# Sakurai 1.17 - Operators and Complete Eigenkets

1. Dec 16, 2008

### 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 $$A_1,A_2$$.

2. Dec 16, 2008

### Domnu

come on 25 views and no post? :(

3. Dec 16, 2008

### Dick

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

4. Dec 16, 2008

### 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?

5. Dec 16, 2008

### 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.