Why isn't commutation transitive?

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Discussion Overview

The discussion revolves around the concept of commutation in quantum mechanics, specifically addressing why commutation is not transitive. Participants explore the implications of non-abelian groups and the diagonalizability of operators in relation to the Hamiltonian.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the transitivity of commutation, citing the case of the Hamiltonian and non-abelian groups.
  • Another participant clarifies that a basis is determined by a complete set of commuting observables, indicating that multiple bases can exist.
  • A participant confirms that two operators commuting with the Hamiltonian can split its eigen-subspaces differently, leading to a lack of shared diagonalizability.
  • It is noted that while additional operators may be diagonalizable together, not all eigenvectors of one operator are necessarily eigenvectors of the other, particularly in the context of non-abelian symmetries.

Areas of Agreement / Disagreement

Participants generally agree on the non-transitive nature of commutation and the implications for diagonalizability, but there is ongoing exploration of specific cases and conditions under which these properties hold.

Contextual Notes

The discussion highlights the dependence on definitions of diagonalizability and the structure of the Hilbert space, as well as the role of non-abelian symmetries in the context of commuting operators.

metroplex021
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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?!
 
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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 Jz.
 
Bill_K said:
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 Jz.

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?
 
metroplex021 said:
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.
 
Awesome. Thanks people!
 

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