
#1
Dec1212, 09:57 PM

P: 1

In quantum optics or laser physics, while solving an ordinary differential equation (ODE) using the integrating factor, the socalled RateEquation 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) (tz)}F(z)## Assume that the initial conditions die away or are zero. Now, we apply the rateequation 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) (tz)}## 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) (tz)}## I don't understand where this extra factor of ##e^{i\Omega) (tz)}## is coming from. So my question is (as the title suggests) what is the most general form of this rateequation 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. 



#2
Dec1312, 02:35 AM

Sci Advisor
P: 3,378

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.



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