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.

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    $\begingroup$ It means they all commute with one another. $\endgroup$ – alanf Feb 20 '15 at 9:48
  • $\begingroup$ If $\{A_1,\ldots, A_n\}$ is your set then $[A_i,A_j]=0\forall i,j$. $\endgroup$ – ZeroTheHero Jan 13 at 22:14

Mutually commutative means that every operator in the set commutes with every other one. This implies that, if the operators in question are observables, they can all be measured simultaneously.

A complete set of mutually commuting observables is a set of observable, hermitian operators that commute - therefore their eigenvalues can be used to label a state. "Complete" refers to the state being fully determined without degeneracies.

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.

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    $\begingroup$ Another example: if the operators are generators, and they all mutually commute, they form a Cartan subalgebra. $\endgroup$ – JamalS Feb 20 '15 at 10:11
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    $\begingroup$ 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. $\endgroup$ – Phoenix87 Feb 20 '15 at 12:42

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