# Commuting operators

1. Nov 5, 2006

### Fe-56

hello....this might look very stupid but I am totally confused...........

Let have operators A, B, C. Let [A,B]=[A,C]=0 and [B,C] not 0...

When two operators commute, they have the same base in (Hilbert) space.
So base in A representation is the same as in the B representation and also
the basis in A and C representations are the same.

But if B and C do not commute, their basis are different.

There has to be a fatal error in here
thanx:tongue:

2. Nov 5, 2006

### nrqed

As far as I know, this happens only when there is degeneracy.

Let's say that the eigenstates of A are degenerate. Then if [A,B]=0, it means that the eigenstates of B can be written as some linear combination of the eigenstates of A. If [A,C]= 0, the same thing applies for the eigenstates of C (they can be written as a linear combination of the eigenstates of A).

Consider a set of eigenstates of A all corresponding to the same eigenvalue of A, say $A \psi_n(x) = a \psi(x)$ . Then it will be possible to write some of the eigenstates of B (corresponding possibly to different eigenvalues of B) as some linear combinations of these $\psi_n(x)$. And it will be possible to write some of the eigenstates of C as a different linear combination of the eigenstates of A (still corresponding to the same degenerate eigenvalue $a$). So A and B share a common set of basis states, A and C share a common basis of states but B and C don't.

An exception is if there is a state for which applying A, B or C all gives zero. Then all three operators may share the same eigenstate even if they don't all commute.

This is what happens, say, with ${\vec L}^2$ and $L_x$ and $L_z$ for example. We have $[ {\vec L}^2,L_x] = [{\vec L}^2,L_z]=0$ but $[L_x,L_y] \neq 0$ .

Last edited: Nov 5, 2006
3. Nov 5, 2006

### Fe-56

:!!)

YES, thats it!!!!!!! :!!)

so......there must be degeneracy in all eigenvalues of A, yes???

thankx a lot.........

.........and this was not any homework, someone replaced it from QT