Solving time-dependent Schrödinger equation using Lie algebra I am trying to understand this chapter in order to solve the time-dependent Schrödinger equation by using On the time-dependent solutions of the Schrödinger equation (RG).
Example 1:
$$ i \frac{\partial \psi}{\partial t}=\left\{{-\frac{1}{2}\frac{\partial^{2}}{\partial x^{2}}}+E_{0}x \sin \omega t\right\}\psi \tag{1}$$
$$i\frac{\partial \psi}{\partial t}=\left\{\sum_{i=1}^{4} a_{i}H_{i} \right\} \psi \tag{2}$$
where $a_{1}=0$, $H_{1}=1$, $a_{2}=E_{0}\sin\omega t$, $H_2=x$, $a_{3}=0$, $H_{3}=\frac{\partial}{\partial x}$, and $a_{4}=-\frac{1}{2}$, $H_{4}=\frac{\partial^{2}}{\partial x^{2}}$.
Solving above equation will be equal to solving the the evolution operator.
$$i\frac{\partial U}{\partial t}=\dot{g_{1}}H_{1}U+\dot{g_{2}}H_{2}U+\dot{g_{3}}H_{3}U+\dot{g_{4}}H_{4}U \tag{3}$$
where $$\psi(x,t)=U(t)\psi(x,0) \tag{4}.$$
Now the next step was to find $g$s. For that purpose I started from $$\psi (x,t)=e^{g_{1}H_{1}+g_{2}H_{2}+g_{3}H_{3}+g_{4}H_{4}} \psi(x,0) \tag{5}$$
Taking derivative of $(5)$ w.r.t $t$ and then equating to equation $(1)$ as
$\dot{g_{1}}+\dot{g_{2}}x+\dot{g_3} \frac{\partial}{\partial x} +\dot{g_{4}}\frac{\partial^{2}}{\partial x^{2}}=-\frac{1}{2}\frac{\partial^2}{\partial x^2}+E_{0}x\sin \omega t$. By comparing the coefficients came to the conclusion that
$$i\dot{g_{4}}=-\frac{1}{2}$$
$$i\dot{g_{2}}=-E_{0}\sin \omega t$$
$$i\dot{g_{3}}=0$$
$$i\dot{g_{1}}=0$$
However, in the mentioned paper, $\dot{g_{3}}$ and $\dot{g_1}$ is not zero. Where am I wrong?
 A: Your evolution operator (3) is very-very wrong, since $H_3, H_4$ visibly fail to commute with U.  By contrast, your paper's (3) is right,
$$i\frac{\partial U}{\partial t}=i(\dot{g_{1}}H_{1}U+\dot{g_{2}}H_{2}U+ \dot{g_{3}}UH_{3} +\dot{g_{4}}UH_{4}) \tag{3}$$
$$= i(\dot{g_{1}}H_{1} +\dot{g_{2}}H_{2} + \dot{g_{3}}UH_{3} U^{-1}+\dot{g_{4}}UH_{4}U^{-1}) U\\
=i(\dot{g_{1}}H_{1} +\dot{g_{2}}H_{2} + \dot{g_{3}} e^{g_2x} \partial_x  e^{-g_2x}+\dot{g_{4}}e^{g_2x} \partial_x^2 e^{-g_2x}) U .$$
Needless to say, your starting point (5) is completely wrong. The correct starting point  as your reference stresses, is, instead,
$$ 
U(t)= e^{g_1} e^{g_2 x} e^{g_3 \partial_x} e^{g_4 \partial_x^2} .
$$
Can you take it from here to their correct (4)?

NB on request
Recall $$ \partial_x e^{-g_2 x}= e^{-g_2 x}(-g_2 +  \partial_x ),\implies\\
\partial_x^2 e^{-g_2 x}=  e^{-g_2 x}( g_2^2 -2g_2   \partial_x + \partial_x^2).   $$
The term linear in $\partial_x$ must vanish, yielding your source's
$$\dot g_3-2\dot g_4 g_2=0. \tag{4c}
$$

NB on further request
You must have made a mistake. Remember, cf (8), that the previous order of the Hs has effectively been reversed. In explicit terms this second U effectively uses the order of the inverse of the previous one. So your analog of (3) is different, and you have, instead,
$$
\partial_t U= U\Bigl ( U^{-1}(\dot g_1 H_1 +\dot g _2 H_2 )U + \dot{g}_3 H_3+ \dot g_4 H_4 \Bigr ).
$$
They predictably assume that the reader has internalized the strategy and can easily deal with a transposition.
