
#1
May1705, 12:07 AM

P: 56

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 nonspacelike point x,y to make any nonspacelike [\phi(x),\phi(y)]=0. It looks like a paradox. thank you! 



#2
May1705, 01:48 AM

Sci Advisor
HW Helper
P: 1,204

> 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 



#3
May1705, 01:50 AM

P: 53

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
May1705, 01:56 AM

Sci Advisor
HW Helper
P: 1,204

on operator commutation
> 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
May1705, 04:24 AM

Sci Advisor
HW Helper
P: 2,004

[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
May1705, 06:08 AM

P: 56

Thank you all, i see :)



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