How to show that if two operators $A$ and $B$ commute, then simultaneous accurate measurement is possible? I have proved that if two operators commute then their simultaneous accurate measurement is possible using the uncertainty equation but I am unable to do so without using it.
I have tried and reached to the conclusion if two operators $A$ and $B$ commute with each other then they have common eigenstate but how do I show that common eigenstate implies simultaneous accurate measurement is possible?
 A: We need what you already proved: the existence of common eigenstates. So, assume that $A$ and $B$ commute, and therefore we can find states $|a,b\rangle$ such that:
$$A|a,b\rangle=a|a,b\rangle, \quad B|a,b\rangle=b|a,b\rangle.$$
The postulates of quantum mechanics state that, if $A$ is an observable, and if the system is in one of the eigenstates of $A$, the measurement of $A$ yields the correspondent eigenvalue (in our notation $a$) with certainty (probability 1). So, in this case, if the system is in the state $|a,b\rangle$ then when we measure $A$ we get $a$ with probability 1, but when we measure $B$ we get $b$ also with probability 1.
Thus, once you prove that $[A,B]=0$ implies the existence of simultaneous eigenstates, then the existence of simultaneous accurate measurement follows from the postulates of quantum mechanics.
For reference about the postulates you can check Shankar's Principles of Quantum mechanics or Sakurai's Modern Quantun Mechanics. The former reference discusses the postulates explicitly in chapter 4, while the latter does not list the postulates but discuss this part about simultaneous measurements in chapter 1 (section 1.4, page 30 in the second edition).
A: Well, this actually follows from definition. If $A$ and $B$ commute then $AB=BA$.
Thus if if you apply try to measure them then:
$$
\left<\psi\right|AB\left|\psi\right> = \left<\psi\right|BA\left|\psi\right>
$$
The order in which I applied them doesn't change the expected value. Doesn't it suffice?
