# Commutation relations trouble (basic)

1. Aug 10, 2006

### quasar987

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 $[L^2,L_z]=0$, $[L^2,L_x]=0$ and $[L_z,L_x]\neq 0$.

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 $[L_z,L_x]\neq 0$.

Where did I go wrong??

2. Aug 10, 2006

### Physics Monkey

Hi quasar,

Consider the states $$| \ell, \,m_z \rangle$$ which are the usual simultaneous eigenstates of $$L^2$$ and $$L_z$$. To understand your problem, imagine taking superpositions of these states with the same $$\ell$$ but different $$m_z$$ values. For example, the state $$| 1/2,\, 1/2 \rangle + | 1/2, \,-1/2 \rangle$$ is such a superpostion; an equal superposition of spin up and spin down. This state is not an eigenstate of $$L_z$$ (two different $$m_z$$ values), but it is still an eigenstate of $$L^2$$. Thus while it is certainly possible to choose simultaneous eigenstates of $$L_z$$ and $$L^2$$, it is definitely not true that every eigenstate of $$L^2$$ is an eigenstate of $$L_z$$. In fact, the state I just constructed is nothing other than the spin up eigenstate of $$L_x$$. That is, it is a simultaneous eigenstate of $$L^2$$ and $$L_x$$ rather than $$L^2$$ and $$L_z$$. So you see, among all the states with the same value of $$\ell$$, there is a nontrivial freedom to construct states that are eigenstates of $$L_z$$ or $$L_x$$. Finally, the fact that $$L_z$$ and $$L_x$$ don't commute simple means that you won't be able find to find linear combinations which are eigenstates of $$L_z$$ and $$L_x$$.

Hope this helps!

Last edited: Aug 10, 2006
3. Aug 10, 2006

### quasar987

Ah, so I misinterpreted (i). Would a true statement be that if [A,B]=0, then the eigenkets of either one of A or B are eigenkets for the other.

Correct? Can it be made stronger?

4. Aug 10, 2006

### Physics Monkey

The crucial point is that the above statement is definitely not true; eigenstates of A are in no way guaranteed to be eigenstates of B. The guarantee you do have is much weaker. What it says is that if you make a list of linearly independent eigenstates of A with the same eigenvalue, then you can always make from this list states which are also eigenstates of B (provided A and B commute). Again, it does not mean that every state on your list is automatically an eigenstate of B, only that you can find some combinations which are.

5. Aug 10, 2006

### quasar987

Thank you!

6. Aug 11, 2006

### dextercioby

Technically, if $A:D(A)\rightarrow \mbox{Im}(A)$ ,$A:D(B)\rightarrow \mbox{Im}(B)$ AND $D(A) \cap D(B) \neq \emptyset$, then

$$AB-BA=0 \Leftrightarrow A(B\psi)=B(A\psi), \forall \in \psi D(A) \cap D(B) \ \mbox{and} \ B\psi \in D(A) , A\psi \in D(B)$$.

Daniel.

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