I'm pretty sure this is correct, but could someone verify for rigor? Problem Two observables [tex]A_1[/tex] and [tex]A_2[/tex], which do not involve time explicitly, are known not to commute, yet we also know that [tex]A_1[/tex] and [tex]A_2[/tex] both commute with the Hamiltonian. Prove that the energy eigenstates are, in general, degenerate. Are there exceptions? The attempt at a solution Since [tex][H,A_1]=0[/tex], we know that there are complete eigenkets that [tex]A_1[/tex] and [tex]H[/tex] share. The same is true for [tex]A_2[/tex]. Now, generally, two observables that do not commute do not share eigenkets. Thus, we know that there exist two distinct eigenstates such that [tex]H|a\ket = e|a\ket[/tex]; particularly, each [tex]|a\ket[/tex] can be brought from the eigenkets of [tex]A_1,A_2[/tex].