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!
> 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 [tex]A=x, B=y, C=p_y[/tex]. For a counterexample in the Dirac gamma matrices, let [tex]A=\gamma^0, B=\gamma^1\gamma^2, C=\gamma^1\gamma^3[/tex]. 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
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.
> 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
[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.