Why isn't commutation transitive?

[metroplex021]

I know this is really basic, but can anyone explain why commutation isn't transitive? (Eg in the case of invariance of the Hamiltonian under a non-abelian group, all the transformations of the group commute with H but don't all commute with each other.) I thought there was only one basis in which each operator was diagonalizable, hence one basis in which any pair of commuting operators was diagonalizable - so that *all* the operators that commute with an operator such as H should all commute with each other. Where have I gone wrong?!

 

[Bill_K]

I thought there was only one basis in which each operator was diagonalizable

Nope, and you've just given us a counterexample! A basis is uniquely determined by a "complete set of commuting observables". After you've diagonalized H, you've split the Hilbert space into its eigen-subspaces, but these may be further split by diagonalizing another operator, like J_z.

 

[metroplex021]

Nope, and you've just given us a counterexample! A basis is uniquely determined by a "complete set of commuting observables". After you've diagonalized H, you've split the Hilbert space into its eigen-subspaces, but these may be further split by diagonalizing another operator, like J_z.

Thanks... so just to check, is it the case that if I take two operators that commute with H, then they can split the eigen-subspaces of H in two different ways - so that those two operators may not share a basis in which they are both diagonalizable?

 

[cgk]

Thanks... so just to check, is it the case that if I take two operators that commute with H, then they can split the eigen-subspaces of H in two different ways - so that those two operators may not share a basis in which they are both diagonalizable?

Yes, that is possible (that is the general case. Think of H and Sz vs H and Sx vs H and Sy).

It is also possible that the additional operators are diagonalizable together, but that not every eigenvector of one is also automatically an eigenvector of the other one. In that case only certain linear combinations in degenerate subspaces are eigenvectors of both operators at the same time. This generally happens when you have a maximally commuting set of non-abelian point group symmetry generators.

 

[metroplex021]

Awesome. Thanks people!