Mutually Commutative Operators 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.
 A: A pair of selfadjoint operators $A : D(A) \to H$, $B: D(B) \to H$, where $D(A),D(B) \subset H$ are dense subspaces of the Hilbert space $H$, are said to commute if the spectral measures of $A$ and $B$ commute:
$$P^{(A)}_E P_F^{(B)}= P_F^{(B)}P^{(A)}_E \quad \mbox{for every pair of Borel sets $E, F \subset \mathbb{R}$.}\tag{1}$$
The spectral measures (or PVMs) are uniquely determined by the spectral decomposition of $A$ and $B$:
$$A = \int_{\mathbb{R}} \lambda dP^{(A)}(\lambda)\:, \quad B = \int_{\mathbb{R}} \lambda dP^{(B)}(\lambda)\:.$$
Condition (1) for $A$, $B$ selfadjoint, with the said domains, is separately equivalent to each of the  following requirments
$$e^{itA}e^{isB}= e^{isB}e^{itA} \quad \forall s,t \in \mathbb{R}\:,$$
$$e^{itA}B x= Be^{itA}x \quad \forall t \in \mathbb{R},\quad  \forall x \in D(B)$$
$$P^{(A)}_E Bx = BP^{(A)}_Ex\quad \forall x \in D(B), \forall E \subset \mathbb{R} \mbox{$\quad$ Borel.}$$
If $A,B \in \mathfrak{B}(H)$  (the $C^*$-algebra of the bounbed everywhere defined operators in $H$) are selfadjoint, in particular if $H$ is finite dimensional,  then (1) is equivalent to
$$AB=BA\tag{2}\:.$$
If $A^*=A\in  \mathfrak{B}(H)$ and $B=B^*$ is generally unbounded, then (1) is equivalent to
$$ABx = BAx \quad \forall x \in D(B)\:.$$
Let us come to the notion of complete set of mutually commuting observables also known as maximal set of mutually compatible observables with obvious variations of the words compatible, commuting, observables, selfadjoint operators.
Let us assume that $H$ is separable.
Consider a (finite) set of mutually commuting observables (selfadjoint operators)  ${\cal A}:= \{ A_1,\ldots, A_N\}$, so that their spectral measures pairwise commute. In this case it is possible to define a joint spectral measure $P$ defined on the Borel sets of $\mathbb{R}^N$, it is uniquely defined from the requirement  that
$$A_k = \int_{\mathbb{R}^n} \lambda_k dP(\lambda_1,\ldots, \lambda_N)\quad k = 1,\ldots, N\:.$$
${\cal A}$ is said to be complete or maximal if every (bounded) observable ($B =B^*  \in \mathfrak{B}(H)$) which commutes with all $A_k$ is a (bounded) function of these operators:
$$B =  \int_{\mathbb{R}^n} f(\lambda_1,\ldots, \lambda_N) dP(\lambda_1,\ldots, \lambda_N)$$
for some (bounded Borel measurable) function $f: \mathbb{R}^N \to \mathbb{R}$.
It is not difficult to prove the following fact  if all operators $A_k$ have point spectrum $\sigma_p(A_k)= \sigma(A)$ (in particular $H$ is finite dimensional).
Let us define $H^{(A_k)}_{\lambda_k}$ the eigenspace of $A_k$ with eigenvalue $\lambda_k \in \sigma(A_k)$.
If ${\cal A}$ is a maximal set of mutually commuting observables then,
for every choice of the eigenvalues $\lambda_1, \ldots, \lambda_N$,
the intersection of $H^{(A_1)}_{\lambda_1}, \ldots,H^{(A_N)}_{\lambda_N}$ is at most one-dimensional.
(Also the converse fact is true, always referring to a set of mutually commuting selfadjoint operators with point spectrum).
From a physical perspective,  all that implies in particular that
one may "prepare the states" just by measuring a maximal set of commuting observables with point spectrum.
Indeed, if one measures all the observables $A_k$ (with point spectrum) on an initial state vector $\Psi$, the post measurement unit vector  state $\Psi_{\lambda_1,\ldots, \lambda_N}$
is completely fixed (up to a phase as usual) by the sequence of the outcomes $\lambda_1 \in \sigma(A_1),\ldots, \lambda_N \in \sigma(A_N)$.
In the more sophisticated approach based on the notion of von Neumann algebra,
a (finite) set of mutually commuting observables (selfadjoint operators)  ${\cal A}:= \{ A_1,\ldots, A_N\}$ is said to be  maximal if the commutant $\mathfrak{R}'$ of the  von Neumann  algebra $\mathfrak{R}$ generated by ${\cal A}$ (*) is included in $\mathfrak{R}$ itself: $\mathfrak{R}' \subset \mathfrak{R}$.
The generated von Neumann algebra (in this case of $H$ separable) turns out to be
the family of (complex) maps
$$\int_{\mathbb{R}^n} f(\lambda_1,\ldots, \lambda_N) dP(\lambda_1,\ldots, \lambda_N)$$
for some (Borel measurable) bounded function $f: \mathbb{R}^N \to \mathbb{C}$. Therfore this definition is equivalent to the one stated above.
From a physical point of view, the existence of a maximal set of commuting observables is not always guaranteed. It exists referring to elementary systems, but there are important cases where it cannot exist (**). It happens for instance if the physical system admits a non-Abelian gauge group (think of chromodynamics where the group is $SU(3)$ color). In that case one cannot "prepare" the states.

(*) That is, by definition, the family of bounded operators which commute with all bounded operators which commute with all elements $P^{(A_k)}_E$ of the spectral measures of the selfadjoint operators in ${\cal A}$. Equivalently, this family is the weak closure of the set of linear combinations of those operators $P^{(A_k)}_E$.
(**) Here is the reason. (I use standard properties of von Neumann algebras). Suppose that the von Neumann algebra of observables $\mathfrak{R}$ of a certain physical system admits a maximal set of mutually commuting observables ${\cal A}$.  Hence ${\cal A} \subset \mathfrak{R}$ so that, since  ${\cal A}' \subset {\cal A}''$  by definition of ${\cal A}$, we have
$$\mathfrak{R}' \subset {\cal A}' \subset {\cal A}'' \subset \mathfrak{R}'' = \mathfrak{R}\:.$$
This implies in particular that $\mathfrak{R}' \subset \mathfrak{R}$ and thus
$\mathfrak{R}' = \mathfrak{R}\cap \mathfrak{R}'$. Hence the commutant $\mathfrak{R}'$ of the algebra of observables must be Abelian since it coincides with the center of the algebra $ \mathfrak{R}\cap \mathfrak{R}'$. This is not possible if the commutant includes non commuting selfadjoint operators  as it happens if there is a gauge group which is non-Abelian and faithfully acts. The non-commuting selfadjoint operators are the selfadjoint generators of the representation of the group in the Hilbert space.
A: 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.
