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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 $AC=CA$?

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up vote 6 down vote accepted

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.

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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 $[x,p]=\mathrm i\hbar\neq 0$.

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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.

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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.

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