How does commutation between the hamiltonian and angular momentum operator (squared) imply conservation of Angular momentum? So we are looking at central potentials in QM;
The lecturer poses the question, when is $\textbf{L}$ conserved? He then considers the commutator of $\hat{H}$ and $\hat{L^2}$. 
We have;
$$\hat{H}=-\frac{\hbar^2}{2m}\triangledown^2_R + \frac{\hat{L^2}}{2mr^2}+V(\vec{r})$$
where, $\triangledown^2_R=\frac{1}{r^2}\frac{\partial{}}{\partial{r}}(r^2\frac{\partial{}}{\partial{r}})$
$$[\hat{H}, \hat{L^2}]= -\frac{\hbar^2}{2m}[\triangledown^2_R,\hat{L^2}]+\frac{1}{2m}[\frac{\hat{L^2}}{r^2}, \hat{L^2}]+[V(\vec{r}),\hat{L^2}]$$
The first two terms are zero. The last term is zero if $V(\vec{r})$ is ONLY a function of $r$ not $\theta$ and $\phi$.
Thus having a central potential will ensure the commutator yields 0, but how does this imply angular momentum is conserved?
Strictly how does the commutation above imply conservation, if anything it shows that the two operators commute and share eigenstates.

I am aware that angular momentum is conserved if there is a central potential as we will have a central force, $F(r)\hat{r}$ hence:
$$\frac{d\hat{L}}{dt}=\frac{d}{dt}(\hat{r}\times\hat{p})=\frac{d\hat{r}}{dt}\times\hat{p} + \hat{r}\times\frac{d\hat{p}}{dt}=\hat{v}\times m\hat{v}+\hat{r}\times F(r)\hat{r}=0$$
Hence $\hat{L}$ is conserved.
 A: A conserved quantity is one that commutes with the Hamiltonian for the simple reason that $[A,H] = 0$ implies
$$ \frac{\mathrm{d}}{\mathrm{d}t} A = 0$$
in the Heisenberg picture. 
Another way to see that commuting with the Hamiltonian means conservation is to consider that the time evolution operator $U(t) = \exp(-\mathrm{i}Ht)$ is just the exponential of the Hamiltonian, and thus $[A,H] = 0$ implies $[U(t),H]=0$ for all $t$, that is, it makes no difference if you first apply the operator and then evolve the result in time or if you first evolve the state in time and then apply the operator. In particular, eigenstates of $A$ will stay eigenstates with the same eigenvalue forever, and the expectation value of $A$ is constant in time, hence conserved.
A: If you have an eigenstate of an operator $A$:
$$A|Ψ(t=0) \rangle=a|Ψ(t=0)\rangle$$
Then apply $A$ to the time evolution of the wavefunction:
$$A|Ψ(t)\rangle=Ae^{-iHt}|Ψ(t=0)\rangle=e^{-iHt}A|Ψ(t=0)\rangle=e^{-iHt}A|Ψ(t=0)\rangle=a|Ψ(t)\rangle$$
Note that the second to last equality is only true if $[A, H]=0$. So the eigenstate of $A$ remains an eigenstate of $A$! This applies to more general states if we instead talk about the expectation value of A and write the state as a sum of eigenstates. It is in this sense that A is conserved.
A: It can be easily seen using the Heisenberg picture. Take a Hermitian operator $A$ that commutes with the Hamiltonian $H$. Remember that the eigenvalues of $A$ are observables. Then $A$ also commutes with the time evolution operator $U(t) = e^{-i H t}$.
$[A,U(t)] = 0 \quad \rightarrow AU(t) = U(t)A  \rightarrow A = U(t)AU(t)^\dagger $
So the operator $A$  dosen't change under time-evolution, and hence represents a conserved quantity. 
