A two-level system under Ito-calculus I'm considering the evolution of a two-level quantum system given by
$$i\begin{pmatrix}\dot{c}_1\\ \dot{c}_2\end{pmatrix}=\begin{pmatrix}
\frac{1}{2}E & \Delta(t)\\\Delta(t) & -\frac{1}{2}E\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}.$$
If $\,\Delta(t)\,$ is periodic, the model gives Rabi oscillations. But now I'm considering the function $\,\Delta(t)\,$ to be a real Gaussian white noise satisfying
$$\langle\Delta(t)\rangle=0,\quad\langle\Delta(t)\Delta(t')\rangle=\Gamma^2\delta(t-t').$$
In terms of the standard Wiener process $W_t$, the Schrodinger equation becomes
\begin{align}
idc_1&=\frac{1}{2}Ec_1dt+\Gamma c_2dW_t,\\
idc_2&=\Gamma c_1dW_t-\frac{1}{2}Ec_2dt,
\end{align}
where both $c_1dW_t$ and $c_2dW_t$ are in the Ito sense, meaning that $c_1$ and $c_2$ are at $t$ and $dW_t$ is during $[t,t+dt]$. I then go to the interaction picture
$$\tilde{c}_1=c_1e^{\frac{i}{2}Et},\quad\tilde{c}_2=c_2e^{-\frac{i}{2}Et},$$
so that $\tilde{c}_1$ and $\tilde{c}_2$ do not evolve when $\,\Gamma=0$. In general, they satisfy
\begin{align}
id\tilde{c}_1&=\Gamma\tilde{c}_2e^{iEt}dW_t,\\
id\tilde{c}_2&=\Gamma\tilde{c}_1e^{-iEt}dW_t.
\end{align}
Now how do I proceed from here to express $\tilde{c}_1$ and $\tilde{c}_2$ in terms of some integral of $W_t$? Also I'm wondering whether I get decoherence or still just Rabi oscillation because the system selectively absorbs frequency $E\,$ from the noise $W_t$.
 A: To solve these equations
\begin{align}
id\tilde{c}_1&=\Gamma\tilde{c}_2e^{iEt}dW_t,\\
id\tilde{c}_2&=\Gamma\tilde{c}_1e^{-iEt}dW_t.
\end{align}
let's look if we can find a pair of functions $F_1(t,x)$ and $F_2(t,x)$ such that $\tilde{c}_1=F_1(t,W_t)$ and $\tilde{c}_2=F_2(t,W_t)$. Plugging these in the formulas we have
\begin{align}
idF_1(t,W_t)&=\Gamma F_2(t,W_t) e^{iEt}dW_t,\\
idF_2(t,W_t)&=\Gamma F_1(t,W_t) e^{-iEt}dW_t.
\end{align} 
We now apply Itô calculus to the differentials
\begin{align}
i[F_{1t}(t,W_t)dt+F_{1x}(t,W_t)dW_t+1/2F_{1xx}(t,W_t)dt]&=\Gamma F_2(t,W_t) e^{iEt}dW_t,\\
i[F_{2t}(t,W_t)dt+F_{2x}(t,W_t)dW_t+1/2F_{2xx}(t,W_t)dt]&=\Gamma F_1(t,W_t) e^{-iEt}dW_t.\end{align}
In these the letters in the index notation are short hand for partial derivatives w.r.t. the corresponding variables. We hence obtain the following partial differential equations
\begin{align}
F_{1t}(t,x)+1/2F_{1xx}(t,x) & = 0, \\
F_{2t}(t,x)+1/2F_{2xx}(t,x) & = 0, \\
iF_{1x}(t,x)& = \Gamma F_2(t,x) e^{iEt},\\
iF_{2x}(t,x)& = \Gamma F_1(t,x) e^{-iEt}.
\end{align} 
Taking the derivative to $x$ of the third equation and multiplying by $i$ we get
$$-F_{1xx}(t,x) = \Gamma iF_{2x}(t,x) e^{iEt}=\Gamma^2 F_1(t,x)$$
where in the last step I substitute in the fourth equation. Finally substituting this into the first equation we obtain
$$F_{1t}(t,x)=\frac{\Gamma^2}{2}F_1(t,x)$$
which is an easy first order linear partial differential equation in $t$. The solution is 
$$F_1(t,x)=A(x)e^{\Gamma^2 t/2} \; .$$
With $A(x)$ some function to determine from the other equations. For instance, filling the result for $F_1$ in the first formula, we get the following differential equation for $A(x)$:
$$A_{xx}(x)+\Gamma^2 A(x)=0$$
the solutions of which are
$$A(x)=Be^{i\Gamma x}+De^{-i\Gamma x}$$
and thus 
$$F_1(t,x)=(Be^{i\Gamma x}+De^{-i\Gamma x})e^{\Gamma^2 t/2}$$
and likewise 
$$F_2(t,x)=(-Be^{i\Gamma x}+De^{-i\Gamma x})e^{(\Gamma^2/2-iE) t} \; .$$
Thus
\begin{align}
\tilde{c}_1 & =(Be^{i\Gamma W_t}+De^{-i\Gamma W_t})e^{\Gamma^2 t/2} \; , \\
\tilde{c}_2 & =(-Be^{i\Gamma W_t}+De^{-i\Gamma W_t})e^{(\Gamma^2/2-iE) t} \; .
\end{align}
