Physical explanation for power broadening

1. Jul 10, 2014

Carnot

I have been looking into broadening mechanisms and I'm stuck at trying to provide a physical explanation for power broadening. I get how the math shows that at high intenseties the decay rate goes through the roof due to saturation, but how does this increased decay rate manifest in a spread of generated frequencies? Are the electrons reexcited or decaying while between ground and excited states?

2. Jul 10, 2014

Staff: Mentor

The excitation and deexcitation has "less time". Imagine a wavepacket with a shorter length: it has to have a broader frequency distribution (the mathematical "uncertainty principle" for fourier transformations). The same happens here.

3. Jul 12, 2014

M Quack

The precision with which you can define the frequency of a wave depends on the number of cycles. If you have 10 cycles, you can define the wave length or frequency to ~10%, 100 cycles to ~1%, 1000 cycles to ~0.1% and so on.

A strongly damped wave or a short pulse has a small number of cycles. A fast decay means strong damping.

Mathematically, in order to produce a short wave pulse you have to overlay waves with many frequencies. The spread of frequencies increases the shorter the pulse. A single frequency wave would have to be infinitely long in space and in time.

(this is the same thing mfb said, in more words)

4. Jul 13, 2014

Carnot

Thank you for both answers, they helped a lot :-)