- #1
- 4,807
- 32
I am reading the first chapter of Sakurai's Modern QM and from pages 30 and 32 respectively, I understand that
(i) If [A,B]=0, then they share the same set of eigenstates.
(ii) Conversely, if two operators have the same eigenstates, then they commute.
But we know that [itex][L^2,L_z]=0[/itex], [itex][L^2,L_x]=0[/itex] and [itex][L_z,L_x]\neq 0[/itex].
From the first equality, and (i) I gather that L² and L_z have the same eigenkets. From (i) and the second equality, I get that L² and L_x have the same eigenkets. So L_z and L_x have the same eigenkets. Hence, by (ii), they commute, which contradicts [itex][L_z,L_x]\neq 0[/itex].
Where did I go wrong??
(i) If [A,B]=0, then they share the same set of eigenstates.
(ii) Conversely, if two operators have the same eigenstates, then they commute.
But we know that [itex][L^2,L_z]=0[/itex], [itex][L^2,L_x]=0[/itex] and [itex][L_z,L_x]\neq 0[/itex].
From the first equality, and (i) I gather that L² and L_z have the same eigenkets. From (i) and the second equality, I get that L² and L_x have the same eigenkets. So L_z and L_x have the same eigenkets. Hence, by (ii), they commute, which contradicts [itex][L_z,L_x]\neq 0[/itex].
Where did I go wrong??