Evolution operator for time-dependent Hamiltonian When I studied QM I'm only working with time independent Hamiltonians. In this case the unitary evolution operator has the form $$\hat{U}=e^{-\frac{i}{\hbar}Ht}$$ that follows from this equation 
$$
i\hbar\frac{d}{dt}\hat{U}=H\hat{U}.
$$
And in this case, Hamiltonian in Heisenberg picture ($H_{H}$) is just the same as the Hamiltonian in Schrödinger picture ($H_{S}$), i.e. it commutes with $\hat{U}$.
Now I have a Hamiltonian that depends explicitly on time. Specifically,
$$H_{S}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega \hat{q}^2-F_0 \sin(\omega_0t)\hat{q}.$$
And in my problem I need to calculate $H_H$ (Hamiltonian in Heisenberg picture).
I've found that differential equation for $\hat{U}$ (I've mentioned about it above.) has generally solution in the form (with $U(0)=1$)
$$U(t)=1+\xi\int\limits_0^t H(t')\,dt'+ \xi^2\int\limits_0^t H(t')\,dt'\int\limits_0^t' H(t'')\,dt''+\xi^3\int\limits_0^t H(t')\,dt'\int\limits_0^t' H(t'')\,dt''\int\limits_0^t'' H(t''')\,dt'''+...$$
So my questions are:


*

*Is there other ways to calculate $\hat{U}$, could give a link or tell me about them?

*If you know form of the solution for my case, please tell me.

*If you know any articles or papers articles on this topice, please link them to me, too.

 A: Yes, you are on the right track. The series you have there is called Dyson's series.
First note that the $n$'th term looks like
$$
U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n)
$$
The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:
\begin{align}
U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) =
\frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t  dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right]
\end{align}
Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)
Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as
\begin{align}
\mathcal{T}[H(t_1) H(t_2)] = \begin{cases}
H(t_1) H(t_2) & t_2 > t_1\\
H(t_2) H(t_1) & t_2 < t_1
\end{cases}
\end{align}
Putting everything together we have
$$
U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)]
$$
Frequently, this is denoted symbolically as
$$
U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right)
$$
This notation is understood as representing the power series.
