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I have seen stated that ultrashort pulses have a broad bandwidth.

In the link above, it is stated that a "Gaussian pulse with a 1 ps pulse duration(...) has an optical bandwidth of $\approx 0.44$ THz."

But was is the mathematical formula relating these terms?

user7077252
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Fourier analysis: $$ \Delta \omega \Delta \tau \ge \frac{1}{2} $$ where $\Delta \tau$ is the pulse duration and $\Delta \omega$ the bandwidth. It is because to have a wave form whose amplitude goes up and back down again in a time $\Delta \tau$ (and then stays down) you have to add together sinusoidal waveforms with a range of frequencies, so that they reinforce each other during the time $\Delta \tau$ and cancel each other out at other times. For this to be possible they must be getting out of step with one another in a time $\Delta \tau$ so their frequencies cannot all be very similar: they have to be spread apart by separations of order $1/\Delta \tau$ or more.

Andrew Steane
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    Worth to note: few people seem to be aware of the fact that this is actually Heisenberg's uncertainty relation in disguise (well except for the relationship between frequency and energy, which is, of course, de Broglie's relation and as such the very heart of quantum mechanics). – oliver Apr 02 '21 at 18:19
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    Few people are aware that the famous uncertainty is a standard property of waves. – my2cts Apr 02 '21 at 18:39
  • @oliver of course it is a common misconception to think of this as HUR because $t$ is not an operator in QM: see https://physics.stackexchange.com/q/53802/36194 – ZeroTheHero Apr 02 '21 at 19:48