(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?

  • $\begingroup$ The shutter introduces a time dependent interaction. $\endgroup$ Commented Aug 20, 2017 at 21:02
  • $\begingroup$ Did you observe the spread of frequencies or is it a pure theoretical question? $\endgroup$ Commented Aug 21, 2017 at 5:29
  • $\begingroup$ Perhaps you want to read about Existence of monochromatic pulses $\endgroup$ Commented Aug 25, 2017 at 16:04

2 Answers 2


A shutter for a light source is a kind of inverse field generator, that results in nil electric oscillation when it blocks the light. Logically, that means that the shutter is a kind of secondary radiation source, and even if the laser is continuous, your shutter is adding a time-dependent bunch of secondary radiation.

  • 1
    $\begingroup$ These type of considerations are reasonably clear and are about as far as this non-specialist got. But I would love to know if there is a tractable toy model to use for first approximation calculations. $\endgroup$ Commented Aug 20, 2017 at 22:40
  • $\begingroup$ What's "nil"? Do you have anything a bit more quantitative? $\endgroup$ Commented Aug 22, 2017 at 17:37

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.

  • $\begingroup$ But isn't the laser alrady ensuring the beam is collimated and of a definite spot-size? My shutter is >> size of the beam. $\endgroup$ Commented Aug 22, 2017 at 17:38

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