Physical explanation for power broadening

  • Thread starter Carnot
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  • #1
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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?
 

Answers and Replies

  • #2
34,987
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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
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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
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Thank you for both answers, they helped a lot :-)
 

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