# On operator commutation

1. May 17, 2005

### wangyi

Hi, I have a question,
As it is said in QM, if two operators commute, they have so many common eigenstates that they form a basis. And the inverse is right.
Now there is the question,
if A,B,C are operators, [A,B]=0, [A,C]=0,
then is "[B,C]=0" also right?

If we simply say A and B, A and C both have common eigenstates, so B and C have common eigenstates, so [B,C]=0, it seems to be right.

But in QFT, if x,y spacelike, then [\phi(x),\phi(y)]=0,
if the above is right, then we can find a point z which is spacelike according to two non-spacelike point x,y to make any non-spacelike [\phi(x),\phi(y)]=0. It looks like a paradox.

thank you!

2. May 17, 2005

### CarlB

> If A,B,C are operators, [A,B]=0, [A,C]=0,
> then is "[B,C]=0" also right?

No, it's not right. For a counterexample in the usual QM variables, let $$A=x, B=y, C=p_y$$.

For a counterexample in the Dirac gamma matrices,
let $$A=\gamma^0, B=\gamma^1\gamma^2, C=\gamma^1\gamma^3$$.

For a counterexample in QFT, replace the gamma matrices with your favorite four anticommuting field variables.

In each of these counterexamples, A commutes with B and A commutes with C, but B and C do not commute.

Carl

3. May 17, 2005

### PhilG

No. The angular momentum operators give a counterexample: A = L^2, B = L_x, and C = L_y. Then [A,B] = [A,C] = 0. But [B,C]= [L_x, L_y] = ih L_z.

It is true, however, that [B, C] commutes with A. This can be seen from the jacobi identity

[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0.

4. May 17, 2005

### CarlB

> If we simply say A and B, A and C both have
> common eigenstates, so B and C have common
> eigenstates, so [B,C]=0, it seems to be right.

If A has no degeneracy in its eigenvalues, then your logic works. In the presence of degeneracy, A can arrange to share a different set of eigenstates with B than it shares with C.

Carl

5. May 17, 2005

### Galileo

[A,B]=0 means you can find a set of eigenstates common to A and B.
[A,C]=0 means you can find a set of eigenstates common to A and C.

That doesn't imply these two sets are the same, so it will in general not give a set of eigenstates common to B and C.

6. May 17, 2005

### wangyi

Thank you all, i see :)