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On operator commutation

by wangyi
Tags: commutation, operator
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wangyi
#1
May17-05, 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 non-spacelike point x,y to make any non-spacelike [\phi(x),\phi(y)]=0. It looks like a paradox.

thank you!
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CarlB
#2
May17-05, 01:48 AM
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> 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
PhilG
#3
May17-05, 01:50 AM
P: 53
Quote Quote by wangyi
Now there is the question,
if A,B,C are operators, [A,B]=0, [A,C]=0,
then is "[B,C]=0" also right?
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.

CarlB
#4
May17-05, 01:56 AM
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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
Galileo
#5
May17-05, 04:24 AM
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[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.
wangyi
#6
May17-05, 06:08 AM
P: 56
Thank you all, i see :)


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