Sakurai 1.17 - Operators and Complete Eigenkets

[Domnu]

I'm pretty sure this is correct, but could someone verify for rigor?

Problem
Two observables A_1 and A_2, which do not involve time explicitly, are known not to commute, yet we also know that A_1 and A_2 both commute with the Hamiltonian. Prove that the energy eigenstates are, in general, degenerate. Are there exceptions?

The attempt at a solution
Since [H,A_1]=0, we know that there are complete eigenkets that A_1 and H share. The same is true for A_2. Now, generally, two observables that do not commute do not share eigenkets. Thus, we know that there exist two distinct eigenstates such that H|a\ket = e|a\ket; particularly, each |a\ket can be brought from the eigenkets of A_1,A_2.

 

[Domnu]

come on 25 views and no post? :(

 

[Dick]

Impatience is not a virtue. Can't A1 and A2 commute on a subspace without commuting on the whole space?

 

[Domnu]

Heh, it was more of a bump since it was already on the second page :) Yes, they can, but the question asks "generally," and "generally" this doesn't happen, right?

 

[Dick]

Ok, considered as a teasing bump. But the question asks, are there exceptions? I.e. could there be energy eigenstates that aren't degenerate? At least that's how I read it.