TITLE: Commutability of two physical quantity matrices
QUESTION [1 upvotes]: Suppose that two matrices $A$ and $B$, representing real($\mathbb{R}$) physical quantity, can be multiplied commutatively with each other; i.e. $AB =BA$. However, each matrix cannot be multiplied commutatively with other matrices all the time (obviously). If two matrices represent physical quantity, can they be stated to have definite values togeether still? Or as two matrices cannot be multiplied commutatively with other matrices, is it not possible for two matrices to have definite values together?
Also, suppose $C=A+B$. If A and B can be multiplied commutatively with each other, does $C$ have definite values? If $A$ and $B$ cannot be multiplied commutatively with each other, is there any way $C$ can have definite values?

REPLY [1 votes]: If two real (i.e. self-adjoint) operators $A$ and $B$ do not commute on the intersection of their domain, they don't have a basis of common eigenvectors, hence they do not have a joint probability distribution. In this case, $C=A+B$ doesn't need to be selfadjoint, hence does not always represent an observable (though it is always Hermitian). One needs some domain compatibility condition for the selfadjointness of $C$. (For finite-dimensional matrices, Hermitian implies selfadjoint, though.)
But if two self-adjoint operators $A$ and $B$ commute on a common dense domain, they have a joint probability distribution, and in a state where $A$ and $B$ have definite values $a$ and $b$, any function $f(A,B)$ with finite $f(a,b)$ has the definite value $f(a,b)$ in such a state. In particular, this holds for $C=A+B$.
{"set_name": "stack_exchange", "score": 1, "question_id": 38477}