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In the double slit experiment, when we place a detector on one of the slits, the electrons will start to act like particles. When we unplug the detector they return to acting like waves.

What if our detector was turning on and off with a high frequency in a way that we could not tell for all the electrons if during the time they were passing in the slits our detector was on or off, how would the interference pattern look then?

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You would not observe the interference pattern. In fact, when you consider 'interference pattern' and 'electron information', this situation would be a loss-loss, because you don't get both (on the contrary, plugging or unplugging completely would still get you an loss-win and win-loss).

So, why does this happen? Let's say you have a glitchy detector, which switches on and off every other second. And let's have a few electrons passing through the slit at a rate of say $h$ electrons per second. Now at $t=1s$, the first $h$ electrons pass through the slit, and the detector is off, so the electrons land according to the probabilities determined by the interference of their wavefunctions. Now, at $t=2s$, the detector turns on. But the corresponding batch of $h$ electrons have now been measured, so they hit the screen at random spots, adding a random pattern in front of the ordered interference pattern.

And this goes on. The next batch forms an interference pattern, but now you wont be even able to tell it apart from the random arrangement. And finally, after the experiment, you will see random arrangements disrupting the interference pattern, which would be what you obtain had the detector remained switched off.

If you had a normal detector (one that stayed on), you would at least get the information of which slit the electron passed through. But, in this case, you can't even get that. This was the loss-loss situation I was referring to.

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  • $\begingroup$ In other words, it'd be a classical superposition of a double-slit interference pattern and two diffraction patterns, each centered around the corresponding slit. $\endgroup$
    – user87745
    Commented May 27, 2020 at 12:51
  • $\begingroup$ Not exactly, the electrons which are once measured wont actually diffract, they would form a random distribution. It would be a combination of double-slit pattern and random patterns. $\endgroup$
    – PNS
    Commented May 27, 2020 at 12:56
  • $\begingroup$ @PNS so what you are saying is that even that the detector is working part of the time it will still not be able to detect the path that the electrons are taking. In such case the nature needs to somehow realize our system and give us an interference pattern, or is she? $\endgroup$ Commented May 27, 2020 at 14:06
  • $\begingroup$ In such a case, the electron is not measured, so it's wavefunction can interfere with itself giving us the interference pattern. But when the detector momentarily turns on, the electron is measured i.e. it now has, a particular path, and it would just randomly hit the screen at any point. So it all boils down to whether the electron is measured or not. $\endgroup$
    – PNS
    Commented May 27, 2020 at 14:13
  • $\begingroup$ It has nothing to do with Nature realizing whether we are looking or not, but rather about whether we are making a measurement or not. $\endgroup$
    – PNS
    Commented May 27, 2020 at 14:14

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