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?

Similar question: Two operators commute with the Hamiltonian, but do not commute with each other

  • $\begingroup$ Why is the example an exception? $\endgroup$ Commented Jul 10, 2023 at 18:40
  • $\begingroup$ 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. $\endgroup$
    – JohnA.
    Commented Jul 10, 2023 at 19:01
  • 1
    $\begingroup$ The point is that there must be degeneracy somewhere, but not every eigenvalue need be degenerate. $\endgroup$
    – march
    Commented Jul 10, 2023 at 19:51

1 Answer 1


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.


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