Full time-derivative of a function and Schrodinger equation From Hamiltonian formalism there is well known equation,
$$
\frac{d F}{dt} = \frac{\partial F}{\partial t} + \{F, H\}_{PB},
$$
where $ \{H, F\}_{PB}$ is the Poisson bracket.
After using Hamiltonian formalism in quantum mechanics it transforms into 
$$
\frac{d \hat {F}}{dt} = \frac{\partial \hat {F}}{\partial t} + \frac{i}{\hbar }[\hat {H}, \hat {F}]
$$
in the Heisenberg picture, where $[\hat {H}, \hat {F}]$ is the commutator.
How can I get Schroedinger equation from this expression?
 A: The link you're looking for is hidden by the fact that Heisenberg kets $| \psi \rangle_H$ are not Schroedinger kets $|\psi \rangle _S$. The tranlation from the Heisenberg to the Schroedinger pictures is done via the time-evolution operator $U(t - t_0) = \exp\{-i H * (t - t_0)/\hbar\}$ (I do not explicity put hats on operators, since it should be pretty obrvious what's an operator and what's not).
In the Heisenberg picture, kets $|\psi \rangle_H \equiv |\psi (t_0) \rangle_H$ are not time-dependent. Heisenbergs idea was that the state of a system is not time-dependent, but the observables are. So an operator in the Heisenberg picture at a time $t$ is
$$ F(t) = U^\dagger(t - t_0) F(t_0) U(t-t_0)$$
The Heiseberg equation follows from this:
$$ \frac{d}{dt} F(t) = U^\dagger(t-t_0) (\partial_t F(t_0)) U(t-t_0) +  (\partial_t U^\dagger (t-t_0)) F(t_0) U(t-t_0)  + + U^\dagger(t-t_0) F(t_0) (\partial_t U(t-t_0))$$
The Leibniz terms give exacly the commutator from the Heisenberg equation.
In the Schroedinger picture, the operators are constant $F(t) \equiv F(t_0)$, but the state changes $| \psi \rangle_S = | \psi(t) \rangle_S  $. As all physics needs to be equal in both pictures, we can look at expectation values:
\begin{aligned}\langle \psi | F | \psi \rangle 
&= {}_H\langle \psi | F(t) | \psi \rangle_H &&= \langle \psi | U^\dagger(t-t_0) F(t_0) U(t-t_0) | \psi \rangle \\
& = {}_S \langle \psi(t) | F(t_0) | \psi(t) \rangle_S &&= \langle \psi(t_0) | U^\dagger(t-t_0) F(t_0) U(t-t_0) | \psi(t_0) \rangle
\end{aligned}
where we take the basis of our Hilbert space to conincide for both pictures at $t_0$.
Thus, assuming the Heisenberg euqation is correct, the equation for the time-evolution of a state in the Schroedinger picture is
$$ | \psi(t) \rangle_S = U(t-t_0) | \psi(t_0) \rangle_S$$
The time derviative then gives you the Schroedinger equation
$$ \frac{d}{dt} |\psi(t)\rangle = \Big(\frac{d}{dt} U(t-t_0)\Big) |\psi(t_0)\rangle = -\frac{i}{\hbar} H | \psi(t_0) \rangle $$
A: In the Heisenberg representation, the "equivalent" of the Schrodinger equation is :
$$\hat  H(t) = \frac{\hat P^2(t)}{2m} + V(\hat X(t))$$
with $[\hat P(t), \hat X(t)] = i\hbar$
If you are looking at eigenstates and eigenvalues of the hamiltonian, you will look for a constant Hamiltonian.
For instance, for the harmonic oscillator, you will have : 
$$Constant ~ operator ~ \hat H(t) = \frac{\hat P^2(t)}{2m} + \frac{m \omega^2}{2} \hat X^2(t)$$
For the harmonic oscillator, solutions are : 
$\hat X(t) = \sqrt{\frac{\hbar}{m \omega}}  (a e^{i \omega t} + a^+e^{-i \omega t})$, with $Constant ~ operator ~ \hat H(t) = (a^+a + 1/2) \hbar \omega$
Here $a$ and $a^+$ are constant operators such as $[a,a^+] = 1$, explicitely the only non-null members of the operator $a$ are $a_{n+1,n} = \sqrt{n}$
Explicitely, $\hat H(t)$ is a diagonal operator, with $H_{nn} = (n+1/2) \hbar \omega$
In this representation, the "wavefunction" is simply a vector in the vector basis upon which these operators act.
