# Short pulse of monochromatic light?

(Potentially connected to this question, but could not find the answer to my particular question there.)

The frequency spread and time duration of a pulse are related by:

$$\Delta \omega \Delta t \approx 2 \pi,$$

from which perfectly monochromatic radiation ($\Delta \omega$ = 0) would require an infinite "pulse", $\Delta t \rightarrow \infty$.

Now: let's think of a (locked) CW laser, emitting a stable frequency with a linewidth of ~10s of kHz. Actually let's even assume 0 linewidth, let's assume it's ideal.

I have a shutter (or some other sort of switch) in the beam path, that goes ON and then OFF in a very short amount of time (100s of µs). Because of the finite duration of the pulse, I now have a spread in frequencies, following the Fourier relations.

So there are photons with a little bit more and a little bit less energy than originally. How? What's the interaction that allowed the reshuffling in energy?

• The shutter introduces a time dependent interaction. – Count Iblis Aug 20 '17 at 21:02
• Did you observe the spread of frequencies or is it a pure theoretical question? – HolgerFiedler Aug 21 '17 at 5:29
• Perhaps you want to read about Existence of monochromatic pulses – HolgerFiedler Aug 25 '17 at 16:04

Well, when you bring photons into the question, you should also bring quantum mechanics. The shutter constrains the photon's position, thereby introducing uncertainty in its momentum [equivalently, energy]. Note that this is just $\hbar$ times your original statement.