it would commute with it). In the hypothesis there must be only one symmetry... $\endgroup$ – Valter Moretti Jul 3 '17 at 15:50
  • $\begingroup$ I see, so the symmetries need not commute with each other. I missed that point. Thanks for clarifying. $\endgroup$ – Mikael Fremling Jul 3 '17 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.