Most General Form of the Rate-Equation Approximation

[AdvPhBird]

Most General Form of the "Rate-Equation Approximation"

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.