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Suppose we perform a double-slit experiment with a detector placed at a position of minimum intensity (maximum destructive interference), off-center where the path lengths differ by half a wavelength. The light source is alternately turned on and off (or blocked and unblocked near the source) and the intensity over time is recorded. I interpret the uncertainty principle to mean that there will be a peak in intensity at the times when the switch is flipped (whether on-to-off or off-to-on). i.e., it will look something like this (in ASCII art):

 __________'-----'__________'-----'__________

Is this correct? I have had trouble convincingly explaining my reasons for thinking so. What will be the measured intensity over time and why?

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  • $\begingroup$ You're assuming the flipping of the switch makes the light shut off instantaneously, which I suspect is physically impossible if you're turning off a laser, and probably extremely hard to accomplish if you're blocking the light. $\endgroup$ – Peter Shor Mar 17 '11 at 10:43
  • $\begingroup$ My understanding is that the source can be anything: a hot cathode, an LED, or a even a low-frequency AC. Its energy can be constrained by some means and then focused to a plane-wave by an intermediate single-slit. I don't know very much about lasers. I think the peaks should exist because if a photon/electron is detected near the switching time, then we have information about which slit it passed through, therefore there is less interference than in the case when that information is absent. I am seeking a qualitative explanation, assuming the switching frequency is small. $\endgroup$ – Dan Brumleve Mar 17 '11 at 11:22
  • $\begingroup$ Dan, where is the which-slit information coming from? The switching is happening at the source (ie pre the slits) and the detection happens (if at all) at the screen. There is nothing at the slits to do (or cause) any detecting as I understand this scenario. $\endgroup$ – Roy Simpson Mar 17 '11 at 11:32
  • $\begingroup$ If we detect a photon immediately after the light is turned off, then it probably came from the more distant slit. If we detect it immediately after the light is turned on, then it probably came from the nearer slit. If the light has been on for some time (and will be on for some time) we have no such information and the probabilities are equal and we see maximum destructive interference. $\endgroup$ – Dan Brumleve Mar 17 '11 at 11:46
  • $\begingroup$ I might amend my answer, but the question would be clearer if it had been stated that it was the reasoning of the last two comments which was to be analysed. Just to note for now, that the first sentence raises several issues, including the "instantaneousness" assumption noted by Peter Shor. $\endgroup$ – Roy Simpson Mar 17 '11 at 14:08
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Dear Dan, first of all, you shouldn't use the term "uncertainty principle" if you're talking about "light sources" and light may be explained by ordinary - classical (non-quantum) - electrodynamics where no uncertainty principle applies.

This is just an exercise in the propagation of waves.

Second, when you flip the switch, there may be temporary variations of the intensity, but they're not necessary, either. For example, you may find a minimum such that the number of wave peaks on the two trajectories (coming from the two slits) differs by 13.5 - one arm is 13.5 wavelengths longer than the other one. It will mean that the destructive interference only occurs when the beams from both slits are synchronized, and there will always be a period lasting about 13 periods after each flip of the switch when only one beam is coming to the detector. That will indeed eliminate the destructive interference, and give you the "apostrophes" in your ASCII art.

The precise shape of the graph depends on the character of the switches, geometry of the experiment, and other things.

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