# Understanding exception to: Two non-commuting Hermitian operators commute with the hamiltonian implies degenerate energy eigenvalues

For context, I am working through the exercises in Modern Quantum Mechanics by Sakurai and Napolitano Second Ed. I have previously completed (years ago in undergrad) the Griffiths 3rd ed. Introduction to Quantum Mechanics.

I am having trouble understanding part of Problem 1.17 in Sakurai and Napolitano. The problem statement is the following:

Two observables $$A_1, A_2$$, which do not involve time explicitly, are known not to commute. We also know they each commute with $$H$$. Prove that the energy eigenstates are, in general, degenerate. Are there exceptions? As an example, you may think of the central-force problem $$H = p^2/2m +V(r)$$ with $$A_1\to L_z$$, $$A_2 \to L_x$$.

So, I didn't really understand the "exceptions" part. My attempt at the solution is the following:

1. Assume that $$H$$ has no degeneracy (as a proof by contradiction).
2. Therefore, the dimension of each eigenspace of $$H$$ is 1.
3. Therefore the size of basis of each eigenspace is a single vector.
4. This implies any other basis that diagonalizes $$H$$ is just a reordering of the basis formed by collecting the vectors that span each size $$1$$ eigenspace (each vector up to a phase, if we assume they are all normal).
5. This implies $$A_1$$ and $$A_2$$ are simultaneously diagonalizable. Therefore, they commute.
6. This is a contradiction so there must be degeneracy on $$H$$.

The example given in the problem statement seems to be an exception of this. However, I am not sure where in the proof there is room for an exception. Where am I incorrect, or what am I misunderstanding?

• Why is the example an exception? Commented Jul 10, 2023 at 18:40
• I am not sure, I suppose because of the phasing of the question, I assumed it was. Let's see: each operator commutes with H and they do not commute with each other. Does H have degeneracy? According to chapter 3 of the book (page 207 is the discussion on central potentials which I just skimmed), there is degeneracy in H. I am not sure then why is this specific hamiltonian raised here. Commented Jul 10, 2023 at 19:01
• The point is that there must be degeneracy somewhere, but not every eigenvalue need be degenerate. Commented Jul 10, 2023 at 19:51

## 1 Answer

Your argument functions to say that there is "some" degeneracy in $$H$$, but does not exclude the possibility that some non-degenerate state exists. To be clear, the "exception" is not that there exists a combination of $$A_1, A_2, H$$ satisfying the conditions such that every energy eigenstate is non-degenerate, your proof demonstrates that isn't possible.

However, the central force problem has states (S states) for which the non-commuting angular momentum operators all have simultaneous eigenstates with eigenvalue zero. You should consider why an eigenvalue of zero for the non-commuting operators enables this.