Prove conservation law in quantum mechanics

The most elegant way under my perspective is try to transform (1) to an analogous equation given by the Heisenberg equation, but this can lead to confusion.

In order to keep it simple, let suppose that $H$ and $O$ are not time dependent (in case they were we can, in principle move the time dependence to the states instead of the operators, c.f. Scrödinger picture), therefore

$$ partial_t O = 0quad partial_t H = 0, $$

where $partial_tcdot = frac{partialcdot}{partial t}$. Assume $O$ is an observable, and let $psi(t,vec{r})equivpsi$ be an eigenstate of $O$, i.e. $Opsi = lambdapsi$. Then:

$$ [H,O]psi = HOpsi-OHpsi == H(Opsi)-O(Hpsi) == Hlambdapsi-Oi hbarpartial_tpsi == lambda Hpsi-i hbarpartial_t Opsi == lambda ihbarpartial_tpsi-i hbarpartial_t lambdapsi == ihbarlambda partial_tpsi-i hbarlambdapartial_t psi =0quad blacksquare $$

But as stated in the comments, there are several issueas in this concept: this relations holds only under the statements considered and, although is related to the concept of conserved quantities in QM, must be taken carefully.

(2) says that if $psi$ is an eigenvector of $hat{O}$ at some time $t=0$, it stays at eigenvector of $hat{O}$ thereafter. For that to be true, $psi$ must also be an eigenvector of $hat{H}$ since, infinitesimally,

$$psi(t+delta t) = psi(t) + delta t frac{partial}{partial t} psi(t).$$

From (1), you know that this means

$$psi(t+delta t) = psi(t) + (delta t) frac{hat{H}}{i hbar} psi(t) $$

and so $hat{H} psi = E psi$. Now you have proven there are simultaneous eigenvectors of $hat{H}$ and $hat{O}$, which is the same as $[hat{H},hat{O}] = 0$