Commuting with the Hamiltonian operators are conserved quantities In this question What does "commuting with the Hamiltonian" mean?. I've read that if an operator commutes with the Hamiltonian it is a conserved quantity, this means that the average value of that observable does not vary in the time namely: $\frac{d \hat O}{dt} \ne 0$.
The first question is: what does this mean? When I studied this subject, I read that in order to obtain the average value of an observable, we have to imagine this experiment: I prepare in the laboratory $n$ systems all equal and measure the same observable $n$ times obtaining $n$ results with a certain frequency. Then I make an average of the results obtained and I obtain the mean value of the observable on that system.
But what does it mean to measure the mean value at different times?
Do I do the operation I mentioned, let the eigenstates obtained evolve over time and repeat the operation?
The second question is: in general, if I have an operator that commutes with $H$, do its eigenstates not change over time? That is, if $O$ commutes with $H$ and I have an eigenstate of $O$ relative to a certain eigenvalue, as time passes, if I measure $O$ on this eigenstate, do I continue to obtain the same eigenvalue? What if the operator $O$ does not commute with $H$? In general, do the eigenstates of an operator that does not commute with $H$ evolve over time by changing? So that if I make a measurement of $O$ on one of its eigenstates at different times I might get different results?
 A: To answer your first question let's fix what we mean by a time evolution (I'll use Shrodinger picture, but heisenberg one is equivalent.). A state of a quantum system is some vector $\psi \in \mathcal H$ of some Hilbert space. Quantum mechanics tells that if we start with a system in the state $\psi$ then we can know what state the system will be on any time $t$, namely $\psi(t)$. Now, what mean by the expected value is the average of measured values of a system in a given state (so there no mention of time here). Thus what it means for the expected value to be time invariant that the expected value of an operator $A$ that commutes with the Hamiltonian of a system in the state $\psi_0$ is the same as that of a system in the state $\psi(t)$.
To answer the second question we'll dive into the formalities. To any self-adjoint operator $A$ we can associate an one parameter group of unitary transformations $U_A(t)=e^{itA}$. For the Hamiltonian, we get $e^{itH}$ and this gives the time evolution of the system. That is, if in the time $0$ a system was in a state $\psi_0 \in \mathcal H$, the in a time $t$ the system will be on the state $\psi(t)=e^{-iHt}\psi_0$.
Now, suppose that $\psi_0$ is an eigenvector of $A$ and that $A$ commutes with $H$. We know that $[A,e^{-iHt}]=0$, so that $A$ commutes with $e^{-iHt}$ for all $t$. Now, note that
$$A(\psi(t))=A(e^{-iHt}\psi_0)=e^{iHt}A\psi_0= \lambda e^{-iHt}\psi_0=\lambda\psi(t).$$
Thus we conclude that the time evolution of eigenvectors of $A$ continue to be eigenvectors of A with the same associated eigenvalues. More generally, the probability of a measure of $A$ to lie on some subset $E \subset \mathbb R$ will not change over time.
