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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.

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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.

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