Time derivative of $\hat p$ under time varying Hamiltonian How does one show that
$$ \frac {d\hat p}{dt} =  \frac 1{i\hbar}[\hat p, \hat H]$$ is valid even when Hamiltonian is time dependent explicitly? I can see that this is true when $\hat H$ is time independent where we use $|\psi(t) \rangle= e^{-\frac{i}{\hbar}Ht}|\psi(t=0) \rangle$. How is this still true even when $|\psi(t) \rangle= \hat U|\psi(t=0) \rangle$ where $\hat U$ is supposed to be a unitary operator (when $\hat H$ is time varying) from what I read.
 A: If we are in the Schrödinger picture, we have
$$i\hbar\frac{d}{dt}\psi(t)=H_S(t)\psi(t), \qquad\psi(0)=\psi_0$$
Then, we can describe the time evolution via a unitary propagator $U(t,0)$, such that
$$\psi(t)=U(t,0)\psi_0$$
Substituting, leads to
\begin{gather}&i\hbar\frac{d}{dt}U(t,0)\psi_0=H_S(t)U(t,0)\psi_0&\implies\\
&\frac{d}{dt}U(t,0)=\frac{1}{i\hbar}H_S(t)U(t,0)\end{gather}
Then, in the Heinseberg picture, since $A_H(t)=U(t,0)^*A_SU(t,0)$, we have
\begin{align}\frac{d}{dt}A_H(t)&=\frac{d}{dt}\left[U(t,0)^*A_SU(t,0)\right]\\
&=\left[\frac{d}{dt}U(t,0)\right]^*A_SU(t,0)+U(t,0)^*A_S\frac{d}{dt}U(t,0)\\
&=\left[\frac{1}{i\hbar}H_S(t)U(t,0)\right]^*A_SU(t,0)+U(t,0)^*A_S\frac{1}{i\hbar}H_S(t)U(t,0)\\
&=-\frac{1}{i\hbar}U(t,0)^*H_S(t)A_SU(t,0)+\frac{1}{i\hbar}U(t,0)^*A_SH_S(t)U(t,0)\\
&=\frac{1}{i\hbar}U(t,0)^*\left[A_SU(t,0)U(t,0)^*H_S(t)-H_S(t)U(t,0)U(t,0)^*A_S\right]U(t,0)\\
&=\frac{1}{i\hbar}[A_H(t),H_H(t)]\end{align}
For the existence of the unitary propagator $U(t,s)$ via a Dyson series satisfying the Schrödinger equation for some Hamiltonians, see Simon Vol. II, section X.12.
