- #1

wangyi

- 56

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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!