# Is commutation relation an equivalence relation?

I'm now learning quantum mechanics with Liboff. In the book it deals with "a compete set of mutually compatible observables" in order to make a state maximally informative. How can one find such set? It seems very hard unless commutation relation is an equivalence relation. Is commutation relation an equivalence relation? That is, if $A, B, C$ are Hermitian operators, then does $AB=BA, BC=CB$ imply $AC=CA$?

Commuting is not an equivalence relation. All components of angular momentum commute with $J^2$ but they don't commute with each other.

How to find a complete set of mutually commuting observables is a difficult problem and I don't think you can give an algorithmic answer. It depends very much on the specific problem. An observable that commutes with the Hamiltonian is conserved, this can be a good starting point. For example, angular momentum is conserved when the Hamiltonian is rotationally symmetric.

No, it does not! Let me give you a counterexample:

Consider the Hermitian operators $\mathsf{1}$ (identity operator), $p$ (momentum) and $x$ (position) in 1D.

Now, the trivial commutation relations $[\mathsf{1},x]=0$ and $[\mathsf{1},p]=0$ do not imply $[x,p]=0$ as the correct relation is $[x,p]=\mathrm i\hbar\neq 0$.

As everyone points out commutation is not a shorthand for equivalence, as, given your relations, the Jacobi identity, [A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0 dictates that when the first two terms vanish, the third must too, so that B must commute with [C,A], non vanishing in general, as remarked repeatedly.

Lie algebra commutators do, nevertheless, parameterize conjugacy, that is $~A^{-1} B A - B= A^{-1} [ B,A]$, so an observable commuting with everything collapses to its very own conjugacy class.

Commutation does become transitive, and thus an equivalence relation (reflexive and symmetric are trivial), when you impose an extra condition: nondegeneracy.

If $A$, $B$, $C$ are Hermitian operators, and each of them has only unique eigenvalues, then $AB\! =\! BA\, \cap\, BC\! =\! CB$ implies $AC = CA$.

Proof: for a nondegenerate operator, the eigenbasis is well-defined, so if $A$ and $B$ share an eigenbasis $E^{AB} = \{|\Psi^{AB}_i\rangle\}$ (as commutating operators do), and $B$ and $C$ share $E^{BC} = \{|\Psi^{BC}_j\rangle\}$, then $E^{AB} = E^{BC}$, and it is a shared eigenbasis of $A$ and $C$. Therefore, $A$ and $C$ commute.

Just to avoid confusion: Of course, that does not mean a system of degenerate operators never commutes; for instance, consider projectors onto 3 orthogonal states $|\Phi_A\rangle$, $|\Phi_B\rangle$, $|\Phi_C\rangle$, i.e.

$$A |\psi\rangle = |\Phi_A\rangle \langle\Phi_A| \psi \rangle$$ etc.. Because the states are orthogonal, $AB = BA = AC = CA = BC = CB = 0$, so the operators trivially commute though they all have the degenerate eigenvalue 0.