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Is $\frac{d }{dt} e^{H(t)}=H(t)' e^{H(t)}$ the minimum condition for time evolution operator to be written as $U(t,t_0)=e^{-\frac{i}{\hbar} \int_{t_0}^t H(t') dt'}$?

Further, what's the minimum condition for time evolution operator to be written as $U(t,t_0)=e^{-\frac{i}{\hbar} E_n(t-t_0)}$?(Actually, what I think I meant was $U(t,t_0)=[$ matrix with diagonal element of $e^{-\frac{i}{\hbar} E_n(t-t_0)}]$)? The same as the previous question, or something extra?

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The most general "solution" to the Schrödinger equation

$$ i\hbar\frac{d|\psi\rangle}{dt} = \hat{H}(t)|\psi\rangle $$

is given by the time ordered exponential

$$ | \psi(t) \rangle = \mathcal{T}\left\{ e^{-\frac{i}{\hbar}\int_{t_0}^{t} dt'\ \hat{H}(t') } \right\} | \psi(t_0)\rangle, $$

where $\mathcal{T}$ is the time ordering operator. That is, the general time evolution operator is

$$ U(t, t_0) = \mathcal{T}\left\{ e^{-\frac{i}{\hbar}\int_{t_0}^{t} dt'\ \hat{H}(t') } \right\}. $$

Therefore, the minimum condition to be able to "remove" the time ordering symbol and hence obtain

$$ U(t, t_0)= e^{-\frac{i}{\hbar}\int_{t_0}^{t} dt'\ \hat{H}(t') } $$

is that the hamiltonian commutes with itself at any two times. That is, the condition is given by

$$ U(t, t_0)= e^{-\frac{i}{\hbar}\int_{t_0}^{t} dt'\ \hat{H}(t') } \iff [H(t), H(t')]=0\ \ \ \forall t, t'. $$

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