Mutually Commutative

What is the definition of a Mutually Commutative set of operators? I've found articles describing a complete set of mutually commutative operators, but I can't actually find what mutually commutative means. I ask because I'm asked to prove that a particular set of operators is mutually commutative.

• It means they all commute with one another. – alanf Feb 20 '15 at 9:48

As an example: The most famous set are the quantum numbers labeling the Hydrogen orbitals, corresponding to the set of observables $${\mathcal H, \vec J^2, J_z, \vec L^2, \vec S^2}$$ With the five eigenvalues of these operators the state of the electron in the hydrogen atom can be uniquely determined and all these values can be measured simultaneously because every operator in the set commutes with all others.
• I'm not sure that your definition of "complete" is the correct one. A set $F$ of pair-wise self-adjoint operators from an Abelian von Neumann algebra $M$ is complete if every other element in $M$ is an $L^\infty$ function of the elements in $F$ and every function in $L^\infty$ of the elements in $F$ gives an element in $M$. Such $M$ usually arises as the bicommutant of some commutative C*-algebra of observables. – Phoenix87 Feb 20 '15 at 12:42