How to solve differential equation with operator as coefficient? I have been working on a problem in Quantum Mechanics and I have encountered a equation as given below.
$$\frac{d\hat A(t)}{dt} = \hat F(t)\hat A(t)$$
Where ^ denotes it is an operator 
How will this differential equation be solved? Will the usual rules for linear homogeneous first order differential with variable coefficients apply here?
 A: You can solve it by iteration (assuming convergence). Assuming that you are interested in the solution with the initial condition $\hat A(0)= I$, the iterative solution reads
$$\hat A(t) = I +\int_0^t\hat F(t_1)\,dt_1 + \int_0^t\int_0^{t_1}\hat F(t_1) \hat F(t_2)\,dt_1\,dt_2 + \cdots \tag{1}$$
For convenience, one might introduce the concept of the ordered exponential. With that the solution assumes the compact form
$$\hat A(t) = \mathcal{T} \left\{\exp\left[ \int_0^t \hat F(t')\,dt'\right] \right\}$$
where $\mathcal{T}$ indicates that when expanding the exponential, the $\hat F$ in the individual terms should be ordered according to their time argument (and thus reproducing (1)).
A: Let us introduce an evolution operator $\hat U(t_1, t_0)$ such that
$\hat A(t_1) = \hat U(t_1, t_0) \hat A(t_0).$
It satisfies 
$$\frac{\partial}{\partial t_1} \hat U(t_1, t_0) = \hat F(t_1) \, \hat U(t_1, t_0).$$
$\newcommand{\prodint}{{\prod}}$
We can use a product integral to express $\hat U$:
$$\hat U(t_1, t_0) = \prodint_{t_0}^{t_1} e^{\hat F(s) \, ds}$$
where $t$ increases from right to left in the product since we have
$$\begin{align}
\hat U(t_1 + \Delta t_1, t_0) 
& = \prodint_{t_0}^{t_1 + \Delta t_1} e^{\hat F(s) \, ds} \\
& = \left( \prodint_{t_1}^{t_1 + \Delta t_1} e^{\hat F(s) \, ds} \right)
\left( \prodint_{t_0}^{t_1} e^{\hat F(s) \, ds} \right) \\
& \approx e^{\hat F(t_1) \, \Delta t_1}
\left( \prodint_{t_0}^{t_1} e^{\hat F(s) \, ds} \right) \\
& \approx \left( 1 + \hat F(t_1) \, \Delta t_1 \right) \, \hat U(t_1, t_0) \\
\end{align}$$
so that 
$$
\frac{\partial}{\partial t_1} \hat U(t_1, t_0) 
= \lim_{\Delta t_1 \to 0} \frac{\hat U(t_1 + \Delta t_1, t_0) - \hat U(t_1, t_0)}{\Delta t_1}
= \hat F(t_1) \, \hat U(t_1, t_0)
$$
Before we continue, let us study what happens if $\hat F$ depends on parameter and we differentiate $\hat U$ with respect to this parameter. We add an index $\lambda$ to $\hat F$ and $\hat U$, approximate the product integral with an finite product, differentiate, and take limits:
$$
\frac{\partial}{\partial\lambda} \hat U_\lambda(t_1, t_0)
\approx \frac{\partial}{\partial\lambda} \prod_{k=1}^{n} e^{\hat F_\lambda(s_k) \, \Delta s_k} \\
= \sum_{k=1}^{n} 
\left( \prod_{l=k+1}^{n} e^{\hat F_\lambda(s_l) \, \Delta s_l} \right) 
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(s_k) \, \Delta s_k \, e^{\hat F_\lambda(s_k) \, \Delta s_k} \right)
\left( \prod_{l=1}^{k-1} e^{\hat F_\lambda(s_l) \, \Delta s_l} \right) \\
\to \int_{t_0}^{t_1} 
\left( \prodint_{t}^{t_1} e^{\hat F_\lambda(s) \, ds} \right) 
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\left( \prodint_{t_0}^{t} e^{\hat F_\lambda(s) \, ds} \right) 
\, dt \\
= \int_{t_0}^{t_1} 
\hat U_\lambda(t_1, t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\hat U_\lambda(t, t_0)
\, dt
$$
The second derivative becomes
$$
\frac{\partial^2}{\partial\lambda^2} \hat U_\lambda(t_1, t_0)
= \frac{\partial}{\partial\lambda} \int_{t_0}^{t_1} 
\hat U_\lambda(t_1, t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\hat U_\lambda(t, t_0)
\, dt \\
= \int_{t_0}^{t_1} 
\frac{\partial \hat U_\lambda}{\partial\lambda}(t_1, t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\hat U_\lambda(t, t_0)
\, dt +
\int_{t_0}^{t_1} 
\hat U_\lambda(t_1, t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\frac{\partial \hat U_\lambda}{\partial\lambda}(t, t_0)
\, dt \\
= \int_{t_0}^{t_1} 
\left( 
\int_{t}^{t_1} 
\hat U_\lambda(t_1, t')
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t') \right)
\hat U_\lambda(t', t)
\, dt' 
\right)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\hat U_\lambda(t, t_0)
\, dt \\
+
\int_{t_0}^{t_1}
\hat U_\lambda(t_1, t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\left( 
\int_{t_0}^{t} 
\hat U_\lambda(t, t')
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t') \right)
\hat U_\lambda(t', t_0)
\, dt' 
\right)
\, dt \\
= \int_{t_0}^{t_1} 
\int_{t}^{t_1} 
\hat U_\lambda(t_1, t')
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t') \right)
\hat U_\lambda(t', t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\hat U_\lambda(t, t_0)
\, dt' 
\, dt \\
+
\int_{t_0}^{t_1}
\int_{t_0}^{t} 
\hat U_\lambda(t_1, t)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t) \right)
\hat U_\lambda(t, t')
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t') \right)
\hat U_\lambda(t', t_0)
\, dt' 
\, dt \\
= \int_{t_0}^{t_1} 
\int_{t_0}^{t_1} 
\hat U_\lambda(t_1, t_+)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t_+) \right)
\hat U_\lambda(t_+, t_-)
\left( \frac{\partial \hat F_\lambda}{\partial\lambda}(t_-) \right)
\hat U_\lambda(t_-, t_0)
\, dt' 
\, dt
$$
where $t_+ = \max(t,t')$ and $t_- = \min(t,t').$
Often one can write $\hat F(t) = \hat F_0 + \lambda \hat F_i(t)$ where $\hat F_0$ is constant and corresponds to no interaction. We can then expand $\hat U_\lambda(t_1, t_0)$ as a Taylor series in $\lambda$:
$$\begin{align}
U_\lambda(t_1, t_0) 
& = U_0(t_1, t_0) \\
& + \lambda \int_{t_0}^{t_1} U_0(t_1, t) \, \hat F_i(t) \, U_0(t, t_0) \, dt \\
& + \frac12 \lambda^2 \int_{t_0}^{t_1} \int_{t_0}^{t_1} U_0(t_1, t_+) \, F_i(t_+) \, U_0(t_+, t_-) \, F_i(t_-) \, U(t_-, t_0) \, dt' \, dt \\
& + \cdots
\end{align}$$
The terms can be seen as a no interaction, one interaction, two interactions, and so on.
