Most General Form of the "Rate-Equation Approximation"

AdvPhBird

In quantum optics or laser physics, while solving an ordinary differential equation (ODE) using the integrating factor, the so-called Rate-Equation Approximation is used. I have come across different sources implementing it differently. For example, if I have an equation say:

##\dot{\alpha}(t) = (-i\omega-\Gamma) \alpha(t)+F(t)##

where ##F(t)## is a function slowly varying with respect to time. Solving for ##\alpha(t)## (using the integrating factor technique) we get

##\alpha(t) = \int_0^t dz\,e^{-(i\omega+\Gamma) (t-z)}F(z)##

Assume that the initial conditions die away or are zero. Now, we apply the rate-equation approximation by claiming that ##F(z)## is slow compared to ##e^{-i\omega z}##. Therefore, ##e^{-i\omega z}## "effectively" acts as a sampling function on ##F(z)## and we can pull it out of the integral to get

##\alpha(t) = F(t) \int_{-\infty}^t dz\,e^{-(i\omega+\Gamma) (t-z)}##

The lower limit was taken to be ##-\infty## because most of the physics is occurring at ##t## and it is also computationally convenient since the integrand vanishes in this limit. However, I have also come across implementations of this type of approximation where we also have an ODE in ##F(t)##. Say it has the form

##\dot{F}(t) = -i\Omega F(t) + \beta \alpha(t)##

where ##\beta## is some constant. In this case, when you pull ##F(z)## out of the integral you would get something like

##\alpha(t) = F(t) \int_{-\infty}^t dz\,e^{-(i(\omega-\Omega)+\Gamma) (t-z)}##

I don't understand where this extra factor of ##e^{i\Omega) (t-z)}## is coming from. So my question is (as the title suggests) what is the most general form of this rate-equation approximation?

Note: I am aware that this is simply a set of coupled ODEs which can be solved exactly using standard techniques. But I am specifically interested in the application of this approximation.

DrDu

In the second example, F is not slowly varying, but is the product of a slowly varying function and exp(i Omega t). The second factor is therefore left in the integral.