Hermitian operators and cummutators problem

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


@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.
 
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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
 
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No, from the Jacobi identity one obtains

[[C,A],B] = 0
 


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
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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