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Physical explanation for power broadening 
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#1
Jul1014, 11:53 AM

P: 19

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
Jul1014, 04:23 PM

Mentor
P: 11,869

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
Jul1214, 02:42 AM

P: 667

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
Jul1314, 03:36 AM

P: 19

Physical explanation for power broadening
Thank you for both answers, they helped a lot :)



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