Hermitian operators and cummutators problem

[nakbuchi]

A,B and C are three hermitian operators such that [A,B]=0, [B,C]=0.
Does A necessarily commutes with C?

 

[sgd37]

yes since if $$[A,B] \psi = 0 , [B,C] \psi = 0$$ that means that there exists a psi which is an eigenstate of all three and all operators commute

or you could try the Jacobi identity

 

[nakbuchi]

@sgd37:
A and B have some common eigenvectors, and so does B and C; but it doesn't necessarily mean that A and C will have common eigenvectors.

 

[sgd37]

Sure. But in that case you have some kind of algebra going on. I was naive in thinking they belonged to the same set. But no. If you consider the components of the angular momentum operator and the L^2 operator

 

[dextercioby]

No, from the Jacobi identity one obtains

$$[[C,A],B] = 0$$

 

[Dickfore]

No, let A = x, B = y, C = px

[A, B] = [x, y] = 0

[B, C] = [y, px] = 0

But,

[A, C] = [x, px] = $i \hbar \neq 0$