Do Commuting Operators Always Share a Common Basis of Eigenvectors?

[Funzies]

Hey guys,
I'm studying some quantum physics at the moment and I'm having some problems with understanding the principles behind the necessary lineair algebra:

1) If two operators do NOT commutate, is it correct to conclude they don't have a similar basis of eigenvectoren? Or are there more conditions to be verified to conclude this?

2) If two operators DO commutate, you MAY find a similar basis of eigenvectoren? If this is correct, what are the conditions for it to be absolutely true?

The context to which I am asking these question are the ladder operators for angular momentum states and especially the fact that Lx and Ly do not commutate and don't have a similar basis of eigenvectors, whereas L^2 and Lx/Ly/Lz do commutate and do have a basis of eigenvectors.

Thanks in advance!

 

[dextercioby]

In my ancient lecture notes from college I remember having the proof of a strong theorem:

If A and B are 2 bounded and self-adjoint linear operators with discrete spectrum on a separable Hilbert space, they commute if and only if the Hilbert space has an orthonormal basis made up of common eigenvectors of A and B.

This theorem applies to your question, with the minor notice that the orbital angular momentum ops are generally unbounded in the Hilbert space $L^2 \left(\mathbb{R}^3, dx\right)$, so one speaks about the closures of the these operators when referring to self-adjointness and the common dense domain involved is not the domain of the closure, but of the original operator.

 

[arkajad]

If A and B are 2 bounded and self-adjoint linear operators with discrete spectrum on a separable Hilbert space, they commute if and only if the Hilbert space has an orthonormal basis made up of common eigenvectors of A and B.

In the standard QM setup it will be not too dangerous to assume that the following holds:

If A and B are hermitian operators then they commute if and only if there is a common basis of eigenvectors.

This basis can be discrete or continuous or partly discrete and partly continuous - whatever it may mean.