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?
On
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.
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)).